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190200 | https://ocw.mit.edu/courses/22-105-electromagnetic-interactions-fall-2005/055c4b3f7f730d6270e4e2664a589286_chap4.pdf | Chapter 4 Radiation By Moving Charges 4.1 Potentials and Fields of a moving point charge The general solution (4.1) Looks as if it will give the result for a point charge directly in the same way as the static solution. For a stationary point charge p = q6 (x - r), where r is the charge position, 4 = U L For brevity let's write R = x - r. One might think for a moving charge 47x0 X - r . $(x,t) = &but this is incorrect. We haven't taken care with derivatives etc. of Figure 4.1: Vector coordinates of charge and field point retarded quantities. Let's go carefully! The charge density is p (x, t) = q6 (x - r (t)) where r is now allowed to vary with time so we want -4 / 6 (x' - r (t')) d3d (4.2) 4rto x - r (t') 1 where we make use of the fact that the delta function is non-zero only where its argument is zero, so all the contribution to the integral comes from the place where x' = r(t1), which is where the particle is at retarded time i.e. [This requires self-referential notation which is one reason we write it .] Now we have to do the integral J6 (x' - r (t)) d 3 . This is unity because x' appears inside r(t') as well as in x'. The delta function is defined such that but now its argument is y = x' - r ( t ) . We need to relate d3y to d3z' for the integral we want. Consider the gradient of one component: y = V ( X - r (t)) = [ [ x- rill [since ri is a function of t but not x' directly.] Choose axes such that x' = (xi,x:, x : ) with component 1 in the R = x - x' direction. Then the second term is present only for the xl component not the other two (because they are 1to x - x ) . Also (V'X~) = Jij. [i.e. 1 iff 2 = 1 . Thus Consequently Let's write Then And finally "x, t) = ' 1 ' 1 471to nR (4.10) By exactly the same process we can obtain the correct value for each component of A and in total A (x,t) = [y] 47r nR - - Figure 4.2: Integral of charge density over a square-shaped moving charge at retarded time. (v = drldt) ,j = qv6. These expressions are called the "Libnard-Wiechert" potentials of a moving point charge. Since the K correction factor is so important and the scientific literature is strewn with papers that get it wrong, let's obtain the result graphically. The retarded integral J d3x' can be viewed as composed of contributions from a spherical surface S which sweeps inward towards the observation (field) point x, at the speed of light, arriving at time t. The charge that we integrate is the value of p when the surface S sweeps past. If we are dealing with a localized charge density, such as illustrated, the surface can be approximated as planar at the charge. If the charge region is moving rigidly at speed v towards x, then its influence or contribution to the integral is increased because by the time the surface S has swept from front to back, the charge has moved. Consequently, the volume of the contribution (in x') is larger by the ratio of the charge v 0 1 -e How + charge much is this? When does S reach 'front' of charge? At the moment S reaches the front, Figure 4.3: Snapshots as the integration surface, S, crosses the back and the front of the charge. So A t = ( c P v ) L and L 1 =-C L = -l L (4.13) C-v I - " C Thus L' 1 1 - - -- -(4.14) L 1 " K as before. Notice that transverse velocity does nothing, and that approximations implicit in taking S to be planar become exact for a point charge, with spatial extent 0. The i quantity ti can also be seen to relate intervals of time, dt, to the corresponding retarded time intervals, dt'. So dt 1 dR' - -I t -. dt' c dt' But dR' d 1 d - = x -x = { ( x - x ' ) . ( x x i ) ) 112 dt' dt' dt' Hence dt v.R' - = I - -= t i dt' cR' Strictly speaking, it is the value of ti at retarded time, when the surface S passes the particle, that is required here if v is changing. 4.2 Potential of a Point Charge in Uniform Motion An important special case is when v = r =const. From the retarded solution Lorentz derived his transformation, which is the basis of special relativity. Take axes such that v = v x . We need to calculate the potential at x = ( x ,y , z ) and we'll suppose that the particle is at the origin at time of interest, (t = 0). Tricky part is just to calculate the retarded time t' and position x'. By definition Substitute (-t')v = -x' Figure 4.4: Coordinates of a uniformly moving charge at x ( t ) . Gather terms d2($ - 1j+ 2ir' - (i2 + tJ2+ i2) = O solution ( ; -l ) x ' = -x j ; w (where the - sign must be taken). And x + \i'.2 + (xi +v2+ z2)($ - 1) 1 C R = x ' = C -v v c2 - 1 212 we also need the retarded value of K i.e. 1 - (R'/R').(v/c). Substituting for R ' we get This is the value at time t = 0.At any other time t,the particle is at the position x = vt instead of at the origin, J: = 0. Our formula was developed for the particle at the origin. So to use it we must move the origin to x = vt,which means we simply have to replace x in this formula with x - vt. So finally, substituting the general result for n'R' into the Liknard-Wiechert formula we get See how we have the beginnings of relativity. We get electromagnetic potential dependence on spatial coordinates that can only be consistent with the formula in the frame of reference in which the particle is at rest: - - if coordinates transform as This is the (spatial part of the) Lorentz transformation, incorporating the Fitzgerald con- traction in the direction of motion. Now we also need to recognize there is a vector potential So the electric field has both contributions: To evaluate these, denote by R" the quantity in the denominator of $ and A: [Note that this is not R', the retarded radius]. Its derivatives are dR" - x - vt 1 dR" - y dR" - z dR" - v (x - vt) -- ---- --- ---(4.33) a x 1- "" R" ' 8 R" ' dZ R" ' a t (1 - $) R" c2 Consequently v-1 = 1 Rll -R U 2 giving 1 x - vt i and a A 4 1 1 (-v2/c2) (x - vt) 2 2 a t 4r.o 4 -R 1 ( (1-/C ) . o,o) SO E = - $ - A = -4 1 ( 1 v t ,y,z) (4.37) 4rto RU3 This is a remarkable result. It shows that despite the fact that contributions to E arise from the retarded position of the particle, the direction of E is actually radially outward from the instantaneous (i.e. non retarded) position. The E field at t = 0 is along the radius vector (x, y , ~ ) . The electric field is not just the same as for a stationary charge. The field is not Retarded Particle Position Position \0 \ \ E field \ Weak strong Figure 4.5: Electric field lines of a charge in uniform motion point outward from the instan-taneous (not retarded) position but the field strength is not symmetric. spherically symmetric, since it is proportional to which makes it stronger in the perpendicular direction and weaker in the parallel direction. The magnetic field may be obtained from B = VAA by recognizing VA(fv) = v A V f , if v is constant. Hence, using A = v$/c2, [The latter form uses the fact that (A and) 2 are parallel to v so v A 2 = 01. This expression for the magnetic field can also be rewritten, by noticing that E is in the direction of R, R' A R = (t - t')v A R and t - t' = R'lc; so v A E = (R'cIR') A E. To summarize: E = -4 x - vt (4.40) 4rto 3 1 2 and 1 1 B = V A E = R ' A E (4.41) c2 cR' A helpful way to think of the result that the electric field is still radial but with a non-spherically-symmetric distribution, is to think about what happens to the field lines when viewed in the lab frame of reference [components (x,y, z)] compared with a frame of reference in which the particle is at rest [components (xl,yl,zl)]. It turns out that the electric field we have calculated is exactly that which would be obtained by assuming that the spherically symmetric distribution of field-lines in the rest-frame is simply compressed together with the rest of space in the x-direction through the coordinate transform of eq 4.29. This contraction is illustrated in figure 4.6. Stationary Charge Figure 4.6: Contraction of space which gives the electric field-line distribution of a moving charge. For a purely geometric compression in one dimension like this, the angles between the direction of R and v (for the two cases) are related by where y = 1/J(l - v2/c2). Consequently Now the element of solid angle corresponding to an angle increment dx is dfl = 27rsinx dx and So the relationship between corresponding solid-angles is where R = x2+ y2+z2. Therefore if the field-lines are compressed in this purely geometrical way, the number of field-lines per unit solid angle, which is proportional to the electric field intensity, in the lab-frame is equal to the value in the rest-frame times the factor dCll/dCl =yR3/RU3.
Thus the geometric compression would lead to an electric field: This is precisely what we calculated directly from the equations of the fields. In other words, we can regard the non-symmetric electric field of eq 4.40 as arising from a compression of space corresponding to the Lorentz transformation (eq 4.29). We are not here invoking the Lorentz transformation based on an understanding of special relativity. In fact the opposite is the historic situation. Lorentz's transform was part of the prior basis for the discovery of relativity. See Jackson 1998 pp. 514-518 for a discussion of electromagnetism as the historic foundation of relativity. Maxwell's equations are already fully relativistic. They don't need to be corrected for relativistic effects, the way Newton's laws require correction for example. Of course the point is stronger than that: Maxwell's equations can only be consistent when special relativity applies (i.e. Lorentz, not Galilean transformations). We don't have time to cover relativity but we don't have to make a special point of it since EM equations already are relativisitc. 4.3 Fields of a Generally-Moving Charge The Lienard Wiechert potentials give the general potential solution. From them we can obtain the general E and B fields from a particle moving with arbitrary velocity: not just uniform v . Since both potentials and fields depend only on the values at retarded time, our calculation will be almost the same as for the uniform motion with the exception that we must use the value of v at that retarded time and we must account for possible time- derivatives of v. Our derivations of $ and A go through exactly as before except that the origin of coordinates is at a point x' +v't' along the projected path of the particle if it were to continue past the retarded time with constant speed v'. [Here we are putting prime on v to remind that it is the retarded value we require.] A (x,t) = 4 v' (4.48) 47rtoc2J(x - V I ~ )+ ( ~ 2 + ~ 2 ) (1 -g) Now we need to get the fields by differentiation. We get exactly the same terms as before plus extra terms arising from the time derivative of v. We could do this directly by taking into account all the contributions. Instead, let's do a vector calculation starting with the Lienard-Wiechert forms: Again, extreme care must be taken with the differentials. For any function f(x, t), This is the same situation as we had before. There we had V' i.e. gradient with respect to retarded position, x', keeping x and t fixed. Here we are talking about gradient w.r.t. x keeping t fixed. Apply the above equation to the function x x ' which is, strictly speaking, = -(4.53) Then returning to the general identity: In the same way Substituting R = x - r for f shows and, since Ok, now we have the tools to evaluate E: With everything inside the retardation operator, it is safe to proceed with algebra as if x' is fixed. In particular, where dot denotes &. The terms in E are then Gathering terms together, and denoting R = g,we get or alternatively, using vector triple product identities, The magnetic field is B = V A A which is by an identity directly analogous to the one we showed for gradient. Also Hence 47rtoc2 l a 1 = ; [ [ ~ A E , (4.74) by comparison with our expression (4.63) for E. Summarizing our results, the fields due to a point charge q moving with variable velocity v such that the radius vector from the charge to the field-point is R may be expressed using A n = 1 - R.v/c as: There are several different forms of these expressions, useful to illustrate different aspects of the fields of a moving point charge. See Jackson and Feynman for discussion of some of these. 4.4 Radiation from Moving Charges 4.4.1 Near Field and Radiation Terms The form for E that we obtained was exhibited in a way that had 2 separate terms. The first of those terms does not contain v while the second is proportional to v. Therefore the first term is exactly what would be obtained for uniform motion v = 0 (although this is not obvious when comparing with our earlier formula expressed in coordinates). Also, everything inside the brackets is dimensionless (R,z) except & and A:. These factors decide the behaviour of their respective terms at large field-point distances, R. The 'static' (constant v) term is K & but the v term is K i.Consequently, the Poynting vector is respectively. If we ask about the total EM power flux across a spherical surface far from the charge, that value scales like the surface area 47rR2 tims E A H. Thus power flux K $ for the constant v term, and K 1 for the v term. We see then, that the constant-v term gives rise to vanishingly small power flux far from the charge but the v term gives rise to finite power flux even at infinity. This distinction requires us to regard these two terms as the "near field" term: 1 E K V R2 (4.78) and "radiation" term: 1 E K V R . (4.79) A charged particle radzates only if it accelerates. 4.4.2 Radiation into a Specific Solid-angle Having identified just the 1/R term as the radiation term, we will drop the other, near field, term from consideration. Imagine, then, a sphere of radius R surrounding the retarded position of the particle. The Poynting vector of the radiation term there is where the last form recognizes that the radiation term has E perpendicular to [R]. Radiated energy thus crosses the sphere, normal to its surface with a local intensity (energylunit arealunit time) E2/cPo, with E given by the second term of eq 4.71. One is very often interested in the power radiated per unit solid angle, fl,, subtended by the area at the point of radiation. By definition of solid angle, a small area of the sphere, A, subtends a solid angle AIR2. Consequently the power per unit solid angle is R2E2/cPo. The extra term R2 cancels the R2 occurring in E2,leaving an expression independent of the radius, R, of the sphere. By convention we can write the power per unit solid angle using the notation 4.4.3 Radiation from Non-relativistic Particles: Dipole Approxi- mation Considerable algebraic simplifications occur when v/c << 1 and so we can approximate ( R v/c) -. R, and K = 1. Then where a is the angle between R, the direction of the solid angle (propagation), and ii, the acceleration. An integration of the total radiated power over the entire sphere (all solid- angles) can readily be done. Taking the direction of v to be the polar direction, the integral is such that dfl, = 27r sin a d a . (4 33) - -- So, noting that " 87r sin2a27rsinada=27r 1 c o s 2 a sinada=27r ) 0 3 ' (434) we get (4.85) This expression for the total radiation from a non-relativistic accelerated charge is known as Larmor's formula. The non-relativistic expressions for P and dPldCl, are often referred to as the "dipole approximation" because they are exactly what is obtained for the radiation from a stationary oscillating dipole electric distribution when the electric dipole moment p, is such that p = qv (4.86) Thus this radiation pattern and intensity is what is obtained also from dipole antennas that are much smaller than the radiation wavelength. 4.5 Radiation from Relativistic Particles The general expression for radiation by an accelerated particle, without invoking approxi- mations requiring v << c, is given by eq (4.81). However an important distinction must be drawn in discussions of energy per unit time between expressions based on time-at-field- point, t, such as eq (4.81), and expressions referring to time-at-particle, retarded time t'. If we want to know how much energy a particle is radiating per unit time-at-particle, which is what we do want if, for example, we want to calculate how rapidly the particle is losing energy, or indeed if we want to calculate the total energy radiated per unit volume by adding up the energy radiated by all the particles in that volume, then we must multiply expressions for energy per time-at-field-point by the ratio dtldt' = n. This conversion lowers the power of n in the denominator by one. We shall work henceforth with such expressions of energy per unit time-at-particle and will indicate this by a prime on the power: P'. Even so, we still have a factor n5 in the denominator of dP'ldCl,. This factor is the most important effect. Since n = 1 - R.V/C = 1 - P c o s ~ ' , when we are dealing with particles moving near the speed of light, n becomes extremely small when 6' -. 0,that is for radiation in the direction along the particle's velocity. As a result, the radiation is greatly enhanced in this forward direction, an effect that is sometimes called the relativistic "headlight" effect. 4.5.1 Acceleration Parallel to v The simplest case algebraically is when v and v are parallel. The radiation is then rota- tionally symmetric about this direction, having the n factor as its only alteration, from the dipole formula: dP' -q2 v2 sin26' dCl, 47rto 47rc3 (1 -P cos 6')5 Figure 4.7: Polar plots of the radiation intensity as a function of direction, with acceleration parallel to v, for different values of = vlc. Velocity is in the x-direction. The radiation in the exactly forward direction 6' = 0 is zero because of the sin26' term. The maximum radiation is in the direction 8 , -. 1/(2y) when - 1. Here y is the relativistic factor (1 - v2/~2)-1/2. Moreover the intensity in this direction becomes extremely large as p gets close to one. 4.5.2 Acceleration Perpendicular t o v Figure 4.8: Definition of the angles for radiation when v is perpendicular to v. An even more important case is when v and v are perpendicular. One can then obtain dP' -q2 v2 sin26' cos2$ (1- P2) - --I [ I -(4.88) dfl, 47rto 47rc3 (1 -cos 6')3 (1- p cos where 6' is the angle of R with respect to v and $ is the polar angle of R about v measured with respect to v as zero. This distribution is likewise highly peaked in the forward direction for - 1, having a typical half-angle extent of approximately $. Figure 4.9: Polar plots of the radiation intensity as a function of direction, with acceleration perpendicular to v, for the case where R lies in the plane of v and ii A v . 4.5.3 Total Radiated Power The general expression (4.81) can be integrated over solid angles by elementary but tedious methods to obtain the first form of which was obtained by Lienard (1898). These two alternate forms are convenient for obtaining the power when v is parallel to v : and v is perpendicular to v : / q2 2 v2 p = y4 (4.91) 4 r t o 3 C C2 These expressions give important quantitative information about the rate of energy loss by a charge undergoing acceleration. The first thing we can see is that a charge could never be accelerated through the velocity c, because y i oo at 0 i 1 and so infinite amounts of radiation would be emitted. This remark is quite independent of Einstein's theory of relativity which shows that the mass becomes infinite as 0 1. Thus in 1898 when Lienard i obtained his expression he already could have deduced that a charge could not be accelerated past v = c. Second, let us compare the rate of radiative energy loss to the energy gain from an accelerating electrostatic force. Write the field as equivalent to the field a distance r from a charge Zq so that the acceleration is Zq2 1 2) = --(4.92) 4 r t o r2moy ' Figure 4.10: Comparing the radiative energy loss to the energy gained from force during acceleration due to a nearby charge. accounting for the relativistic mass increase. Supposing v to be parallel v, the rate of radiative loss is / q2 2 z2q4 1 -f4 p = ----(4.93) ' 4rto 3C3 ( 4 ~ t ~ ) ~ r4mg This will equal the rate of gain of energy due to acceleration, namely when or The left-hand side, here, is the potential energy of the charge and the right-hand side is a square-root factor times the rest-mass of the charge (expressed as an energy). For mod- estly relativistic particles, when we can take the square-root factor to be of order unity, we therefore see that radiation would begin to have an important effect relative to the parallel acceleration only when an electron (for example) is in a potential well at a depth - moc 2 = 511 keV. Remembering that the binding energy of a hydrogen atom is only 13.6 eV this could happen only in the most exotic of situations (e.g. inner shells of heavy elements). Of course those situations would really have to be treated by quantum mechanics. Moreover these immensely strong electric fields (-loz0 V/m) are never even approached in present accelerators. So radiation caused by acceleration parallel to v, such as in a linac, is never a serious consideration. If v is perpendicular to v, however, the lowest order energy gain by the acceleration is zero. Compared with this the radiation may well be important. In the atomic force-field and the classical kinetic energy in a circular orbit at radius r is , Classical Orbit ... . - - ~ - - ~ ~ ~ Figure 4.11: Radiation from a particle moving in a circular orbit arises from its perpendicular acceleration. Hence the orbital energy is radiated with a characteristic time constant where I is the binding energy of the particle in this circular orbit. For a "classical" hydrogen atom circular orbit, I = 13.6eV, Z=1, and r = a. = 5.29 x 10-l1 m (the Bohr radius) we get T = 1.5 x 10-gs. Thus the rate of loss of energy by an electron in a "classical" Bohr orbit is such that the electron would spiral into the nucleus in a few nanoseconds. This, of course, was one of the key problems with classical electrodynamics that physics faced in the early 1900s, which prompted the eventual discovery of quantum mechanics. As an immediately practical matter, we can also ask how fast a particle radiates energy because of being accelerated by a magnetic field, in the circular orbit of a cyclotron, for example. In this case, the acceleration is v = v 2 / d , where d is the orbit radius. The power radiated is then, from eq (4.91), For a relativistic particle (0 -. 1) the power therefore increases proportional to the fourth power of the energy (y4), and the energy loss per orbit for electrons moving with radius of curvature r can be written numerically in the form This amounts to a major limitation for electron storage rings and accelerators above a few GeV energy. Jackson (p 668) cites the Cornell electron synchrotron with r = 100 meters having a loss of 8.8 MeV per turn at 10 GeV. The MIT Bates accelerator storage ring is designed for up to 1 GeV energy. With a bend radius of 9.1 m the loss is 9.8 keV per turn which is compensated by an accelerating stage within the ring. 4.6 Scattering of Electromagnetic Radiation 4.61 Thomson Scattering We have seen that a non-relativistic accelerated charge radiates according to (eq 4.82), where a is the angle between the direction of radiation and the direction of the acceleration, v. If the acceleration arises from an electric field Ei, then Therefore the power radiated per unit solid angle from a single electron can be written: The combination of parameters arising in the last form of this equation, has the dimensions of length, and is called the classical electron radius. A steady electric field will not give rise to radiation that is particularly interesting, but if the electric field is oscillating, it will give rise to radiation that is at a corresponding frequency. Figure 4.12: Schematic illustration of the process of Thomson Scattering. The most elementary case one might consider is when the electric field varies sinusoidally with angular frequency w. This is exactly the situation that arises if a charged particle such as an e1ect)ron experiences the oscillating electric field of an incident electromagnetic wave at frequency w. In this situation we speak of "scattering" of the incident wave by the electron. This process of acceleration of a free electron by an incident wave and reradiation of a wave into other directions is known as Thomson scattering. Now the instantaneous power per unit area of the incident wave is given by the Poynting vector whose magnitude is and we evaluate it at retarded time t' (i.e. at the time necessary to give rise to radiation at the field point at later time t). Therefore the scattered power per unit solid angle from a single electron can be written: 2 .
2 dp = re sin a si . (4.106) The differential (energy) scattering cross-section is the ratio of dP/dCl, to the incident power density si. One can rapidly verify that this definition is in accord with the standard definition of a cross-section: that it should be such that the number of collisions per unit length is equal to the product of the cross-section and the density of targets. In this case the "projectiles" are represented by the incident energy of the wave. The projectiles can be considered to have a flux density proportional the wave power flux density, si. An alternative view of this cross-section is to regard it as the area across which the incident power flux density would have to flow in order to give rise to the power scattered. The cross-section is 2 .
2 = re sin a , (4.107) dfls where a is the angle between the scattering direction and the electric field (i.e. the polar- ization direction) of the incident wave. Integrated over all scattering angles this expression yields the total Thomson scattering cross-section 0 = r 8~ e 2 . (4.108) 3 If the electron is stationary apart from the oscillation that the wave imparts to it, then the scattered radiation will have exactly the same frequency (in this classical approximation) as the incident wave. However if the electron is moving prior to its perturbation by the incident wave, then there will be a Doppler shift of the scattered frequency both because the moving electron will experience the incident wave at a different frequency and because its radiation will be Doppler shifted at the observer. These two effects give a scattered frequency w, that is related to the incident frequency wi by 1 - ki.vo/c W, = wi + (k, - ki).vo= wi A (4.109) 1- k,.vo/c ' where ki and k, are the wave-vectors of the incident and scattered waves respectively, whose magnitudes are ki = wi/c and k, = w,/c, and hats indicate unit vectors. The numerator and denominator of the fractional form for w, represent the two Doppler shifts just referred to. This one-to-one relationship between the scattered frequency and the component of the electron velocity along the direction k, - ki is extremely helpful in plasma diagnostic applications. The velocity distribution of the electrons is directly revealed in the spectrum of Thomson scattered light. 4.6.2 Compton Scattering One approximation implicit in our treatment of Thomson scattering is that all the incident wave does to the electron is to cause it to oscillate and that this oscillatory motion is added to an otherwise unperturbed prior motion. In other words, after the scattering has happened, the electron remains either stationary or moving at the same velocity as it had before. [In this section we will henceforward take the electron to be stationary prior to the scattering for simplicity.] But this cannot really be right, even on a classical picture, because we know that electromagnetic fields carry momentum. So if the wave is scattered, changing its momentum, then the electron's momentum must also be changed so as to conserve total momentum. The classical effect can easily be calculated. By the symmetry of the sin2 a angular distribution of scattering, the scattered radiation has zero momentum on average. Therefore the momentum imparted to the electron is just that of the incident radiation. We saw in section 3.2.3that the momentum density of electromagnetic fields is equal to l/c2 times the energy flux density. The force exerted by the incident radiation on the electron is equal to the total cross-section times the momentum flux density, which is c times the momentum densitv. So this force is In this classical picture there is a radiation pressure, applying over an area equal to the Thomson cross-section of the electron, which steadily pushes it in the direction of the incident radiation Quantum mechanics teaches us, however, that electromagnetic radiation is not smooth and infinitely divisible. Instead it takes the form of photons whose energy is 5w when the angular frequency of the radiation is w. If the size of the photon, the quantum of energy, is much less than the other energy scales in the problem, then the classical limit discussed above, can apply. If the photon energy is large, it cannot. Actually, the crucial question here is the momentum of the photon but this can be related to energy and compared with the rest energy of the electron (moc2 = 511 keV) as we shall see. The quantum picture, then, is that each individual photon may, on encountering a free electron, bounce off in a scattering event. When it does so, the photon's momentum is changed, and the electron's momentum changes also so as satisfy conservation. As a consequence, for energetic (large momentum) photons, even an initially stationary electron recoils from a scattering event with substantial momentum. This recoil leads to a downshift in the energy (and hence frequency) of the scattered photon that will depend on the direction in which it is scattered. The kinematics of the problem, momentum and energy conservation, are all that is needed to relate the energy shift to the angle of scattering. Scattering takes place in a scattering plane. We will suppose that the photon is scattered through an angle 6 ' and the electron recoils in a direction at an angle 4 to the initial direction of the photon. We have to do the problem relativistically and we appeal to the general relativistic relationship relating energy and momentum: 4 -& = (4.111) We denote the final momentum of the electron by p, the photon energy by & before and &' after the scattering collision. Then the momentum of the photon is &/c (from the energy Incident Photon Recoi>' h Photon i-Figure 4.13: Compton scattering geometry in the scattering plane relationship above or from our knowledge about the relationship between energy flux and mo- mentum of electromagnetic fields). Then we write down the two components of momentum conservation uarallel and perpendicular &' 0 = s i n B + p s i n 4 (4.113) C to the incident photon, and the energy conservation: We eliminate 4 by separating the 4 terms in eqs 4.112 and 4.113 squaring and adding to get: And then we eliminate the momentum p by squaring the square-root term of eq 4.114 to get and subtracting from c2 times the previous equation to get 0 = &&'(I- cos 8) - moc2(& - &') . (4.117) This is the equation that relates the photon energy downshift to the angle of photon scatter- ing. It is most often written in a form governing the photon wavelength X = 2.rrclw = he/& and using 1- cos 8 = 2 sin2 812, which expresses the "Compton Shift" of wavelength in terms of the "Compton Wavelength", A , = 2.426 x 10-l2 m, of the electro~~. h/moc = A photon's wavelength equals the Compton wavelength when its energy is equal to the rest Inass ol l l ~ r rlectron, moc2 = 511 keV. Therefore the Compton shift is important only for very energetic x-rays and for 7-rays. The energy of the scattered photon is and the energy lost by the photon, and hence gained as kinetic energy by the electron is 1- cos 6' & (4.120) 1 - cos 6' + moc2/& Figure 4.14: Compton scattering cross-section angular variation. [a= &/m,c2] The cross section for this scattering must reduce to the Thomson cross-section at low photon energy. It was first calculated using relativistic quantum mechanics (1928) by Klein and Nishina shortly after Dirac's formulation of the relativistic quantum equations for the electron, predicting spin and negative energy states. The agreement of the Klein-Nishina cross-section with experiments was one of the early triumphs of Dirac's theory. For unpo- larized radiation, the differential cross-section for photon scattering (which is different from the energy scattering cross-section by virtue of the photon energy shift) per unit solid angle In this form the reduction to the Thomson cross-section at low photon energy, so that &'/& i 1, can be verified by integration of the Thomson formula over all possible incident radiation polarization directions. At high photon energy, & > m,c2, forward or small angle scattering tends to dominate the cross-section, because the (&'/&)2 term becomes small at larger angles; although for those photons that are back-scattered 6' z z 18O0, they lose practically all their energy to the electrons and retain only &' i m,c2/2. Figure 4.14 shows polar plots of the cross-section at different energies. Image removed due to copyright restrictions.
The Compton scattering process is a dominant attenuation mechanism in the 1to 4 MeV photon energy range. It is sometimes helpful to distinguish between the cross-section for scattering of a photon, given above, and the cross-section for removal of energy from a photon beam, which is equal to the product of the scattering cross-section and the ratio of energy loss to initial photon energy. This later is sometimes called the Compton "absorption" cross- section since it represents the rate at which energy is transfered from photons to Compton scattered electrons. In either case, the attenuation of a photon stream of intensity I is governed by a differential equation: where o is the cross-section per electron, and the fact that the Z electrons are bound to each atom is ignored since the photon energy is so much higher than the electron binding energy. The solutions to this equation are exponential (K exp(-n,o!)) with inverse decay length n,o, which is called the "attenuation coefficient". Figure 4.15: Photon attenuation coefficients for lead. [From Evans] Since the ratio of mass to charge of most nuclei is very similar, between 2 and 2.8, the greatest attenuation arises from the greatest electron density, which corresponds to the greatest mass density. Hence lead, for example, has one of the largest attenuation coefficients. Figure 4.15 shows the total (angle-integrated) Compton attenuation coefficients together with the photoelectric absorption and pair production coefficients. These latter processes will be discussed later. The Compton attenuation, since it is simply the product n,o can be scaled to any other material by multiplying by the ratio of electron densities, that is (for elements) Image removed due to copyright restrictions.
by the quantity Alead potherzother pleadzlead Aother where p is mass density, A is atomic weight, and Z is atomic number |
190201 | https://www.sciencing.com/calculate-phat-8384855/ | © 2025 Static Media. All Rights Reserved
How To Calculate P-Hat
Math
How To Calculate P-Hat
By Chris Deziel Updated
bizoo_n/iStock/GettyImages
In statistics, the letter "p" denotes the probability of a certain event occurring or a certain parameter being true for a certain population, but when a population is large, it may be impractical or impossible to measure it directly. As an alternative, statisticians take a sample that they can measure, and they denote the result as "p-hat," which is written as a p with a triangular hat over it ( ^). This sampling strategy is common in political polls that seek to determine how many people in the country agree with a certain policy or approve of the job a government official, such as the president, is doing.
Calculating P-hat
Calculating P-hat
The actual calculation of p-hat is not challenging. To do it, you need two numbers. One is the sample size (n) and the other is the number of occurrences of the event or parameter in question (X). The equation for p-hat is p-hat = X/n. In words: You find p-hat by dividing the number of occurrences of the desired event by the sample size.
An example helps clarify this:
A poll wishes to determine how any Americans agree with the policies of the current president. Pollsters contact 1,000 voters and ask the question: "Do you approve of the president's policies?" The poll produces 175 yes answers and 825 no answers, so p-hat for the poll is 175/1,000 = 0.175. The results are usually reported as a percentage, which in this case would be 0.175 x 100 = 17.5 percent.
The Significance of P-hat in Polls
The Significance of P-hat in Polls
While it's possible to determine p-hat, the value of p remains unknown, and the degree to which it's possible to trust p-hat as being an accurate representation of p is known as the confidence level. P-hat is a reliable representation of p only if the sample is large enough and is truly random. While political pollsters make efforts to ensure random samples, it's often difficult to do in practice, and the results are often skewed. Skewing can be countered by taking larger samples or by repeating the poll in different parts of the country.
Another factor that influences the confidence level of p-hat is the number of respondents in a poll who actually answer the question. Many will decline to answer and opt to remain undecided, and the more that do so, the less pollsters can meaningfully relate p-hat to p. One way to counter this is to ask simple questions that require yes or no answers.
References
University of Northern Iowa: Proportions
University of Florida: Next Sampling Distribution of the Sample Proportion, p-hat
Cite This Article
MLA
Deziel, Chris. "How To Calculate P-Hat" sciencing.com, 13 March 2018.
APA
Deziel, Chris. (2018, March 13). How To Calculate P-Hat. sciencing.com. Retrieved from
Chicago
Deziel, Chris. How To Calculate P-Hat last modified March 24, 2022. |
190202 | https://fiveable.me/microbio/unit-9/2-oxygen-requirements-microbial-growth/study-guide/0d3nmhcM3ewVHDg8 | Oxygen Requirements for Microbial Growth | Microbiology Class Notes | Fiveable | Fiveable
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9.2 Oxygen Requirements for Microbial Growth
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Oxygen requirements shape microbial growth and survival. From obligate aerobes thriving in oxygen-rich environments to anaerobes that can't tolerate it, microbes have diverse needs. Understanding these requirements is crucial for studying microbial ecology and pathogenesis.
Carbon dioxide also plays a role, with some microbes needing higher levels for growth. These gas requirements influence where microbes can live and how they interact with their environment, impacting everything from human health to environmental processes.
Oxygen Requirements for Microbial Growth
Graphs of microbial gas requirements
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Oxygen concentration
Aerobic microbes require higher oxygen levels for optimal growth thrive in oxygen-rich environments
Anaerobic microbes grow best in environments with little to no oxygen cannot tolerate high oxygen concentrations
Facultative anaerobes can grow in both high and low oxygen environments adapt to varying oxygen levels
Carbon dioxide concentration
Capnophiles require higher carbon dioxide levels for growth typically 5-10% CO2 for optimal growth
Most microbes tolerate a range of carbon dioxide concentrations but may have specific preferences
Growth curves depict microbial population changes over time in different conditions
Lag phase: microbes adapt to new environment, minimal growth preparing for exponential growth
Log phase: exponential growth, optimal conditions for the microbe population doubles at a constant rate
Stationary phase: growth rate equals death rate, nutrient depletion or waste accumulation limiting factors balance population
Death phase: death rate exceeds growth rate, population declines due to unfavorable conditions or resource exhaustion
Categories of oxygen requirements
Obligate aerobes
Require oxygen for growth and survival cannot survive without oxygen
Use oxygen as the terminal electron acceptor in aerobic respiration to generate ATP
Utilize the electron transport chain for efficient energy production
Microaerophiles
Require low levels of oxygen for growth (2-10%) thrive in environments with limited oxygen
Sensitive to high oxygen concentrations, which can be toxic due to oxidative stress
Facultative anaerobes
Grow in both aerobic and anaerobic environments versatile and adaptable
Prefer aerobic conditions but can switch to fermentation or anaerobic respiration when oxygen is limited
Aerotolerant anaerobes
Do not require oxygen for growth but can tolerate its presence oxygen does not harm them
Lack the ability to use oxygen in energy-generating processes rely on fermentation or anaerobic respiration
Obligate anaerobes
Cannot tolerate oxygen and are killed by its presence oxygen is toxic to these microbes
Lack enzymes to detoxify oxygen radicals which cause cellular damage
Can be cultured using an anaerobic jar to create oxygen-free conditions
Capnophiles
Require higher levels of carbon dioxide (5-10%) for optimal growth CO2 is essential for their metabolism
Often found in environments with high carbon dioxide, like the human body (respiratory tract, gastrointestinal tract)
Examples of oxygen-dependent microbes
Obligate aerobes
Pseudomonas aeruginosa: opportunistic pathogen, causes infections in immunocompromised patients (pneumonia, sepsis)
Mycobacterium tuberculosis: causative agent of tuberculosis infects lungs and other tissues
Microaerophiles
Helicobacter pylori: causes gastric ulcers and is linked to stomach cancer survives in the mucus layer of the stomach
Campylobacter jejuni: common cause of bacterial foodborne illness associated with undercooked poultry
Facultative anaerobes
Escherichia coli: normal gut flora, some strains cause intestinal and extraintestinal infections (UTIs, meningitis)
Staphylococcus aureus: skin commensal, can cause various infections like pneumonia and sepsis also forms biofilms
Aerotolerant anaerobes
Streptococcus mutans: primary cause of dental caries ferments sugars to produce acid that erodes tooth enamel
Lactobacillus acidophilus: probiotic species found in yogurt and supplements helps maintain healthy gut flora
Obligate anaerobes
Clostridium tetani: produces the neurotoxin that causes tetanus spores are highly resistant to environmental conditions
Bacteroides fragilis: most common anaerobic pathogen, causes abdominal infections often involved in peritonitis and abscesses
Capnophiles
Neisseria gonorrhoeae: causative agent of gonorrhea sexually transmitted infection
Haemophilus influenzae: can cause meningitis, pneumonia, and otitis media common in young children
Oxygen-related microbial processes and tests
Oxidase test: determines the presence of cytochrome c oxidase in bacteria, important for identifying certain species
Catalase test: detects the enzyme catalase, which breaks down hydrogen peroxide and is present in many aerobic and facultative anaerobic bacteria
Reactive oxygen species: harmful byproducts of oxygen metabolism that can damage cellular components
Redox potential: measure of the tendency of a chemical species to acquire electrons, influencing microbial growth in different environments
Oxygen toxicity: damaging effects of excessive oxygen on microbial cells, particularly relevant for anaerobic organisms
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190203 | https://math.stackexchange.com/questions/4252046/winning-strategy-for-a-combinatorial-game | Skip to main content
Winning strategy for a combinatorial game
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There are 107 stones on the desk. Two people take turns at taking out stones and they can take out any number of stones as long as the number is the format of Pn. Here P is any prime number, and n is any nonnegative integer. The person who takes the last stone wins. Is there a winning strategy for the first player?
I have tried the following:
(1) Per the Pn rule, each player can take out 1, 2, 3, 4, or 5, but not 6 stones every time (the player can also take 7, 9, 11, 13, … stones).
(2)If there are just 1, 2, 3, 4, or 5 stones left, the player who moves next wins simply by taking all the stones.
(3) With 6 stones left, the player who moves next must leave 1, 2, 3, 4, or 5 stones in the pile and the current player (the one just moved) will be able to win. So leaving 6 stones to the opponent is a winning strategy for the current player.
(4) With 12 stones left, the player who moves next must leave 1, 3, 5, 7, 8, 9, 10, or 11 stones in the pile by taking 1, 2, 3, 4, 5, 7, 9, or 11 stones. The current player can then win by taking all stones (if there are less than 6 stones left) or by moving to the position with 6 stones left.
(5) So to win the game, we like to move into the target positions of leaving 0, 6 or 12 stones.
(6) How can I theorize that the winning strategy is to leave a multiple of 6 stones?
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asked Sep 16, 2021 at 14:06
TanakaTanaka
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Hint
In order to prove that the winning strategy is to always leave a multiple of six stones, you need to show two things:
If the current number of stones is not a multiple of six, then there exists a move which makes it a multiple of six.
If the current number of stones is a multiple of six, then every move will leave a heap which is not a multiple of six.
This shows that a player who follows the strategy is never without a move, and therefore wins.
I think you should be able to prove both of these facts on your own. I've included a proof of the first bullet behind a spoiler block in case you get stuck. For the second bullet, all you need is some light number theory.
The number of stones can be written in the form 6q+r for some positive integers q,r with r∈{1,…,5} (we exclude r=0 since we assumed the number is not a multiple of six). The current player can leave 6q stones by removing r stones, which is legal since every number in {1,…,5} is a prime power.
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answered Sep 16, 2021 at 14:18
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Nosocomial Infections (Healthcare-Associated Infections)
Nosocomial infections — also called healthcare-associated infections (HAIs) — are infections you can get while in a healthcare facility. HAIs may occur after a medical or surgical procedure. They can be mild or life-threatening. You can avoid most nosocomial infections with diligent infection prevention efforts.
ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionLiving With
Overview
What is nosocomial infection?
Nosocomial (naa-suh-KOW-mee-uhl) infections (also called healthcare-associated infections, hospital-acquired infections or HAIs) are illnesses you can get at a medical facility while you’re getting treatment for another condition. These infections can cause serious and sometimes life-threatening conditions. HAIs can occur in a variety of settings like hospitals, surgical centers, dialysis clinics and long-term care facilities.
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By definition, HAIs are infections that happen within:
Forty-eight hours of arrival or hospital admission.
Three days after discharge from a hospital or surgical center.
Thirty days of a surgical procedure.
Anyone receiving care at a healthcare facility can get nosocomial infections. But they’re most common in people with compromised immune systems. You can get HAIs from these pathogens (germs):
Bacteria.
Viruses.
Funguses.
Most HAIs are avoidable. Yet every year in the United States, more than a half-million patients and healthcare providers get them. To stay safe, it’s important for medical teams and families to work together.
Types of healthcare-associated infections (HAIs)
There are several types of nosocomial infections and many reasons why they occur:
C. diff (Clostridioides difficile) infection: C. diff is one of the most common HAIs, causing nearly half a million infections in the U.S. each year. It causes diarrhea and colitis (inflammation of the colon). Over 16% of people who get a C. diff infection will get another one within two to eight weeks.
Catheter-associated urinary tract infection (CAUTI):CAUTIs make up 32% of all HAIs. These urinary tract infections can develop due to long-term catheter use. A catheter is a tube that goes from your bladder through your urethra to allow urine to drain from your body into a collection bag.
Central line-associated bloodstream infection (CLABSI): Bloodstream infections make up 14% of all HAIs. They can happen in people who need a central line (central venous catheter). A provider places the central venous catheter into a vein near your heart so they can give medications and take blood without multiple injections.
MRSA (methicillin-resistant Staphylococcus aureus): MRSA is one of the most common HAIs. It’s a staph infection that’s resistant to certain antibiotics like penicillin, amoxicillin and methicillin.
Surgical site infection (SSI):About 22% of all HAIs are surgical site infections. SSIs can affect incision wounds. They can also affect deeper tissues, organs and even surgical implants (like pacemakers or artificial joints).
Ventilator-associated pneumonia (VAP):Ventilator-associated pneumonia accounts for 15% of all HAIs. This condition can develop in people who need mechanical ventilation — a type of treatment that breathes for you when you can’t breathe on your own. Infection happens if germs enter the lungs through a breathing tube.
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Symptoms and Causes
What are the symptoms of nosocomial infections?
People with HAIs can develop different symptoms depending on the type of infection. Some of the most common nosocomial infection symptoms include:
Fever.
Chills.
Shortness of breath (dyspnea).
Cough.
Abdominal pain.
Heart palpitations.
Diarrhea.
Peeing more than usual (polyuria).
Painful urination (dysuria).
Change in mental state, like confusion or irritability.
What causes nosocomial infections?
A variety of pathogens (germs) can cause nosocomial infections. Common examples include:
Gram-negative bacteria, which can cause blood infections, pneumonia or meningitis.
Clostridioides difficile (C. diff), a bacterium (singular form of bacteria) that can cause colon inflammation and a very contagious form of diarrhea.
Methicillin-resistant Staphylococcus aureus(MRSA), a type of bacterial infection resistant to common antibiotics like penicillin, amoxicillin and methicillin.
Hepatitis, a group of easily transmittable viral infections that affect the liver.
How do nosocomial infections spread?
Nosocomial infections can spread through respiratory droplets when a person sneezes or coughs. They can also occur when healthcare providers don’t follow proper infection control procedures (like sterilization).
What are the risk factors for nosocomial infections?
A risk factor is something that increases your chance of getting an illness. You’re more likely to get a nosocomial infection if your treatment involves:
Catheters (drainage tubes).
Ventilation (mechanical breathing).
Injections.
Surgery.
What are the complications of nosocomial infections?
Complications depend on the type and severity of the nosocomial infection. Listed below are the types of HAIs and some of the possible complications:
Central line-associated bloodstream infection (CLABSI)
Endocarditis.
Thrombophlebitis.
Osteomyelitis.
Abscess.
Sepsis.
Surgical site infection (SSI)
Slow wound healing.
Repeat surgery.
Rejection of implanted medical devices.
Body cavity infections.
Sepsis.
Catheter-associated urinary tract infection (CAUTI)
Urinary tract infections.
Sepsis.
Ventilator-associated pneumonia (VAP)
Respiratory failure.
Empyema.
Sepsis.
Diagnosis and Tests
How are nosocomial infections diagnosed?
Most of the time, healthcare providers can tell a person has an HAI when they develop symptoms like the ones mentioned above. Early signs may include a skin rash or redness around an incision. Your provider might need to order urinalyses (pee tests) or blood tests to confirm or rule out a diagnosis.
Management and Treatment
How are nosocomial infections treated?
Treatment for HAIs depends on the type of infection. Healthcare providers may use:
Antibiotics.
Antivirals.
Antifungals.
Challenges like antibiotic resistance, antiviral resistance and antifungal resistance can sometimes make HAI treatment less effective. Resistance is when pathogens evolve (change) so much that the drugs designed to treat them no longer work. This is one reason HAIs are so difficult to treat.
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Many people with HAIs make a full recovery with early diagnosis and treatment. But they also have longer hospital stays and additional obstacles to overcome.
In some cases, nosocomial infections can cause serious health complications or death. According to the Centers for Disease Control and Prevention (CDC), about 99,000 people die from healthcare-associated infections each year in American hospitals alone.
Prevention
Can nosocomial infections be prevented?
You can’t prevent every single instance of hospital-acquired infections. But you and your family can take the following steps to significantly reduce your risk:
Keep your hands clean. Regular handwashing is one of the best ways to prevent spreading germs and avoid getting sick.
Ask your healthcare providers about their infection control procedures. Don’t be afraid to speak up if you have concerns.
If you have a catheter, ask each day if it’s necessary.
Take antibiotics exactly as prescribed and only when your provider says you need them.
Ask your healthcare provider to run tests to ensure they prescribe the appropriate antibiotic.
Ask your healthcare provider how they prevent infection during and after surgery. Ask how you can prepare for surgery to help reduce your risk.
Recognize the signs of skin infection. Redness, draining, or pain around incisions or catheter insertion sites could indicate infection.
Stay up to date on your flu shot and other recommended vaccinations.
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You should tell your healthcare provider any time you develop concerning or worsening symptoms — especially if you’re currently under medical care or had recent surgery. If something doesn’t seem right, let your provider know right away so they can recommend appropriate treatment.
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190205 | https://www.encyclopedia.com/manufacturing/news-wires-white-papers-and-books/hockey-puck | Hockey Puck
Hockey Puck
Background
Hockey pucks are flat, solid, black disk-shaped objects made of vulcanized rubber. Regulation National Hockey League (NHL) pucks are black, 3 in (7.6 cm) in diameter, 1 in (2.54 cm) thick, and weighing 5.5-6 oz (154-168 g). The edge has a series of "diamonds," slightly raised bumps or grooves. The diamonds give a taped hockey stick something to grip when the puck is shot. The blue pucks used in junior hockey are sometimes only 4 oz (143 g).
During a game, each team keeps a supply of pucks in a freezer at all times. When a professional hockey team receives their supply of pucks for a season, they are rotated so that the older pucks are used first. During games, pucks are kept frozen in an icepacked cooler, which usually sits on the officials' bench. All pucks are frozen to reduce the amount of bounce.
Though no one knows exactly how the hockey puck got its name, many believe that it was named for the character in William Shakespeare's A Midsummer's Night Dream. Like the impish flighty Puck, the hockey disk moves very quickly, sometimes in unexpected directions.
History
Played in Europe for several hundred years, field hockey is a predecessor of ice hockey, which sprang up in Great Britain during the 1820s. The game blossomed in the British protectorate of Canada in the second half of the nineteenth century. In Canada where long, cold winters are a certainty, ice hockey soon became the national game. Hockey also became popular in the northern parts of the United States during the same time period.
At first, amateurs dominated hockey and the rules were ever changing. The first professional league was organized in 1904 and called the International Hockey League. It only lasted three years. In 1917, the National Hockey League (NHL) was created, and is still the top level of professional hockey played in North America today. With the establishment of the NHL came codified rules and regularization of the game. Today, hockey is played by all ages, both men and women, throughout North America and many parts of the world.
In the early years, c. 1860-1870s, a rubber ball was the object used in hockey. Because the ball bounced too much, a block of wood was sometimes used instead. The modern hockey puck was invented around 1875. There are two different versions of its origination. One story claims that in 1875, students at Boston University sliced a rubber ball in half to make a puck. Another version places the evolution in Montreal, Quebec, Canada. The owner of one of the first indoor ice rinks, Victoria Rink, also allegedly sliced a rubber ball in half. In any case, the first recorded use of a flat disk was in Montreal in March 1875.
Early pucks were made by gluing two pieces of rubber together (sometimes from recycled tires). Because of this construction, the pucks could split when they hit the goal post. During the 1931-1932 season, a puck with beveled (sloped) edges was used. By midseason, complaints by players and teams led to the return of the original puck. Though there was no official NHL puck until the
1990-1991 season, the basic construction from the early 1900s remained the same.
The FoxTrax puck
During the 1995-1996 NHL season, a slightly different puck was introduced. While the outside of the puck remained the same, the inside and effect was totally different. That year, the Fox television network obtained the rights to air the NHL All-Star Game and the Stanley Cup playoffs. Fox believed that to attract new viewers to the game, the network had to make the smalllooking puck easier to follow on television. To that end, they developed an enhanced puck called the FoxTrax puck. It contained a computer board and battery at its center and 20-pin holes all over the puck (12 on the edges, four on top, and four on the bottom) that guided infrared emitters, each beeping approximately 30 pulses per minute. These emitters communicated with 16 sensoring devices placed around the rink to follow the puck's movement. The sensoring devices were linked by fiber optics to computers outside in the "Puck Truck."
When processed by computer, the FoxTrax puck had a completely different look to the television audience. It had a translucent blue halo, which was supposed to make the puck more visible on a small screen. When a player shot the puck at speeds exceeding 50 mph (80 kph), a red tail appeared on television. If the puck reached speeds over 75 mph (120 kph), the tail was green. When put into play, each FoxTrax was remotely activated by a wireless controller. Unlike standard pucks, which were used until they went into the stands or otherwise damaged, FoxTrax pucks could only be used for about 10 minutes before the battery ran out. While Fox-Trax pucks weigh about the same as NHL regulation pucks, they cost much more to make. Each puck had a value of about $400.
From its first use, players complained that the FoxTrax puck did not move the same way a normal puck did. The FoxTrax puck also did not hold the cold as well. FoxTrax pucks became bouncy much more quickly than their regulation counterparts. When the Fox network declined to renew its contract to air the NHL All-Star Game and play-offs after the 1998-1999 season, the FoxTrax puck was no longer used or manufactured.
Raw Materials
A hockey puck is made of vulcanized rubber. The top and bottom of some pucks are decorated with team and/or league logos. These logos are silk-screened on to the rubber. The silkscreen process uses a rubber-based ink and four-color processing.
In addition to the rubber and silk-screened ink, the FoxTrax puck included several computer components—a lithium battery; 20 infrared emitters; a ceramic oscillator; an accelerometer; CMOS logic and switching; a four-layer, silver-dollar sized circuit board; surface mount parts; and a flexible epoxy to pot the board.
Design
The design of NHL regulation pucks was regularized in 1940 by Art Ross. Though pucks remained basically the same, Ross's innovation was a puck that was easy to manufacture and acted with some consistency when used in play.
Logos that are silk-screened on the puck are designed by the various professional hockey leagues (including the NHL) and individual teams.
The Manufacturing Process
Currently, hockey pucks are only made in four countries: Canada, Russia, China, and the Czech Republic. There are two kinds of manufacturing processes for pucks. One is for practice and souvenir pucks. The other is for regulation NHL and other professional league pucks that are used in games.
Practice/souvenir pucks
Regulation NHL pucks
FoxTrax pucks
Blue (junior hockey) pucks
Quality Control
Pucks are checked for the regulation size and weight. If regulation pucks do not meet prescribed standards, they are recycled and the rubber is reused to make pucks. After regulation pucks are made, certain specimens are frozen for 10 days, then bounced. The tester ensures that the pucks bounce the same ways as those in previous batches. A consistent product is important in the production of pucks. Every puck must act the same way on the ice.
During the silk-screen process, the ink can be affected by the moisture in the air, dust particles, and hair. The pucks are checked for the effect of any of these qualities. Any effected pucks are washed with paint thinner and go through the silk-screening process again.
In the past, Russian-made pucks sometimes had metal fragments in them. These pucks were rejected for use by North American markets. Pucks with air bubbles or softer rubber in the middle were rejected for similar reasons.
Wayne Douglas Gretzky was born on January 26, 1961, in Brantford, Ontario. Able to skate at two, he signed with World Hockey Association's Indianapolis Racers at 17. After eight games, Gretzky was sold to the Edmonton Oilers, who were admitted to the National Hockey League (NHL) in 1979. In his nine seasons with the Oilers, from 1980 to 1988, Gretzky scored 583 goals and handed out 1,086 assists. For six of those years he averaged 73 goals and 130 assists a season and led the Oilers to four Stanley Cup championships. Gretzky led Team Canada to win Canada's Cup in 1987. He was traded to the Los Angeles Kings in August of 1988 and in his first year scored 54 goals and passed for 114 assists. The next season, Gretzky broke Gordie Howe's all-time scoring record of 1,850 points. In February 1996, he was traded to the St. Louis Blues, and the next season signed with the New York Rangers as a free agent.
Gretzky retired on April 16, 1999, having established records of 2,857 points; 1,963 assists; 894 goals; and played in 1,486 games all over his 20 seasons in the NHL. Gretzky was awarded nine league MVPs (Most Valuable Player), three All-Star Game MVPs, and 10 Ross awards. The three-year waiting period was waved, and Gretzky was inducted to the Hockey Hall of Fame on November 22, 1999.
Byproducts/Waste
Any excess rubber from the manufacturing process is collected, re-shred, and used again to make pucks.
The Future
The future does not involve much change to the actual puck, its composition, or manufacture. While a blue-colored puck is currently made for junior hockey, pucks of different colors serve no purpose in the game and are not likely to be manufactured on a large scale. Any improvements to the silk-screen process would result in changes in the decoration of pucks. Plated souvenir pucks might be available in the future.
Where to Learn More
Books
Duplacey, James. Puck. In The Annotated Rules of Hockey. New York: Lyons & Burford, 1996, pp. 52-54.
Periodicals
Modoono, Bill. "Puck's History is Hardly the Stuff of Legends or Lore." Star-Tribune Newspaper of the Twin Cities Minneapolis-St. Paul (December 13, 1992): 3C.
Vizard. "Hockey's Chip Shot." Popular Mechanics (May 1996): 40.
—AnnettePetruso
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190206 | https://en.wikipedia.org/wiki/Magnetic_field | Published Time: Fri, 19 Sep 2025 02:14:24 GMT
Magnetic field - Wikipedia
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1 DescriptionToggle Description subsection
1.1 The B-field
1.2 The H-field
1.3 Measurement
1.4 Visualization
2 Magnetic field of permanent magnetsToggle Magnetic field of permanent magnets subsection
2.1 Magnetic pole model
2.2 Amperian loop model
3 Interactions with magnetsToggle Interactions with magnets subsection
3.1 Force between magnets
3.2 Magnetic torque on permanent magnets
4 Interactions with electric currentsToggle Interactions with electric currents subsection
4.1 Magnetic field due to moving charges and electric currents
4.2 Force on moving charges and current
4.2.1 Force on a charged particle
4.2.2 Force on current-carrying wire
5 Relation between H and BToggle Relation between H and B subsection
5.1 Magnetization
5.2 H-field and magnetic materials
5.3 Magnetism
6 Stored energy
7 Appearance in Maxwell's equationsToggle Appearance in Maxwell's equations subsection
7.1 Gauss' law for magnetism
7.2 Faraday's Law
7.3 Ampère's Law and Maxwell's correction
8 Formulation in special relativity and quantum electrodynamicsToggle Formulation in special relativity and quantum electrodynamics subsection
8.1 Relativistic electrodynamics
8.1.1 As different aspects of the same phenomenon
8.1.2 Magnetic vector potential
8.1.3 Propagation of Electric and Magnetic fields
8.1.4 Magnetic field of arbitrary moving point charge
8.2 Quantum electrodynamics
9 Uses and examplesToggle Uses and examples subsection
9.1 Earth's magnetic field
9.2 Rotating magnetic fields
9.3 Hall effect
9.4 Magnetic circuits
9.5 Largest magnitude magnetic fields
10 Common formulæ
11 HistoryToggle History subsection
11.1 Early developments
11.2 Mathematical development
11.3 Modern developments
12 See alsoToggle See also subsection
12.1 General
12.2 Mathematics
12.3 Applications
13 Notes
14 References
15 Further reading
16 External links
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From Wikipedia, the free encyclopedia
Distribution of magnetic force
For other uses, see Magnetic field (disambiguation).
A permanent magnet, a piece of magnetized metal alloy
A solenoid (electromagnet), a coil of wire with an electric current through it
The shape of the magnetic fields of a permanent magnet and an electromagnet are revealed by the orientation of iron filings sprinkled on pieces of paper
A magnetic field (sometimes called B-field) is a physical field that describes the magnetic influence on moving electric charges, electric currents,: ch1 and magnetic materials. A moving charge in a magnetic field experiences a force perpendicular to its own velocity and to the magnetic field.: ch13 : 278 A permanent magnet's magnetic field pulls on ferromagnetic materials such as iron, and attracts or repels other magnets. In addition, a nonuniform magnetic field exerts minuscule forces on "nonmagnetic" materials by three other magnetic effects: paramagnetism, diamagnetism, and antiferromagnetism, although these forces are usually so small they can only be detected by laboratory equipment. Magnetic fields surround magnetized materials, electric currents, and electric fields varying in time. Since both strength and direction of a magnetic field may vary with location, it is described mathematically by a function assigning a vector to each point of space, called a vector field (more precisely, a pseudovector field).
In electromagnetics, the term magnetic field is used for two distinct but closely related vector fields denoted by the symbols B and H. In the International System of Units, the unit of B, magnetic flux density, is the tesla (in SI base units: kilogram per second squared per ampere),: 21 which is equivalent to newton per meter per ampere. The unit of H, magnetic field strength, is ampere per meter (A/m).: 22 B and H differ in how they take the medium and/or magnetization into account. In vacuum, the two fields are related through the vacuum permeability, B/μ 0=H{\displaystyle \mathbf {B} /\mu _{0}=\mathbf {H} }; in a magnetized material, the quantities on each side of this equation differ by the magnetization field of the material.
Magnetic fields are produced by moving electric charges and the intrinsic magnetic moments of elementary particles associated with a fundamental quantum property, their spin.: ch1 Magnetic fields and electric fields are interrelated and are both components of the electromagnetic force, one of the four fundamental forces of nature.
Magnetic fields are used throughout modern technology, particularly in electrical engineering and electromechanics. Rotating magnetic fields are used in both electric motors and generators. The interaction of magnetic fields in electric devices such as transformers is conceptualized and investigated as magnetic circuits. Magnetic forces give information about the charge carriers in a material through the Hall effect. The Earth produces its own magnetic field, which shields the Earth's ozone layer from the solar wind and is important in navigation using a compass.
Description
The force on an electric charge depends on its location, speed, and direction; two vector fields are used to describe this force.: ch1 The first is the electric field, which describes the force acting on a stationary charge and gives the component of the force that is independent of motion. The magnetic field, in contrast, describes the component of the force that is proportional to both the speed and direction of charged particles.: ch13 The field is defined by the Lorentz force law and is, at each instant, perpendicular to both the motion of the charge and the force it experiences.
There are two different, but closely related vector fields which are both sometimes called the "magnetic field" written B and H.[note 1] While both the best names for these fields and exact interpretation of what these fields represent has been the subject of long running debate, there is wide agreement about how the underlying physics work. Historically, the term "magnetic field" was reserved for H while using other terms for B, but many recent textbooks use the term "magnetic field" to describe B as well as or in place of H.[note 2] There are many alternative names for both (see sidebars).
The B-field
Finding the magnetic force
A charged particle that is moving with velocity v in a magnetic field B will feel a magnetic force F. Since the magnetic force always pulls sideways to the direction of motion, the particle moves in a circle.
Since these three vectors are related to each other by a cross product, the direction of this force can be found using the right hand rule.
| Alternative names for B |
| Magnetic flux density: 138 Magnetic induction Magnetic field (ambiguous) |
The magnetic field vector B at any point can be defined as the vector that, when used in the Lorentz force law, correctly predicts the force on a charged particle at that point:: 204
Lorentz force law (vector form, SI units)
F=q E+q(v×B){\displaystyle \mathbf {F} =q\mathbf {E} +q(\mathbf {v} \times \mathbf {B} )}
Here F is the force on the particle, q is the particle's electric charge, E is the external electric field, v, is the particle's velocity, and × denotes the cross product. The direction of force on the charge can be determined by a mnemonic known as the right-hand rule (see the figure).[note 3] Using the right hand, pointing the thumb in the direction of the current, and the fingers in the direction of the magnetic field, the resulting force on the charge points outwards from the palm. The force on a negatively charged particle is in the opposite direction. If both the speed and the charge are reversed then the direction of the force remains the same. For that reason a magnetic field measurement (by itself) cannot distinguish whether there is a positive charge moving to the right or a negative charge moving to the left. (Both of these cases produce the same current.) On the other hand, a magnetic field combined with an electric field can distinguish between these, see Hall effect below.
The first term in the Lorentz equation is from the theory of electrostatics, and says that a particle of charge q in an electric field E experiences an electric force: F electric=q E.{\displaystyle \mathbf {F} _{\text{electric}}=q\mathbf {E} .}
The second term is the magnetic force:F magnetic=q(v×B).{\displaystyle \mathbf {F} _{\text{magnetic}}=q(\mathbf {v} \times \mathbf {B} ).}
Using the definition of the cross product, the magnetic force can also be written as a scalar equation:: 357 F magnetic=q v B sin(θ){\displaystyle F_{\text{magnetic}}=qvB\sin(\theta )} where F magnetic, v, and B are the scalar magnitude of their respective vectors, and θ is the angle between the velocity of the particle and the magnetic field. The vector B is defined as the vector field necessary to make the Lorentz force law correctly describe the motion of a charged particle. In other words,: 173–4
[T]he command, "Measure the direction and magnitude of the vector B at such and such a place," calls for the following operations: Take a particle of known charge q. Measure the force on q at rest, to determine E. Then measure the force on the particle when its velocity is v; repeat with v in some other direction. Now find a B that makes the Lorentz force law fit all these results—that is the magnetic field at the place in question.
The B field can also be defined by the torque on a magnetic dipole, m.: 174
Magnetic torque (vector form, SI units)
τ=m×B{\displaystyle {\boldsymbol {\tau }}=\mathbf {m} \times \mathbf {B} }
The SI unit of B is tesla (symbol: T).[note 4] The Gaussian-cgs unit of B is the gauss (symbol: G). (The conversion is 1 T ≘ 10000 G.) One nanotesla corresponds to 1 gamma (symbol: γ).
The H-field
| Alternative names for H |
| Magnetic field intensity Magnetic field strength: 139 Magnetic field Magnetizing field Auxiliary magnetic field |
The magnetic H field is defined:: 269 : 192 : ch36
Definition of the H field(vector form, SI units)
H≡1 μ 0 B−M{\displaystyle \mathbf {H} \equiv {\frac {1}{\mu _{0}}}\mathbf {B} -\mathbf {M} }
where μ 0{\displaystyle \mu _{0}} is the vacuum permeability, and M is the magnetization vector. In a vacuum, B and H are proportional to each other. Inside a material they are different (see H and B inside and outside magnetic materials). The SI unit of the H-field is the ampere per metre (A/m), and the CGS unit is the oersted (Oe).: 286
Measurement
Main article: Magnetometer
An instrument used to measure the local magnetic field is known as a magnetometer. Important classes of magnetometers include using induction magnetometers (or search-coil magnetometers) which measure only varying magnetic fields, rotating coil magnetometers, Hall effect magnetometers, NMR magnetometers, SQUID magnetometers, and fluxgate magnetometers. The magnetic fields of distant astronomical objects are measured through their effects on local charged particles. For instance, electrons spiraling around a field line produce synchrotron radiation that is detectable in radio waves. The finest precision for a magnetic field measurement was attained by Gravity Probe B at 5 aT (5×10−18 T).
Visualization
Main article: Field line
Visualizing magnetic fields
Left: the direction of magnetic field lines represented by iron filings sprinkled on paper placed above a bar magnet.
Right: compass needles point in the direction of the local magnetic field, towards a magnet's south pole and away from its north pole.
The field can be visualized by a set of magnetic field lines, that follow the direction of the field at each point. The lines can be constructed by measuring the strength and direction of the magnetic field at a large number of points (or at every point in space). Then, mark each location with an arrow (called a vector) pointing in the direction of the local magnetic field with its magnitude proportional to the strength of the magnetic field. Connecting these arrows then forms a set of magnetic field lines. The direction of the magnetic field at any point is parallel to the direction of nearby field lines, and the local density of field lines can be made proportional to its strength. Magnetic field lines are like streamlines in fluid flow, in that they represent a continuous distribution, and a different resolution would show more or fewer lines.
An advantage of using magnetic field lines as a representation is that many laws of magnetism (and electromagnetism) can be stated completely and concisely using simple concepts such as the "number" of field lines through a surface. These concepts can be quickly "translated" to their mathematical form. For example, the number of field lines through a given surface is the surface integral of the magnetic field.: 237
Various phenomena "display" magnetic field lines as though the field lines were physical phenomena. For example, iron filings placed in a magnetic field form lines that correspond to "field lines".[note 5] Magnetic field "lines" are also visually displayed in polar auroras, in which plasma particle dipole interactions create visible streaks of light that line up with the local direction of Earth's magnetic field.
Field lines can be used as a qualitative tool to visualize magnetic forces. In ferromagnetic substances like iron and in plasmas, magnetic forces can be understood by imagining that the field lines exert a tension, (like a rubber band) along their length, and a pressure perpendicular to their length on neighboring field lines. "Unlike" poles of magnets attract because they are linked by many field lines; "like" poles repel because their field lines do not meet, but run parallel, pushing on each other.
Magnetic field of permanent magnets
Main article: Magnetic moment §Models
Permanent magnets are objects that produce their own persistent magnetic fields. They are made of ferromagnetic materials, such as iron and nickel, that have been magnetized, and they have both a north and a south pole.
The magnetic field of permanent magnets can be quite complicated, especially near the magnet. The magnetic field of a small[note 6] straight magnet is proportional to the magnet's strength (called its magnetic dipole momentm). The equations are non-trivial and depend on the distance from the magnet and the orientation of the magnet. For simple magnets, m points in the direction of a line drawn from the south to the north pole of the magnet. Flipping a bar magnet is equivalent to rotating its m by 180 degrees.
The magnetic field of larger magnets can be obtained by modeling them as a collection of a large number of small magnets called dipoles each having their own m. The magnetic field produced by the magnet then is the net magnetic field of these dipoles; any net force on the magnet is a result of adding up the forces on the individual dipoles.
There are two simplified models for the nature of these dipoles: the magnetic pole model and the Amperian loop model. These two models produce two different magnetic fields, H and B. Outside a material, though, the two are identical (to a multiplicative constant) so that in many cases the distinction can be ignored. This is particularly true for magnetic fields, such as those due to electric currents, that are not generated by magnetic materials.
A realistic model of magnetism is more complicated than either of these models; neither model fully explains why materials are magnetic. The monopole model has no experimental support. The Amperian loop model explains some, but not all of a material's magnetic moment. The model predicts that the motion of electrons within an atom are connected to those electrons' orbital magnetic dipole moment, and these orbital moments do contribute to the magnetism seen at the macroscopic level. However, the motion of electrons is not classical, and the spin magnetic moment of electrons (which is not explained by either model) is also a significant contribution to the total moment of magnets.
Magnetic pole model
See also: Magnetic monopole
The magnetic pole model: two opposing poles, North (+) and South (−), separated by a distance d produce a H-field (lines).
Historically, early physics textbooks would model the force and torques between two magnets as due to magnetic poles repelling or attracting each other in the same manner as the Coulomb force between electric charges. At the microscopic level, this model contradicts the experimental evidence, and the pole model of magnetism is no longer the typical way to introduce the concept.: 258 However, it is still sometimes used as a macroscopic model for ferromagnetism due to its mathematical simplicity.
In this model, a magnetic H-field is produced by fictitious magnetic charges that are spread over the surface of each pole. These magnetic charges are in fact related to the magnetization field M. The H-field, therefore, is analogous to the electric fieldE, which starts at a positive electric charge and ends at a negative electric charge. Near the north pole, therefore, all H-field lines point away from the north pole (whether inside the magnet or out) while near the south pole all H-field lines point toward the south pole (whether inside the magnet or out). Too, a north pole feels a force in the direction of the H-field while the force on the south pole is opposite to the H-field.
In the magnetic pole model, the elementary magnetic dipole m is formed by two opposite magnetic poles of pole strength q m separated by a small distance vector d, such that m = q md. The magnetic pole model predicts correctly the field H both inside and outside magnetic materials, in particular the fact that H is opposite to the magnetization field M inside a permanent magnet.
Since it is based on the fictitious idea of a magnetic charge density, the pole model has limitations. Magnetic poles cannot exist apart from each other as electric charges can, but always come in north–south pairs. If a magnetized object is divided in half, a new pole appears on the surface of each piece, so each has a pair of complementary poles. The magnetic pole model does not account for magnetism that is produced by electric currents, nor the inherent connection between angular momentum and magnetism.
The pole model usually treats magnetic charge as a mathematical abstraction, rather than a physical property of particles. However, a magnetic monopole is a hypothetical particle (or class of particles) that physically has only one magnetic pole (either a north pole or a south pole). In other words, it would possess a "magnetic charge" analogous to an electric charge. Magnetic field lines would start or end on magnetic monopoles, so if they exist, they would give exceptions to the rule that magnetic field lines neither start nor end. Some theories (such as Grand Unified Theories) have predicted the existence of magnetic monopoles, but so far, none have been observed.
Amperian loop model
Main article: Magnetic dipole
See also: Spin magnetic moment and Micromagnetism
The Amperian loop model
A current loop (ring) that goes into the page at the x and comes out at the dot produces a B-field (lines). As the radius of the current loop shrinks, the fields produced become identical to an abstract "magnetostatic dipole" (represented by an arrow pointing to the right).
In the model developed by Ampere, the elementary magnetic dipole that makes up all magnets is a sufficiently small Amperian loop with current I and loop area A. The dipole moment of this loop is m = IA.
These magnetic dipoles produce a magnetic B-field.
The magnetic field of a magnetic dipole is depicted in the figure. From outside, the ideal magnetic dipole is identical to that of an ideal electric dipole of the same strength. Unlike the electric dipole, a magnetic dipole is properly modeled as a current loop having a current I and an area a. Such a current loop has a magnetic moment of m=I a,{\displaystyle m=Ia,} where the direction of m is perpendicular to the area of the loop and depends on the direction of the current using the right-hand rule. An ideal magnetic dipole is modeled as a real magnetic dipole whose area a has been reduced to zero and its current I increased to infinity such that the product m = Ia is finite. This model clarifies the connection between angular momentum and magnetic moment, which is the basis of the Einstein–de Haas effectrotation by magnetization and its inverse, the Barnett effect or magnetization by rotation. Rotating the loop faster (in the same direction) increases the current and therefore the magnetic moment, for example.
Interactions with magnets
Force between magnets
Main article: Force between magnets
Specifying the force between two small magnets is quite complicated because it depends on the strength and orientation of both magnets and their distance and direction relative to each other. The force is particularly sensitive to rotations of the magnets due to magnetic torque. The force on each magnet depends on its magnetic moment and the magnetic field[note 7] of the other.
To understand the force between magnets, it is useful to examine the magnetic pole model given above. In this model, the H-field of one magnet pushes and pulls on both poles of a second magnet. If this H-field is the same at both poles of the second magnet then there is no net force on that magnet since the force is opposite for opposite poles. If, however, the magnetic field of the first magnet is nonuniform (such as the H near one of its poles), each pole of the second magnet sees a different field and is subject to a different force. This difference in the two forces moves the magnet in the direction of increasing magnetic field and may also cause a net torque.
This is a specific example of a general rule that magnets are attracted (or repulsed depending on the orientation of the magnet) into regions of higher magnetic field. Any non-uniform magnetic field, whether caused by permanent magnets or electric currents, exerts a force on a small magnet in this way.
The details of the Amperian loop model are different and more complicated but yield the same result: that magnetic dipoles are attracted/repelled into regions of higher magnetic field. Mathematically, the force on a small magnet having a magnetic moment m due to a magnetic field B is:: Eq. 11.42
F=∇(m⋅B),{\displaystyle \mathbf {F} ={\boldsymbol {\nabla }}\left(\mathbf {m} \cdot \mathbf {B} \right),}
where the gradient∇ is the change of the quantity m · B per unit distance and the direction is that of maximum increase of m · B. The dot productm · B = mB cos(θ), where m and B represent the magnitude of the m and B vectors and θ is the angle between them. If m is in the same direction as B then the dot product is positive and the gradient points "uphill" pulling the magnet into regions of higher B-field (more strictly larger m · B). This equation is strictly only valid for magnets of zero size, but is often a good approximation for not too large magnets. The magnetic force on larger magnets is determined by dividing them into smaller regions each having their own m then summing up the forces on each of these very small regions.
Magnetic torque on permanent magnets
Main article: Magnetic torque
If two like poles of two separate magnets are brought near each other, and one of the magnets is allowed to turn, it promptly rotates to align itself with the first. In this example, the magnetic field of the stationary magnet creates a magnetic torque on the magnet that is free to rotate. This magnetic torque τ tends to align a magnet's poles with the magnetic field lines. A compass, therefore, turns to align itself with Earth's magnetic field.
Torque on a dipole
In the pole model of a dipole, an H field (to right) causes equal but opposite forces on a N pole (+q) and a S pole (−q) creating a torque.
Equivalently, a B field induces the same torque on a current loop with the same magnetic dipole moment.
In terms of the pole model, two equal and opposite magnetic charges experiencing the same H also experience equal and opposite forces. Since these equal and opposite forces are in different locations, this produces a torque proportional to the distance (perpendicular to the force) between them. With the definition of m as the pole strength times the distance between the poles, this leads to τ = μ 0 m H sin θ, where μ 0 is a constant called the vacuum permeability, measuring 4π×10−7V·s/(A·m) and θ is the angle between H and m.
Mathematically, the torque τ on a small magnet is proportional both to the applied magnetic field and to the magnetic moment m of the magnet:
τ=m×B=μ 0 m×H,{\displaystyle {\boldsymbol {\tau }}=\mathbf {m} \times \mathbf {B} =\mu _{0}\mathbf {m} \times \mathbf {H} ,\,}
where × represents the vector cross product. This equation includes all of the qualitative information included above. There is no torque on a magnet if m is in the same direction as the magnetic field, since the cross product is zero for two vectors that are in the same direction. Further, all other orientations feel a torque that twists them toward the direction of magnetic field.
Interactions with electric currents
Currents of electric charges both generate a magnetic field and feel a force due to magnetic B-fields.
Magnetic field due to moving charges and electric currents
Main articles: Electromagnet, Biot–Savart law, and Ampère's law
Right hand grip rule: a current flowing in the direction of the white arrow produces a magnetic field shown by the red arrows.
All moving charged particles produce magnetic fields. Moving point charges, such as electrons, produce complicated but well known magnetic fields that depend on the charge, velocity, and acceleration of the particles.
Magnetic field lines form in concentric circles around a cylindrical current-carrying conductor, such as a length of wire. The direction of such a magnetic field can be determined by using the "right-hand grip rule" (see figure at right). The strength of the magnetic field decreases with distance from the wire. (For an infinite length wire the strength is inversely proportional to the distance.)
A Solenoid with electric current running through it behaves like a magnet.
Bending a current-carrying wire into a loop concentrates the magnetic field inside the loop while weakening it outside. Bending a wire into multiple closely spaced loops to form a coil or "solenoid" enhances this effect. A device so formed around an iron core may act as an electromagnet, generating a strong, well-controlled magnetic field. An infinitely long cylindrical electromagnet has a uniform magnetic field inside, and no magnetic field outside. A finite length electromagnet produces a magnetic field that looks similar to that produced by a uniform permanent magnet, with its strength and polarity determined by the current flowing through the coil.
The magnetic field generated by a steady current I (a constant flow of electric charges, in which charge neither accumulates nor is depleted at any point)[note 8] is described by the Biot–Savart law:: 224 B=μ 0 I 4 π∫w i r e d ℓ×r^r 2,{\displaystyle \mathbf {B} ={\frac {\mu {0}I}{4\pi }}\int {\mathrm {wire} }{\frac {\mathrm {d} {\boldsymbol {\ell }}\times \mathbf {\hat {r}} }{r^{2}}},} where the integral sums over the wire length where vector dℓ is the vector line element with direction in the same sense as the current I, μ 0 is the magnetic constant, r is the distance between the location of dℓ and the location where the magnetic field is calculated, and r̂ is a unit vector in the direction of r. For example, in the case of a sufficiently long, straight wire, this becomes: |B|=μ 0 2 π r I{\displaystyle |\mathbf {B} |={\frac {\mu {0}}{2\pi r}}I} where _r = |r|. The direction is tangent to a circle perpendicular to the wire according to the right hand rule.: 225
A slightly more general[note 9] way of relating the current I{\displaystyle I} to the B-field is through Ampère's law: ∮B⋅d ℓ=μ 0 I e n c,{\displaystyle \oint \mathbf {B} \cdot \mathrm {d} {\boldsymbol {\ell }}=\mu {0}I{\mathrm {enc} },} where the line integral is over any arbitrary loop and I enc{\displaystyle I_{\text{enc}}} is the current enclosed by that loop. Ampère's law is always valid for steady currents and can be used to calculate the B-field for certain highly symmetric situations such as an infinite wire or an infinite solenoid.
In a modified form that accounts for time varying electric fields, Ampère's law is one of four Maxwell's equations that describe electricity and magnetism.
Force on moving charges and current
Force on a charged particle
Main article: Lorentz force
A charged particle moving in a B-field experiences a sideways force that is proportional to the strength of the magnetic field, the component of the velocity that is perpendicular to the magnetic field and the charge of the particle. This force is known as the Lorentz force, and is given by F=q E+q v×B,{\displaystyle \mathbf {F} =q\mathbf {E} +q\mathbf {v} \times \mathbf {B} ,} where F is the force, q is the electric charge of the particle, v is the instantaneous velocity of the particle, and B is the magnetic field (in teslas).
The Lorentz force is always perpendicular to both the velocity of the particle and the magnetic field that created it. When a charged particle moves in a static magnetic field, it traces a helical path in which the helix axis is parallel to the magnetic field, and in which the speed of the particle remains constant. Because the magnetic force is always perpendicular to the motion, the magnetic field can do no work on an isolated charge. It can only do work indirectly, via the electric field generated by a changing magnetic field. It is often claimed that the magnetic force can do work to a non-elementary magnetic dipole, or to charged particles whose motion is constrained by other forces, but this is incorrect because the work in those cases is performed by the electric forces of the charges deflected by the magnetic field.
Force on current-carrying wire
Main article: Laplace force
The force on a current carrying wire is similar to that of a moving charge as expected since a current carrying wire is a collection of moving charges. A current-carrying wire feels a force in the presence of a magnetic field. The Lorentz force on a macroscopic current is often referred to as the Laplace force. Consider a conductor of length ℓ, cross section A, and charge q due to electric current i. If this conductor is placed in a magnetic field of magnitude B that makes an angle θ with the velocity of charges in the conductor, the force exerted on a single charge q is F=q v B sinθ,{\displaystyle F=qvB\sin \theta ,} so, for N charges where N=n ℓ A,{\displaystyle N=n\ell A,} the force exerted on the conductor is f=F N=q v B n ℓ A sinθ=B i ℓ sinθ,{\displaystyle f=FN=qvBn\ell A\sin \theta =Bi\ell \sin \theta ,} where i = nqvA.
Relation between H and B
The formulas derived for the magnetic field above are correct when dealing with the entire current. A magnetic material placed inside a magnetic field, though, generates its own bound current, which can be a challenge to calculate. (This bound current is due to the sum of atomic sized current loops and the spin of the subatomic particles such as electrons that make up the material.) The H-field as defined above helps factor out this bound current; but to see how, it helps to introduce the concept of magnetization first.
Magnetization
Main article: Magnetization
The magnetization vector field M represents how strongly a region of material is magnetized. It is defined as the net magnetic dipole moment per unit volume of that region. The magnetization of a uniform magnet is therefore a material constant, equal to the magnetic moment m of the magnet divided by its volume. Since the SI unit of magnetic moment is A⋅m 2, the SI unit of magnetization M is ampere per meter, identical to that of the H-field.
The magnetization M field of a region points in the direction of the average magnetic dipole moment in that region. Magnetization field lines, therefore, begin near the magnetic south pole and ends near the magnetic north pole. (Magnetization does not exist outside the magnet.)
In the Amperian loop model, the magnetization is due to combining many tiny Amperian loops to form a resultant current called bound current. This bound current, then, is the source of the magnetic B field due to the magnet. Given the definition of the magnetic dipole, the magnetization field follows a similar law to that of Ampere's law:∮M⋅d ℓ=I b,{\displaystyle \oint \mathbf {M} \cdot \mathrm {d} {\boldsymbol {\ell }}=I_{\mathrm {b} },} where the integral is a line integral over any closed loop and I b is the bound current enclosed by that closed loop.
In the magnetic pole model, magnetization begins at and ends at magnetic poles. If a given region, therefore, has a net positive "magnetic pole strength" (corresponding to a north pole) then it has more magnetization field lines entering it than leaving it. Mathematically this is equivalent to: ∮S μ 0 M⋅d A=−q M,{\displaystyle \oint {S}\mu {0}\mathbf {M} \cdot \mathrm {d} \mathbf {A} =-q_{\mathrm {M} },} where the integral is a closed surface integral over the closed surface S and q M is the "magnetic charge" (in units of magnetic flux) enclosed by S. (A closed surface completely surrounds a region with no holes to let any field lines escape.) The negative sign occurs because the magnetization field moves from south to north.
H-field and magnetic materials
Comparison of B, H and M inside and outside a cylindrical bar magnet.
See also: Demagnetizing field
In SI units, the H-field is related to the B-field by H≡B μ 0−M.{\displaystyle \mathbf {H} \ \equiv \ {\frac {\mathbf {B} }{\mu _{0}}}-\mathbf {M} .}
In terms of the H-field, Ampere's law is ∮H⋅d ℓ=∮(B μ 0−M)⋅d ℓ=I t o t−I b=I f,{\displaystyle \oint \mathbf {H} \cdot \mathrm {d} {\boldsymbol {\ell }}=\oint \left({\frac {\mathbf {B} }{\mu {0}}}-\mathbf {M} \right)\cdot \mathrm {d} {\boldsymbol {\ell }}=I{\mathrm {tot} }-I_{\mathrm {b} }=I_{\mathrm {f} },} where I f represents the 'free current' enclosed by the loop so that the line integral of H does not depend at all on the bound currents.
For the differential equivalent of this equation see Maxwell's equations. Ampere's law leads to the boundary condition (H 1∥−H 2∥)=K f×n^,{\displaystyle \left(\mathbf {H_{1}^{\parallel }} -\mathbf {H_{2}^{\parallel }} \right)=\mathbf {K} _{\mathrm {f} }\times {\hat {\mathbf {n} }},} where Kf is the surface free current density and the unit normal n^{\displaystyle {\hat {\mathbf {n} }}} points in the direction from medium 2 to medium 1.
Similarly, a surface integral of H over any closed surface is independent of the free currents and picks out the "magnetic charges" within that closed surface: ∮S μ 0 H⋅d A=∮S(B−μ 0 M)⋅d A=0−(−q M)=q M,{\displaystyle \oint {S}\mu {0}\mathbf {H} \cdot \mathrm {d} \mathbf {A} =\oint {S}(\mathbf {B} -\mu {0}\mathbf {M} )\cdot \mathrm {d} \mathbf {A} =0-(-q_{\mathrm {M} })=q_{\mathrm {M} },}
which does not depend on the free currents.
The H-field, therefore, can be separated into two[note 10] independent parts: H=H 0+H d,{\displaystyle \mathbf {H} =\mathbf {H} {0}+\mathbf {H} {\mathrm {d} },}
where H0 is the applied magnetic field due only to the free currents and Hd is the demagnetizing field due only to the bound currents.
The magnetic H-field, therefore, re-factors the bound current in terms of "magnetic charges". The H field lines loop only around "free current" and, unlike the magnetic B field, begins and ends near magnetic poles as well.
Magnetism
Main article: Magnetism
Most materials respond to an applied B-field by producing their own magnetization M and therefore their own B-fields. Typically, the response is weak and exists only when the magnetic field is applied. The term magnetism describes how materials respond on the microscopic level to an applied magnetic field and is used to categorize the magnetic phase of a material. Materials are divided into groups based upon their magnetic behavior:
Diamagnetic materials produce a magnetization that opposes the magnetic field.
Paramagnetic materials produce a magnetization in the same direction as the applied magnetic field.
Ferromagnetic materials and the closely related ferrimagnetic materials and antiferromagnetic materials can have a magnetization independent of an applied B-field with a complex relationship between the two fields.
Superconductors (and ferromagnetic superconductors) are materials that are characterized by perfect conductivity below a critical temperature and magnetic field. They also are highly magnetic and can be perfect diamagnets below a lower critical magnetic field. Superconductors often have a broad range of temperatures and magnetic fields (the so-named mixed state) under which they exhibit a complex hysteretic dependence of M on B.
In the case of paramagnetism and diamagnetism, the magnetization M is often proportional to the applied magnetic field such that: B=μ H,{\displaystyle \mathbf {B} =\mu \mathbf {H} ,} where μ is a material dependent parameter called the permeability. In some cases the permeability may be a second rank tensor so that H may not point in the same direction as B. These relations between B and H are examples of constitutive equations. However, superconductors and ferromagnets have a more complex B-to-H relation; see magnetic hysteresis.
Stored energy
Main article: Magnetic energy
See also: Magnetic hysteresis
Energy is needed to generate a magnetic field both to work against the electric field that a changing magnetic field creates and to change the magnetization of any material within the magnetic field. For non-dispersive materials, this same energy is released when the magnetic field is destroyed so that the energy can be modeled as being stored in the magnetic field.
For linear, non-dispersive, materials (such that B = μH where μ is frequency-independent), the energy density is:
u=B⋅H 2=B⋅B 2 μ=μ H⋅H 2.{\displaystyle u={\frac {\mathbf {B} \cdot \mathbf {H} }{2}}={\frac {\mathbf {B} \cdot \mathbf {B} }{2\mu }}={\frac {\mu \mathbf {H} \cdot \mathbf {H} }{2}}.}
If there are no magnetic materials around then μ can be replaced by μ 0. The above equation cannot be used for nonlinear materials, though; a more general expression given below must be used.
In general, the incremental amount of work per unit volume δW needed to cause a small change of magnetic field δB is:
δ W=H⋅δ B.{\displaystyle \delta W=\mathbf {H} \cdot \delta \mathbf {B} .}
Once the relationship between H and B is known this equation is used to determine the work needed to reach a given magnetic state. For hysteretic materials such as ferromagnets and superconductors, the work needed also depends on how the magnetic field is created. For linear non-dispersive materials, though, the general equation leads directly to the simpler energy density equation given above.
Appearance in Maxwell's equations
Main article: Maxwell's equations
See also: Electromagnetism
Like all vector fields, a magnetic field has two important mathematical properties that relates it to its sources. (For B the sources are currents and changing electric fields.) These two properties, along with the two corresponding properties of the electric field, make up Maxwell's Equations. Maxwell's Equations together with the Lorentz force law form a complete description of classical electrodynamics including both electricity and magnetism.
The first property is the divergence of a vector field A, ∇ · A, which represents how A "flows" outward from a given point. As discussed above, a B-field line never starts or ends at a point but instead forms a complete loop. This is mathematically equivalent to saying that the divergence of B is zero. (Such vector fields are called solenoidal vector fields.) This property is called Gauss's law for magnetism and is equivalent to the statement that there are no isolated magnetic poles or magnetic monopoles.
The second mathematical property is called the curl, such that ∇ × A represents how A curls or "circulates" around a given point. The result of the curl is called a "circulation source". The equations for the curl of B and of E are called the Ampère–Maxwell equation and Faraday's law respectively.
Gauss' law for magnetism
Main article: Gauss's law for magnetism
One important property of the B-field produced this way is that magnetic B-field lines neither start nor end (mathematically, B is a solenoidal vector field); a field line may only extend to infinity, or wrap around to form a closed curve, or follow a never-ending (possibly chaotic) path. Magnetic field lines exit a magnet near its north pole and enter near its south pole, but inside the magnet B-field lines continue through the magnet from the south pole back to the north.[note 11] If a B-field line enters a magnet somewhere it has to leave somewhere else; it is not allowed to have an end point.
More formally, since all the magnetic field lines that enter any given region must also leave that region, subtracting the "number"[note 12] of field lines that enter the region from the number that exit gives identically zero. Mathematically this is equivalent to Gauss's law for magnetism: ∮S B⋅d A=0{\displaystyle \oint {S}\mathbf {B} \cdot \mathrm {d} \mathbf {A} =0} where the integral is a surface integral over the closed surface_S (a closed surface is one that completely surrounds a region with no holes to let any field lines escape). Since dA points outward, the dot product in the integral is positive for B-field pointing out and negative for B-field pointing in.
Faraday's Law
Main article: Faraday's law of induction
A changing magnetic field, such as a magnet moving through a conducting coil, generates an electric field (and therefore tends to drive a current in such a coil). This is known as Faraday's law and forms the basis of many electrical generators and electric motors. Mathematically, Faraday's law is: E=−d Φ d t{\displaystyle {\mathcal {E}}=-{\frac {\mathrm {d} \Phi }{\mathrm {d} t}}}
where E{\displaystyle {\mathcal {E}}} is the electromotive force (or EMF, the voltage generated around a closed loop) and Φ is the magnetic flux—the product of the area times the magnetic field normal to that area. (This definition of magnetic flux is why B is often referred to as magnetic flux density.): 210 The negative sign represents the fact that any current generated by a changing magnetic field in a coil produces a magnetic field that opposes the change in the magnetic field that induced it. This phenomenon is known as Lenz's law. This integral formulation of Faraday's law can be converted[note 13] into a differential form, which applies under slightly different conditions.
∇×E=−∂B∂t{\displaystyle \nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}}
Ampère's Law and Maxwell's correction
Main article: Ampère's circuital law
Similar to the way that a changing magnetic field generates an electric field, a changing electric field generates a magnetic field. This fact is known as Maxwell's correction to Ampère's law and is applied as an additive term to Ampere's law as given above. This additional term is proportional to the time rate of change of the electric flux and is similar to Faraday's law above but with a different and positive constant out front. (The electric flux through an area is proportional to the area times the perpendicular part of the electric field.)
The full law including the correction term is known as the Maxwell–Ampère equation. It is not commonly given in integral form because the effect is so small that it can typically be ignored in most cases where the integral form is used.
The Maxwell term is critically important in the creation and propagation of electromagnetic waves. Maxwell's correction to Ampère's Law together with Faraday's law of induction describes how mutually changing electric and magnetic fields interact to sustain each other and thus to form electromagnetic waves, such as light: a changing electric field generates a changing magnetic field, which generates a changing electric field again. These, though, are usually described using the differential form of this equation given below.
∇×B=μ 0 J+μ 0 ε 0∂E∂t{\displaystyle \nabla \times \mathbf {B} =\mu {0}\mathbf {J} +\mu {0}\varepsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}}
where J is the complete microscopic current density, and ε 0 is the vacuum permittivity.
As discussed above, materials respond to an applied electric E field and an applied magnetic B field by producing their own internal "bound" charge and current distributions that contribute to E and B but are difficult to calculate. To circumvent this problem, H and D fields are used to re-factor Maxwell's equations in terms of the free current densityJf:
∇×H=J f+∂D∂t{\displaystyle \nabla \times \mathbf {H} =\mathbf {J} _{\mathrm {f} }+{\frac {\partial \mathbf {D} }{\partial t}}}
These equations are not any more general than the original equations (if the "bound" charges and currents in the material are known). They also must be supplemented by the relationship between B and H as well as that between E and D. On the other hand, for simple relationships between these quantities this form of Maxwell's equations can circumvent the need to calculate the bound charges and currents.
Formulation in special relativity and quantum electrodynamics
Relativistic electrodynamics
Main article: Relativistic electromagnetism
As different aspects of the same phenomenon
According to the special theory of relativity, the partition of the electromagnetic force into separate electric and magnetic components is not fundamental, but varies with the observational frame of reference: An electric force perceived by one observer may be perceived by another (in a different frame of reference) as a magnetic force, or a mixture of electric and magnetic forces.
The magnetic field existing as electric field in other frames can be shown by consistency of equations obtained from Lorentz transformation of four force from Coulomb's Law in particle's rest frame with Maxwell's laws considering definition of fields from Lorentz force[broken anchor] and for non accelerating condition. The form of magnetic field hence obtained by Lorentz transformation of four-force from the form of Coulomb's law in source's initial frame is given by:: 29–42
B=q 4 π ε 0 r 3 1−β 2(1−β 2 sin 2θ)3/2 v×r c 2=v×E c 2{\displaystyle \mathbf {B} ={\frac {q}{4\pi \varepsilon _{0}r^{3}}}{\frac {1-\beta ^{2}}{(1-\beta ^{2}\sin ^{2}\theta )^{3/2}}}{\frac {\mathbf {v} \times \mathbf {r} }{c^{2}}}={\frac {\mathbf {v} \times \mathbf {E} }{c^{2}}}}
where q{\displaystyle q} is the charge of the point source, ε 0{\displaystyle \varepsilon _{0}} is the vacuum permittivity, r{\displaystyle \mathbf {r} } is the position vector from the point source to the point in space, v{\displaystyle \mathbf {v} } is the velocity vector of the charged particle, β{\displaystyle \beta } is the ratio of speed of the charged particle divided by the speed of light and θ{\displaystyle \theta } is the angle between r{\displaystyle \mathbf {r} } and v{\displaystyle \mathbf {v} }. This form of magnetic field can be shown to satisfy Maxwell's laws within the constraint of particle being non accelerating. The above reduces to Biot-Savart law for non relativistic stream of current (β≪1{\displaystyle \beta \ll 1}).
Formally, special relativity combines the electric and magnetic fields into a rank-2 tensor, called the electromagnetic tensor. Changing reference frames mixes these components. This is analogous to the way that special relativity mixes space and time into spacetime, and mass, momentum, and energy into four-momentum. Similarly, the energy stored in a magnetic field is mixed with the energy stored in an electric field in the electromagnetic stress–energy tensor.
Magnetic vector potential
Main article: Magnetic vector potential
In advanced topics such as quantum mechanics and relativity it is often easier to work with a potential formulation of electrodynamics rather than in terms of the electric and magnetic fields. In this representation, the magnetic vector potentialA, and the electric scalar potentialφ, are defined using gauge fixing such that:
B=∇×A,E=−∇φ−∂A∂t.{\displaystyle {\begin{aligned}\mathbf {B} &=\nabla \times \mathbf {A} ,\\mathbf {E} &=-\nabla \varphi -{\frac {\partial \mathbf {A} }{\partial t}}.\end{aligned}}}
The vector potential, A given by this form may be interpreted as a generalized potential momentum per unit charge just as φ is interpreted as a generalized potential energy per unit charge. There are multiple choices one can make for the potential fields that satisfy the above condition. However, the choice of potentials is represented by its respective gauge condition.
Maxwell's equations when expressed in terms of the potentials in Lorenz gauge can be cast into a form that agrees with special relativity. In relativity, A together with φ forms a four-potential regardless of the gauge condition, analogous to the four-momentum that combines the momentum and energy of a particle. Using the four potential instead of the electromagnetic tensor has the advantage of being much simpler—and it can be easily modified to work with quantum mechanics.
Propagation of Electric and Magnetic fields
Special theory of relativity imposes the condition for events related by cause and effect to be time-like separated, that is that causal efficacy propagates no faster than light.Maxwell's equations for electromagnetism are found to be in favor of this as electric and magnetic disturbances are found to travel at the speed of light in space. Electric and magnetic fields from classical electrodynamics obey the principle of locality in physics and are expressed in terms of retarded time or the time at which the cause of a measured field originated given that the influence of field travelled at speed of light. The retarded time for a point particle is given as solution of:
t r=t−|r−r s(t r)|c{\displaystyle t_{r}=\mathbf {t} -{\frac {\left|\mathbf {r} -\mathbf {r} {s}(t{r})\right|}{c}}}
where t r{\textstyle t_{r}} is retarded time or the time at which the source's contribution of the field originated, r s(t){\textstyle r_{s}(t)} is the position vector of the particle as function of time, r{\textstyle \mathbf {r} } is the point in space, t{\textstyle \mathbf {t} } is the time at which fields are measured and c{\textstyle c} is the speed of light. The equation subtracts the time taken for light to travel from particle to the point in space from the time of measurement to find time of origin of the fields. The uniqueness of solution for t r{\textstyle t_{r}} for given t{\displaystyle \mathbf {t} }, r{\displaystyle \mathbf {r} } and r s(t){\displaystyle r_{s}(t)} is valid for charged particles moving slower than speed of light.
Magnetic field of arbitrary moving point charge
Main article: Liénard–Wiechert potential
The solution of maxwell's equations for electric and magnetic field of a point charge is expressed in terms of retarded time or the time at which the particle in the past causes the field at the point, given that the influence travels across space at the speed of light.
Any arbitrary motion of point charge causes electric and magnetic fields found by solving maxwell's equations using green's function for retarded potentials and hence finding the fields to be as follows:
A(r,t)=μ 0 c 4 π[q β s(1−n s⋅β s)|r−r s|]t=t r=β s(t r)c φ(r,t){\displaystyle {\begin{aligned}\mathbf {A} (\mathbf {r} ,\mathbf {t} )&={\frac {\mu {0}c}{4\pi }}\left[{\frac {q{\boldsymbol {\beta }}{s}}{(1-\mathbf {n} {s}\cdot {\boldsymbol {\beta }}{s})|\mathbf {r} -\mathbf {r} {s}|}}\right]{t=t_{r}}\[1ex]&={\frac {{\boldsymbol {\beta }}{s}(t{r})}{c}}\varphi (\mathbf {r} ,\mathbf {t} )\end{aligned}}}B(r,t)=μ 0 4 π[q c(β s×n s)γ 2(1−n s⋅β s)3|r−r s|2+q n s×(n s×((n s−β s)×β s˙))(1−n s⋅β s)3|r−r s|]t=t r=n s(t r)c×E(r,t){\displaystyle {\begin{aligned}\mathbf {B} (\mathbf {r} ,\mathbf {t} )&={\frac {\mu {0}}{4\pi }}\left[{\frac {qc({\boldsymbol {\beta }}{s}\times \mathbf {n} {s})}{\gamma ^{2}{\left(1-\mathbf {n} {s}\cdot {\boldsymbol {\beta }}{s}\right)}^{3}{\left|\mathbf {r} -\mathbf {r} {s}\right|}^{2}}}+{\frac {q\mathbf {n} {s}\times \left(\mathbf {n} {s}\times \left(\left(\mathbf {n} {s}-{\boldsymbol {\beta }}{s}\right)\times {\dot {{\boldsymbol {\beta }}{s}}}\right)\right)}{{\left(1-\mathbf {n} {s}\cdot {\boldsymbol {\beta }}{s}\right)}^{3}\left|\mathbf {r} -\mathbf {r} {s}\right|}}\right]{t=t{r}}\[1ex]&={\frac {\mathbf {n} {s}(t{r})}{c}}\times \mathbf {E} (\mathbf {r} ,\mathbf {t} )\end{aligned}}}
where φ(r,t){\textstyle \varphi (\mathbf {r} ,\mathbf {t} )}and A(r,t){\textstyle \mathbf {A} (\mathbf {r} ,\mathbf {t} )} are electric scalar potential and magnetic vector potential in Lorentz gauge, q{\displaystyle q} is the charge of the point source, n s(r,t){\textstyle n_{s}(\mathbf {r} ,t)} is a unit vector pointing from charged particle to the point in space, β s(t){\textstyle {\boldsymbol {\beta }}_{s}(t)} is the velocity of the particle divided by the speed of light and γ(t){\textstyle \gamma (t)} is the corresponding Lorentz factor. Hence by the principle of superposition, the fields of a system of charges also obey principle of locality.
Quantum electrodynamics
See also: Standard Model and quantum electrodynamics
The classical electromagnetic field incorporated into quantum mechanics forms what is known as the semi-classical theory of radiation. However, it is not able to make experimentally observed predictions such as spontaneous emission process or Lamb shift implying the need for quantization of fields. In modern physics, the electromagnetic field is understood to be not a classicalfield, but rather a quantum field; it is represented not as a vector of three numbers at each point, but as a vector of three quantum operators at each point. The most accurate modern description of the electromagnetic interaction (and much else) is quantum electrodynamics (QED), which is incorporated into a more complete theory known as the Standard Model of particle physics.
In QED, the magnitude of the electromagnetic interactions between charged particles (and their antiparticles) is computed using perturbation theory. These rather complex formulas produce a remarkable pictorial representation as Feynman diagrams in which virtual photons are exchanged.
Predictions of QED agree with experiments to an extremely high degree of accuracy: currently about 10−12 (and limited by experimental errors); for details see precision tests of QED. This makes QED one of the most accurate physical theories constructed thus far.
All equations in this article are in the classical approximation, which is less accurate than the quantum description mentioned here. However, under most everyday circumstances, the difference between the two theories is negligible.
Uses and examples
Earth's magnetic field
Main article: Earth's magnetic field
A sketch of Earth's magnetic field representing the source of the field as a magnet. The south pole of the magnetic field is near the geographic north pole of the Earth.
The Earth's magnetic field is produced by convection of a liquid iron alloy in the outer core. In a dynamo process, the movements drive a feedback process in which electric currents create electric and magnetic fields that in turn act on the currents.
The field at the surface of the Earth is approximately the same as if a giant bar magnet were positioned at the center of the Earth and tilted at an angle of about 11° off the rotational axis of the Earth (see the figure). The north pole of a magnetic compass needle points roughly north, toward the North Magnetic Pole. However, because a magnetic pole is attracted to its opposite, the North Magnetic Pole is actually the south pole of the geomagnetic field. This confusion in terminology arises because the pole of a magnet is defined by the geographical direction it points.
Earth's magnetic field is not constant—the strength of the field and the location of its poles vary. Moreover, the poles periodically reverse their orientation in a process called geomagnetic reversal. The most recent reversal occurred 780,000 years ago.
Rotating magnetic fields
Main articles: Rotating magnetic field and Alternator
The rotating magnetic field is a common design principle in the operation of alternating-current motors. A permanent magnet in such a field rotates so as to maintain its alignment with the external field.
Magnetic torque is used to drive electric motors. In one simple motor design, a magnet is fixed to a freely rotating shaft and is subjected to a magnetic field from an array of electromagnets. By continuously switching the electric current through each of the electromagnets, thereby flipping the polarity of their magnetic fields, like poles are kept next to the rotor; the resultant torque is transferred to the shaft.
A rotating magnetic field can be constructed using two coils at right angles with a phase difference of 90 degrees between their AC currents. In practice, three-phase systems are used where the three currents are equal in magnitude and have a phase difference of 120 degrees. Three similar coils at mutual geometrical angles of 120 degrees create the rotating magnetic field. The ability of the three-phase system to create a rotating field, utilized in electric motors, is one of the main reasons why three-phase systems dominate the world's electrical power supply systems.
Synchronous motors use DC-voltage-fed rotor windings, which lets the excitation of the machine be controlled—and induction motors use short-circuited rotors (instead of a magnet) following the rotating magnetic field of a multicoiled stator. The short-circuited turns of the rotor develop eddy currents induced by the rotating field of the stator, and these currents in turn produce a torque on the rotor through the Lorentz force.
The Italian physicist Galileo Ferraris and the Serbian-American electrical engineerNikola Tesla independently researched the use of rotating magnetic fields in electric motors. In 1888, Ferraris published his research in a paper to the Royal Academy of Sciences in Turin and Tesla gained U.S. patent 381,968 for his work.
Hall effect
Main article: Hall effect
The charge carriers of a current-carrying conductor placed in a transverse magnetic field experience a sideways Lorentz force; this results in a charge separation in a direction perpendicular to the current and to the magnetic field. The resultant voltage in that direction is proportional to the applied magnetic field. This is known as the Hall effect.
The Hall effect is often used to measure the magnitude of a magnetic field. It is used as well to find the sign of the dominant charge carriers in materials such as semiconductors (negative electrons or positive holes).
Magnetic circuits
Main article: Magnetic circuit
An important use of H is in magnetic circuits where B = μH inside a linear material. Here, μ is the magnetic permeability of the material. This result is similar in form to Ohm's lawJ = σE, where J is the current density, σ is the conductance and E is the electric field. Extending this analogy, the counterpart to the macroscopic Ohm's law (I = V⁄R) is:
Φ=F R m,{\displaystyle \Phi ={\frac {F}{R}}_{\mathrm {m} },}
where Φ=∫B⋅d A{\textstyle \Phi =\int \mathbf {B} \cdot \mathrm {d} \mathbf {A} } is the magnetic flux in the circuit, F=∫H⋅d ℓ{\textstyle F=\int \mathbf {H} \cdot \mathrm {d} {\boldsymbol {\ell }}} is the magnetomotive force applied to the circuit, and R m is the reluctance of the circuit. Here the reluctance R m is a quantity similar in nature to resistance for the flux. Using this analogy it is straightforward to calculate the magnetic flux of complicated magnetic field geometries, by using all the available techniques of circuit theory.
Largest magnitude magnetic fields
This section needs to be updated. Please help update this article to reflect recent events or newly available information.
Last update: October 2018 (July 2021)
As of October 2018[update], the largest magnitude magnetic field produced over a macroscopic volume outside a lab setting is 2.8 kT (VNIIEF in Sarov, Russia, 1998). As of October 2018, the largest magnitude magnetic field produced in a laboratory over a macroscopic volume was 1.2 kT by researchers at the University of Tokyo in 2018. The largest magnitude magnetic fields produced in a laboratory occur in particle accelerators, such as RHIC, inside the collisions of heavy ions, where microscopic fields reach 10 14 T.Magnetars have the strongest known magnetic fields of any naturally occurring object, ranging from 0.1 to 100 GT (10 8 to 10 11 T).
Common formulæ
| Current configuration | Figure | Magnetic field |
---
| Finite beam of current | | B=μ 0 I 4 π x(cosθ 1+cosθ 2){\displaystyle B={\frac {\mu {0}I}{4\pi x}}(\cos \theta {1}+\cos \theta {2})} where I{\displaystyle I} is the uniform current throughout the beam, with the direction of magnetic field as shown. |
| Infinite wire | | B=μ 0 I 2 π x{\displaystyle B={\frac {\mu {0}I}{2\pi x}}} where I{\displaystyle I} is the uniform current flowing through the wire with the direction of magnetic field as shown. |
| Infinite cylindrical wire | | B=μ 0 I 2 π x{\displaystyle B={\frac {\mu {0}I}{2\pi x}}} outside the wire carrying a current I{\displaystyle I} uniformly, with the direction of magnetic field as shown. | B=μ 0 I x 2 π R 2{\displaystyle B={\frac {\mu {0}Ix}{2\pi R^{2}}}} inside the wire carrying a current I{\displaystyle I} uniformly, with the direction of magnetic field as shown. |
| Circular loop | | B=μ 0 I R 2 2(x 2+R 2)3/2 x^{\displaystyle \mathbf {B} ={\frac {\mu {0}IR^{2}}{2(x^{2}+R^{2})^{3/2}}}{\hat {\mathbf {x} }}} along the axis of the loop, where I{\displaystyle I} is the uniform current flowing through the loop. |
| Solenoid | | B=μ 0 n I 2(c o s θ 1+cosθ 2){\displaystyle B={\frac {\mu {0}nI}{2}}(cos\theta {1}+\cos \theta {2})} along the axis of the solenoid carrying current I{\displaystyle I} with n{\displaystyle n}, uniform number of loops of currents per length of solenoid; and the direction of magnetic field as shown. |
| Infinite solenoid | | B=0{\displaystyle \mathbf {B} =0} outside the solenoid carrying current I{\displaystyle I} with n{\displaystyle n}, uniform number of loops of currents per length of solenoid. | B=μ 0 n I{\displaystyle B=\mu {0}nI} inside the solenoid carrying current I{\displaystyle I} with n{\displaystyle n}, uniform number of loops of currents per length of solenoid, with the direction of magnetic field as shown. |
| Circular Toroid | | B=μ 0 N I 2 π R{\displaystyle B={\frac {\mu {0}NI}{2\pi R}}} along the bulk of the circular toroid carrying uniform current I{\displaystyle I} through N{\displaystyle N} number of uniformly distributed poloidal loops, with the direction of magnetic field as indicated. |
| Magnetic Dipole | | B=−μ 0 m 4 π r 3,{\displaystyle \mathbf {B} =-{\frac {\mu {0}\mathbf {m} }{4\pi r^{3}}},} on the equatorial plane, where m{\displaystyle \mathbf {m} } is the magnetic dipole moment. | B=μ 0 m 2 π|x|3,{\displaystyle \mathbf {B} ={\frac {\mu {0}\mathbf {m} }{2\pi {|x|}^{3}}},} on the axial plane (given that x≫R{\displaystyle x\gg R}), where x{\displaystyle x} can also be negative to indicate position at the opposite direction on the axis, and m{\displaystyle \mathbf {m} } is the magnetic dipole moment. |
Additional magnetic field values can be found through the magnetic field of a finite beam, for example, that the magnetic field of an arc of angle θ{\displaystyle \theta } and radius R{\displaystyle R} at the center is B=μ 0 θ I 4 π R{\displaystyle B={\mu {0}\theta I \over 4\pi R}}, or that the magnetic field at the center of a N-sided regular polygon of side a{\displaystyle a} is B=μ 0 N I π a sinπ N tanπ N{\displaystyle B={\mu {0}NI \over \pi a}\sin {\pi \over N}\tan {\pi \over N}}, both outside of the plane with proper directions as inferred by right hand thumb rule.
History
Main article: History of electromagnetic theory
See also: Timeline of electromagnetism and classical optics
One of the first drawings of a magnetic field, by René Descartes, 1644, showing the Earth attracting lodestones. It illustrated his theory that magnetism was caused by the circulation of tiny helical particles, "threaded parts", through threaded pores in magnets.
Early developments
While magnets and some properties of magnetism were known to ancient societies, the research of magnetic fields began in 1269 when French scholar Petrus Peregrinus de Maricourt mapped out the magnetic field on the surface of a spherical magnet using iron needles. Noting the resulting field lines crossed at two points he named those points "poles" in analogy to Earth's poles. He also articulated the principle that magnets always have both a north and south pole, no matter how finely one slices them.[note 14]
Almost three centuries later, William Gilbert of Colchester replicated Petrus Peregrinus' work and was the first to state explicitly that Earth is a magnet.: 34 Published in 1600, Gilbert's work, De Magnete, helped to establish magnetism as a science.
Mathematical development
Hans Christian Ørsted, Der Geist in der Natur, 1854
In 1750, John Michell stated that magnetic poles attract and repel in accordance with an inverse square law: 56 Charles-Augustin de Coulomb experimentally verified this in 1785 and stated explicitly that north and south poles cannot be separated.: 59 Building on this force between poles, Siméon Denis Poisson (1781–1840) created the first successful model of the magnetic field, which he presented in 1824.: 64 In this model, a magnetic H-field is produced by magnetic poles and magnetism is due to small pairs of north–south magnetic poles.
Three discoveries in 1820 challenged this foundation of magnetism. Hans Christian Ørsted demonstrated that a current-carrying wire is surrounded by a circular magnetic field.[note 15] Then André-Marie Ampère showed that parallel wires with currents attract one another if the currents are in the same direction and repel if they are in opposite directions.: 87 Finally, Jean-Baptiste Biot and Félix Savart announced empirical results about the forces that a current-carrying long, straight wire exerted on a small magnet, determining the forces were inversely proportional to the perpendicular distance from the wire to the magnet.: 86 Laplace later deduced a law of force based on the differential action of a differential section of the wire, which became known as the Biot–Savart law, as Laplace did not publish his findings.
Extending these experiments, Ampère published his own successful model of magnetism in 1825. In it, he showed the equivalence of electrical currents to magnets: 88 and proposed that magnetism is due to perpetually flowing loops of current instead of the dipoles of magnetic charge in Poisson's model.[note 16] Further, Ampère derived both Ampère's force law describing the force between two currents and Ampère's law, which, like the Biot–Savart law, correctly described the magnetic field generated by a steady current. Also in this work, Ampère introduced the term electrodynamics to describe the relationship between electricity and magnetism.: 88–92
In 1831, Michael Faraday discovered electromagnetic induction when he found that a changing magnetic field generates an encircling electric field, formulating what is now known as Faraday's law of induction.: 189–192 Later, Franz Ernst Neumann proved that, for a moving conductor in a magnetic field, induction is a consequence of Ampère's force law.: 222 In the process, he introduced the magnetic vector potential, which was later shown to be equivalent to the underlying mechanism proposed by Faraday.: 225
In 1850, Lord Kelvin, then known as William Thomson, distinguished between two magnetic fields now denoted H and B. The former applied to Poisson's model and the latter to Ampère's model and induction.: 224 Further, he derived how H and B relate to each other and coined the term permeability.: 245
Between 1861 and 1865, James Clerk Maxwell developed and published Maxwell's equations, which explained and united all of classical electricity and magnetism. The first set of these equations was published in a paper entitled On Physical Lines of Force in 1861. These equations were valid but incomplete. Maxwell completed his set of equations in his later 1865 paper A Dynamical Theory of the Electromagnetic Field and demonstrated the fact that light is an electromagnetic wave. Heinrich Hertz published papers in 1887 and 1888 experimentally confirming this fact.
Modern developments
In 1887, Tesla developed an induction motor that ran on alternating current. The motor used polyphase current, which generated a rotating magnetic field to turn the motor (a principle that Tesla claimed to have conceived in 1882). Tesla received a patent for his electric motor in May 1888. In 1885, Galileo Ferraris independently researched rotating magnetic fields and subsequently published his research in a paper to the Royal Academy of Sciences in Turin, just two months before Tesla was awarded his patent, in March 1888.
The twentieth century showed that classical electrodynamics is already consistent with special relativity, and extended classical electrodynamics to work with quantum mechanics. Albert Einstein, in his paper of 1905 that established relativity, showed that both the electric and magnetic fields are part of the same phenomena viewed from different reference frames. Finally, the emergent field of quantum mechanics was merged with electrodynamics to form quantum electrodynamics, which first formalized the notion that electromagnetic field energy is quantized in the form of photons.
See also
General
Magnetohydrodynamics– the study of the dynamics of electrically conducting fluids
Magnetic hysteresis– application to ferromagnetism
Magnetic nanoparticles– extremely small magnetic particles that are tens of atoms wide
Magnetic reconnection– an effect that causes solar flares and auroras
Magnetic scalar potential
SI electromagnetism units– common units used in electromagnetism
Orders of magnitude (magnetic field)– list of magnetic field sources and measurement devices from smallest magnetic fields to largest detected
Upward continuation
Moses Effect
Mathematics
Magnetic helicity– extent to which a magnetic field wraps around itself
Applications
Dynamo theory– a proposed mechanism for the creation of the Earth's magnetic field
Helmholtz coil– a device for producing a region of nearly uniform magnetic field
Magnetic field viewing film– Film used to view the magnetic field of an area
Magnetic pistol– a device on torpedoes or naval mines that detect the magnetic field of their target
Maxwell coil– a device for producing a large volume of an almost constant magnetic field
Stellar magnetic field– a discussion of the magnetic field of stars
Teltron tube– device used to display an electron beam and demonstrates effect of electric and magnetic fields on moving charges
Notes
^The letters B and H were originally chosen by Maxwell in his Treatise on Electricity and Magnetism (Vol. II, pp. 236–237). For many quantities, he simply started choosing letters from the beginning of the alphabet. See Ralph Baierlein (2000). "Answer to Question #73. S is for entropy, Q is for charge". American Journal of Physics. 68 (8): 691. Bibcode:2000AmJPh..68..691B. doi:10.1119/1.19524.
^Edward Purcell, in Electricity and Magnetism, McGraw-Hill, 1963, writes, Even some modern writers who treat B as the primary field feel obliged to call it the magnetic induction because the name magnetic field was historically preempted by H. This seems clumsy and pedantic. If you go into the laboratory and ask a physicist what causes the pion trajectories in his bubble chamber to curve, he'll probably answer "magnetic field", not "magnetic induction." You will seldom hear a geophysicist refer to the Earth's magnetic induction, or an astrophysicist talk about the magnetic induction of the galaxy. We propose to keep on calling B the magnetic field. As for H, although other names have been invented for it, we shall call it "the field H" or even "the magnetic field H." In a similar vein, M Gerloch (1983). Magnetism and Ligand-field Analysis. Cambridge University Press. p.110. ISBN978-0-521-24939-3. says: "So we may think of both B and H as magnetic fields, but drop the word 'magnetic' from H so as to maintain the distinction ... As Purcell points out, 'it is only the names that give trouble, not the symbols'."
^An alternative mnemonic to the right hand rule is Fleming's left-hand rule.
^The SI unit of Φ B (magnetic flux) is the weber (symbol: Wb), related to the tesla by 1 Wb/m 2 = 1 T. The SI unit tesla is equal to (newton·second)/(coulomb·metre). This can be seen from the magnetic part of the Lorentz force law.
^The use of iron filings to display a field presents something of an exception to this picture; the filings alter the magnetic field so that it is much larger along the "lines" of iron, because of the large permeability of iron relative to air.
^Here, "small" means that the observer is sufficiently far away from the magnet, so that the magnet can be considered as infinitesimally small. "Larger" magnets need to include more complicated terms in the mathematical expression of the magnetic field and depend on the entire geometry of the magnet not just m.
^Either B or H may be used for the magnetic field outside the magnet.
^In practice, the Biot–Savart law and other laws of magnetostatics are often used even when a current change in time, as long as it does not change too quickly. It is often used, for instance, for standard household currents, which oscillate sixty times per second.: 223
^ The Biot–Savart law contains the additional restriction (boundary condition) that the B-field must go to zero fast enough at infinity. It also depends on the divergence of B being zero, which is always valid. (There are no magnetic charges.)
^A third term is needed for changing electric fields and polarization currents; this displacement current term is covered in Maxwell's equations below.
^To see that this must be true imagine placing a compass inside a magnet. There, the north pole of the compass points toward the north pole of the magnet since magnets stacked on each other point in the same direction.
^As discussed above, magnetic field lines are primarily a conceptual tool used to represent the mathematics behind magnetic fields. The total "number" of field lines is dependent on how the field lines are drawn. In practice, integral equations such as the one that follows in the main text are used instead.
^ A complete expression for Faraday's law of induction in terms of the electric E and magnetic fields can be written as: E=−d Φ d t=∮∂Σ(t)(E(r,t)+v×B(r,t))⋅d ℓ=−d d t∬Σ(t)d A⋅B(r,t){\displaystyle {\mathcal {E}}=-{\frac {d\Phi }{dt}}=\oint {\partial \Sigma (t)}\left(\mathbf {E} (\mathbf {r} ,\ t)+\mathbf {v} \times \mathbf {B} (\mathbf {r} ,\ t)\right)\cdot d{\boldsymbol {\ell }}\ =-{\frac {d}{dt}}\iint {\Sigma (t)}d{\boldsymbol {A}}\cdot \mathbf {B} (\mathbf {r} ,\ t)} where ∂Σ(t) is the moving closed path bounding the moving surface Σ(t), and dA is an element of surface area of Σ(t). The first integral calculates the work done moving a charge a distance dℓ based upon the Lorentz force law. In the case where the bounding surface is stationary, the Kelvin–Stokes theorem can be used to show this equation is equivalent to the Maxwell–Faraday equation.
^His Epistola Petri Peregrini de Maricourt ad Sygerum de Foucaucourt Militem de Magnete, which is often shortened to Epistola de magnete, is dated 1269 C.E.
^During a lecture demonstration on the effects of a current on a campus needle, Ørsted showed that when a current-carrying wire is placed at a right angle with the compass, nothing happens. When he tried to orient the wire parallel to the compass needle, however, it produced a pronounced deflection of the compass needle. By placing the compass on different sides of the wire, he was able to determine the field forms perfect circles around the wire.: 85
^From the outside, the field of a dipole of magnetic charge has exactly the same form as a current loop when both are sufficiently small. Therefore, the two models differ only for magnetism inside magnetic material.
References
^Nave, Rod. "Magnetic Field". HyperPhysics. Retrieved 20 May 2024.
^ abcdefFeynman, Richard P.; Leighton, Robert B.; Sands, Matthew (1963). The Feynman Lectures on Physics. Vol.2. California Institute of Technology. ISBN9780465040858.{{cite book}}: ISBN / Date incompatibility (help)
^Young, Hugh D.; Freedman, Roger A.; Ford, A. Lewis (2008). Sears and Zemansky's university physics: with modern physics. Vol.2. Pearson Addison-Wesley. pp.918–919. ISBN9780321501219.
^Purcell, Edward M.; Morin, David J. (2013). Electricity and Magnetism (3rd ed.). Cambridge University Press. ISBN9781107014022.
^ abcdThe International System of Units(PDF), V3.01 (9th ed.), International Bureau of Weights and Measures, August 2024, ISBN978-92-822-2272-0
^Jiles, David C. (1998). Introduction to Magnetism and Magnetic Materials (2 ed.). CRC. p.3. ISBN978-0412798603.
^John J. Roche (2000). "B and H, the intensity vectors of magnetism: A new approach to resolving a century-old controversy". American Journal of Physics. 68 (5): 438. Bibcode:2000AmJPh..68..438R. doi:10.1119/1.19459.
^ abE. J. Rothwell and M. J. Cloud (2010) Electromagnetics. Taylor & Francis. p. 23. ISBN1420058266.
^ abStratton, Julius Adams (1941). Electromagnetic Theory (1st ed.). McGraw-Hill. p.1. ISBN978-0070621503.{{cite book}}: ISBN / Date incompatibility (help)
^ abcdePurcell, E. (2011). Electricity and Magnetism (2nd ed.). Cambridge University Press. ISBN978-1107013605.
^ abcdGriffiths, David J. (1999). Introduction to Electrodynamics (3rd ed.). Pearson. ISBN0-13-805326-X.
^ abJackson, John David (1998). Classical electrodynamics (3rd ed.). New York: Wiley. ISBN0-471-30932-X.
^ ab"Non-SI units accepted for use with the SI, and units based on fundamental constants (contd.)". SI Brochure: The International System of Units (SI) [8th edition, 2006; updated in 2014]. Bureau International des Poids et Mesures. Archived from the original on 8 June 2019. Retrieved 19 April 2018.
^ abLang, Kenneth R. (2006). A Companion to Astronomy and Astrophysics. Springer. p.176. ISBN9780387333670. Retrieved 19 April 2018.
^"International system of units (SI)". NIST reference on constants, units, and uncertainty. National Institute of Standards and Technology. 12 April 2010. Retrieved 9 May 2012.
^"Gravity Probe B Executive Summary"(PDF). pp.10, 21. Archived(PDF) from the original on 9 October 2022.
^Brown, William Fuller (1962). Magnetostatic Principles in Ferromagnetism. North Holland publishing company. p.12. ASINB0006AY7F8.
^See magnetic moment[broken anchor] and B. D. Cullity; C. D. Graham (2008). Introduction to Magnetic Materials (2 ed.). Wiley-IEEE. p.103. ISBN978-0-471-47741-9.
^E. Richard Cohen; David R. Lide; George L. Trigg (2003). AIP physics desk reference (3 ed.). Birkhäuser. p.381. ISBN978-0-387-98973-0.
^Griffiths 1999, p.438
^ abcGriffiths, David J. (2017). Introduction to Electrodynamics (4th ed.). Cambridge University Press. ISBN9781108357142.
^Griffiths 1999, pp.222–225
^"K. McDonald's Physics Examples - Disk"(PDF). puhep1.princeton.edu. Archived(PDF) from the original on 9 October 2022. Retrieved 13 February 2021.
^"K. McDonald's Physics Examples - Railgun"(PDF). puhep1.princeton.edu. Archived(PDF) from the original on 9 October 2022. Retrieved 13 February 2021.
^Deissler, R.J. (2008). "Dipole in a magnetic field, work, and quantum spin"(PDF). Physical Review E. 77 (3, pt 2): 036609. Bibcode:2008PhRvE..77c6609D. doi:10.1103/PhysRevE.77.036609. PMID18517545. Archived(PDF) from the original on 9 October 2022.{{cite journal}}: CS1 maint: article number as page number (link)
^Griffiths 1999, pp.266–268
^John Clarke Slater; Nathaniel Herman Frank (1969). Electromagnetism (first published in 1947 ed.). Courier Dover Publications. p.69. ISBN978-0-486-62263-7.
^Griffiths 1999, p.332
^ abRJD Tilley (2004). Understanding Solids. Wiley. p.368. ISBN978-0-470-85275-0.
^Sōshin Chikazumi; Chad D. Graham (1997). Physics of ferromagnetism (2 ed.). Oxford University Press. p.118. ISBN978-0-19-851776-4.
^Amikam Aharoni (2000). Introduction to the theory of ferromagnetism (2 ed.). Oxford University Press. p.27. ISBN978-0-19-850808-3.
^M Brian Maple; et al. (2008). "Unconventional superconductivity in novel materials". In K. H. Bennemann; John B. Ketterson (eds.). Superconductivity. Springer. p.640. ISBN978-3-540-73252-5.
^Naoum Karchev (2003). "Itinerant ferromagnetism and superconductivity". In Paul S. Lewis; D. Di (CON) Castro (eds.). Superconductivity research at the leading edge. Nova Publishers. p.169. ISBN978-1-59033-861-2.
^Lieberherr, Martin (6 July 2010). "The magnetic field lines of a helical coil are not simple loops". American Journal of Physics. 78 (11): 1117–1119. Bibcode:2010AmJPh..78.1117L. doi:10.1119/1.3471233.
^Jackson, John David (1975). Classical electrodynamics (2nd ed.). New York: Wiley. ISBN9780471431329.
^ abRosser, W. G. V. (1968). Classical Electromagnetism via Relativity. Boston, MA: Springer. doi:10.1007/978-1-4899-6559-2. ISBN978-1-4899-6258-4.
^Purcell, Edward (22 September 2011). Electricity and Magnetism. Cambridge University Press. doi:10.1017/cbo9781139005043. ISBN978-1-107-01360-5.
^C. Doran and A. Lasenby (2003) Geometric Algebra for Physicists, Cambridge University Press, p. 233. ISBN0521715954.
^E. J. Konopinski (1978). "What the electromagnetic vector potential describes". Am. J. Phys. 46 (5): 499–502. Bibcode:1978AmJPh..46..499K. doi:10.1119/1.11298.
^Griffiths 1999, p.422
^Naber, Gregory L. (2012). The Geometry of Minkowski spacetime: an introduction to the mathematics of the special theory of relativity. Springer. pp.4–5. ISBN978-1-4419-7837-0. OCLC804823303.
^ For a good qualitative introduction see: Richard Feynman (2006). QED: the strange theory of light and matter. Princeton University Press. ISBN978-0-691-12575-6.
^Weiss, Nigel (2002). "Dynamos in planets, stars and galaxies". Astronomy and Geophysics. 43 (3): 3.09 –3.15. Bibcode:2002A&G....43c...9W. doi:10.1046/j.1468-4004.2002.43309.x.
^"What is the Earth's magnetic field?". Geomagnetism Frequently Asked Questions. National Centers for Environmental Information, National Oceanic and Atmospheric Administration. Retrieved 19 April 2018.
^Raymond A. Serway; Chris Vuille; Jerry S. Faughn (2009). College physics (8th ed.). Belmont, CA: Brooks/Cole, Cengage Learning. p.628. ISBN978-0-495-38693-3.
^Merrill, Ronald T.; McElhinny, Michael W.; McFadden, Phillip L. (1996). "2. The present geomagnetic field: analysis and description from historical observations". The magnetic field of the earth: paleomagnetism, the core, and the deep mantle. Academic Press. ISBN978-0-12-491246-5.
^Phillips, Tony (29 December 2003). "Earth's Inconstant Magnetic Field". Science@Nasa. Archived from the original on 1 November 2022. Retrieved 27 December 2009.
^Boyko, B.A.; Bykov, A.I.; Dolotenko, M.I.; Kolokolchikov, N.P.; Markevtsev, I.M.; Tatsenko, O.M.; Shuvalov, K. (1999). "With record magnetic fields to the 21st Century". Digest of Technical Papers. 12th IEEE International Pulsed Power Conference. (Cat. No.99CH36358). Vol.2. pp.746–749. doi:10.1109/PPC.1999.823621. ISBN0-7803-5498-2. S2CID42588549.
^ abDaley, Jason. "Watch the Strongest Indoor Magnetic Field Blast Doors of Tokyo Lab Wide Open". Smithsonian Magazine. Retrieved 8 September 2020.
^Tuchin, Kirill (2013). "Particle production in strong electromagnetic fields in relativistic heavy-ion collisions". Adv. High Energy Phys. 2013: 490495. arXiv:1301.0099. Bibcode:2013arXiv1301.0099T. doi:10.1155/2013/490495. S2CID4877952.{{cite journal}}: CS1 maint: article number as page number (link)
^Bzdak, Adam; Skokov, Vladimir (29 March 2012). "Event-by-event fluctuations of magnetic and electric fields in heavy ion collisions". Physics Letters B. 710 (1): 171–174. arXiv:1111.1949. Bibcode:2012PhLB..710..171B. doi:10.1016/j.physletb.2012.02.065. S2CID118462584.
^Kouveliotou, C.; Duncan, R. C.; Thompson, C. (February 2003). "MagnetarsArchived 11 June 2007 at the Wayback Machine". Scientific American; Page 36.
^Chapman, Allan (2007). "Peregrinus, Petrus (Flourished 1269)". Encyclopedia of Geomagnetism and Paleomagnetism. Dordrecht: Springer. pp.808–809. doi:10.1007/978-1-4020-4423-6_261. ISBN978-1-4020-3992-8.
^ abcdefghijklmnWhittaker, E. T. (1910). A History of the Theories of Aether and Electricity. Dover Publications. ISBN978-0-486-26126-3.{{cite book}}: ISBN / Date incompatibility (help)
^Williams, L. Pearce (1974). "Oersted, Hans Christian". In Gillespie, C. C. (ed.). Dictionary of Scientific Biography. New York: Charles Scribner's Sons. p.185.
^Blundell, Stephen J. (2012). Magnetism: A Very Short Introduction. OUP Oxford. p.31. ISBN9780191633720.
^ abTricker, R. A. R. (1965). Early electrodynamics. Oxford: Pergamon. p.23.
^Erlichson, Herman (1998). "The experiments of Biot and Savart concerning the force exerted by a current on a magnetic needle". American Journal of Physics. 66 (5): 389. Bibcode:1998AmJPh..66..385E. doi:10.1119/1.18878.
^Frankel, Eugene (1972). Jean-Baptiste Biot: The career of a physicist in nineteenth-century France. Princeton University: Doctoral dissertation. p.334.
^Lord Kelvin of Largs. physik.uni-augsburg.de. 26 June 1824
^Huurdeman, Anton A. (2003) The Worldwide History of Telecommunications. Wiley. ISBN0471205052. p. 202
^"The most important Experiments – The most important Experiments and their Publication between 1886 and 1889". Fraunhofer Heinrich Hertz Institute. Retrieved 19 February 2016.
^Networks of Power: Electrification in Western Society, 1880–1930. JHU Press. March 1993. p.117. ISBN9780801846144.
^Thomas Parke Hughes, Networks of Power: Electrification in Western Society, 1880–1930, pp. 115–118
^Ltd, Nmsi Trading; Smithsonian Institution (1998). Robert Bud, Instruments of Science: An Historical Encyclopedia. Taylor & Francis. p.204. ISBN9780815315612. Retrieved 18 March 2013.
^U.S. patent 381,968
^Porter, H. F. J.; Prout, Henry G. (January 1924). "A Life of George Westinghouse". The American Historical Review. 29 (2): 129. doi:10.2307/1838546. hdl:2027/coo1.ark:/13960/t15m6rz0r. ISSN0002-8762. JSTOR1838546.
^"Galileo Ferraris (March 1888) Rotazioni elettrodinamiche prodotte per mezzo di correnti alternate (Electrodynamic rotations by means of alternating currents), memory read at Accademia delle Scienze, Torino, in Opere di Galileo Ferraris, Hoepli, Milano, 1902 vol I pages 333 to 348"(PDF). Archived from the original(PDF) on 9 July 2021. Retrieved 2 July 2021.
Further reading
Jiles, David (1994). Introduction to Electronic Properties of Materials (1st ed.). Springer. ISBN978-0-412-49580-9.
Tipler, Paul (2004). Physics for Scientists and Engineers: Electricity, Magnetism, Light, and Elementary Modern Physics (5th ed.). W. H. Freeman. ISBN978-0-7167-0810-0. OCLC51095685.
External links
Media related to Magnetic fields at Wikimedia Commons
Crowell, B., "ElectromagnetismArchived 30 April 2010 at the Wayback Machine".
Nave, R., "Magnetic Field". HyperPhysics.
"Magnetism", The Magnetic Field (archived 9 July 2006). theory.uwinnipeg.ca.
Hoadley, Rick, "What do magnetic fields look likeArchived 19 February 2011 at the Wayback Machine?" 17 July 2005.
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190207 | https://math.stackexchange.com/questions/1760725/assumptions-for-existence-of-envelope-of-a-family-of-curves | differential geometry - assumptions for existence of envelope of a family of curves - Mathematics Stack Exchange
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assumptions for existence of envelope of a family of curves
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Given a family of curves F(x,y,c)=0 F(x,y,c)=0 for c c in a range of real numbers, the envelope E E is the curve tangent to every member of the family.
I see that it is defined by the solution of F(x,y,c)=0=∂F∂c(x,y,c)F(x,y,c)=0=∂F∂c(x,y,c).
How do I prove this ? I know that if ∂F∂c(x,y,c)≠0∂F∂c(x,y,c)≠0, then we can solve F(x,y,c)=0 F(x,y,c)=0 at least locally for c=c(x,y)c=c(x,y) but then I cannot see how ∂F∂c(x,y,c)=0∂F∂c(x,y,c)=0.
I saw a similar question with an answer that involved some differential geometry and "straightening of fields" that I do not understand. Can I have an answer based on simple analysis that explains
a) How does the solution of the two equations define E E
b) What is the problem if both ∂F∂x=∂F∂y=0∂F∂x=∂F∂y=0
c) What happens if F(x,y,c)=f(x,y)+ϕ(c)F(x,y,c)=f(x,y)+ϕ(c) ?
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asked Apr 27, 2016 at 7:09
me10240me10240
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First of all, you want the individual curves F c(x,y)=F(x,y,c)=0 F c(x,y)=F(x,y,c)=0 for fixed c c to be smooth, so you'd better assume that ∂F∂x∂F∂x and ∂F∂y∂F∂y are never simultaneously zero. Then you want the implicit function theorem to guarantee you that your two equations will define a smooth curve (locally) parametrized by c c, so you want the matrix
⎡⎣∂F∂x∂2 F∂x∂c∂F∂y∂2 F∂y∂c⎤⎦[∂F∂x∂F∂y∂2 F∂x∂c∂2 F∂y∂c]
to be nonsingular at some point (x 0,y 0,c 0)(x 0,y 0,c 0).
Then you have to check that the curve your equations define locally parametrically as (x,y)=ϕ(c)(x,y)=ϕ(c) has the property that it is tangent to the curve F c(x,y)=0 F c(x,y)=0 at ϕ(c)ϕ(c). I'll let you check this yourself by differentiating F(ϕ(c),c)=0 F(ϕ(c),c)=0.
In answer to your third question, you'll get parallel level curves and there will be no envelope. You might want to try doing some examples. Here's one for you: F(x,y,c)=y+c 2 x−c F(x,y,c)=y+c 2 x−c.
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edited Apr 28, 2016 at 1:38
answered Apr 27, 2016 at 20:21
Ted ShifrinTed Shifrin
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If it were a family of surfaces instead, say F c(x,y,z)F c(x,y,z) then the envelope would also be a surface. How would the implicit function theorem work and what would be the condition required ?me10240 –me10240 2016-04-28 14:43:49 +00:00 Commented Apr 28, 2016 at 14:43
You similarly will need to know that at each point the matrix ⎡⎣⎢∂F∂x∂2 F∂x∂c∂F∂y∂2 F∂y∂c∂F∂z∂2 F∂z∂c⎤⎦⎥[∂F∂x∂F∂y∂F∂z∂2 F∂x∂c∂2 F∂y∂c∂2 F∂z∂c] has rank 2 2.Ted Shifrin –Ted Shifrin 2016-04-28 16:24:00 +00:00 Commented Apr 28, 2016 at 16:24
I see. Then would it imply that locally, WLOG, x=x(z,c)x=x(z,c), y=y(z,c),z=z y=y(z,c),z=z. But from there, I am unable to see how the tangency would be proved in this case?me10240 –me10240 2016-04-28 16:37:46 +00:00 Commented Apr 28, 2016 at 16:37
Do the same calculation as in the curve case. Show that the gradient of F c F c is orthogonal to the tangent plane of the parametric surface Φ(z,c)=(x(z,c),y(z,c),z)Φ(z,c)=(x(z,c),y(z,c),z).Ted Shifrin –Ted Shifrin 2016-04-28 16:38:56 +00:00 Commented Apr 28, 2016 at 16:38
The normal to Φ Φ is y,c i−x,c j+(x,c y,z−y,c x,z)k y,c i−x,c j+(x,c y,z−y,c x,z)k where y,c:=∂y∂c y,c:=∂y∂c. But the normal to F c F c isnt parallel based on this calculation. Where am I going wrong ?me10240 –me10240 2016-04-28 19:44:23 +00:00 Commented Apr 28, 2016 at 19:44
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0Necessary and Sufficient Conditions for Existing an Envelope for a Parametric Family of Implicit Surfaces
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190208 | https://www.webqc.org/molecular-weight-of-%28KClO%29.html | (KClO) molar mass
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Enter a chemical formula to calculate its molar mass and elemental composition:
Molar mass of (KClO) is 90.5507 g/mol
Convert between (KClO) weight and moles
| Compound | Moles | Weight, g |
---
| (KClO) | | | | |
Elemental composition of (KClO)
| Element | Symbol | Atomic weight | Atoms | Mass percent |
--- ---
| Potassium | K | 39.0983 | 1 | 43.1784 |
| Chlorine | Cl | 35.453 | 1 | 39.1527 |
| Oxygen | O | 15.9994 | 1 | 17.6690 |
| Computing molar mass step by step |
| First, compute the number of each atom in (KClO): K: 1, Cl: 1, O: 1 Then, lookup atomic weights for each element in periodic table: K: 39.0983, Cl: 35.453, O: 15.9994 Now, compute the sum of products of number of atoms to the atomic weight: Molar mass ((KClO)) = ∑ Count i Weight i = Count(K) Weight(K) + Count(Cl) Weight(Cl) + Count(O) Weight(O) = 1 39.0983 + 1 35.453 + 1 15.9994 = 90.5507 g/mol |
| Mass Percent Composition | Atomic Percent Composition |
--- |
| K Potassium (43.18%) Cl Chlorine (39.15%) O Oxygen (17.67%) | K Potassium (33.33%) Cl Chlorine (33.33%) O Oxygen (33.33%) |
| Mass Percent Composition |
| K Potassium (43.18%) Cl Chlorine (39.15%) O Oxygen (17.67%) |
| Atomic Percent Composition |
| K Potassium (33.33%) Cl Chlorine (33.33%) O Oxygen (33.33%) |
Related compounds
| Formula | Compound name |
--- |
| KClO | Potassium hypochlorite |
| KClO 3 | Potassium chlorate |
| KClO 4 | Potassium perchlorate |
| KClO 2 | Potassium chlorite |
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What Is Competitive Equilibrium?
Competitive equilibrium is a condition in which profit-maximizing producers and utility-maximizing consumers in competitive markets with freely determined prices arrive at an equilibrium price. At this equilibrium price, the quantity supplied is equal to the quantity demanded. In other words, all parties—buyers and sellers—are satisfied that they're getting a fair deal.
Key Takeaways
Competitive equilibrium is achieved when profit-maximizing producers and utility-maximizing consumers settle on a price that suits all parties.
At this equilibrium price, the quantity supplied by producers is equal to the quantity demanded by consumers.
The theory serves many purposes, including as an analytical tool and a benchmark for efficiency in economics.
Understanding Competitive Equilibrium
As discussed in the law of supply and demand, consumers and producers generally want two different things. The former wants to pay as little as possible, while the latter seeks to sell its goods at the highest possible price.
That means when prices are hiked, the quantity that sellers demand tends to fall and the quantity sellers are willing to supply rises—and when prices are slashed, quantity demanded increases and quantity supplied declines.
Whenever these quantities are not in balance, a shortage or surplus occurs on the market. Under these conditions, entrepreneurs have an incentive (in the form of profit opportunities) to engage in arbitrage, or to reallocate real resources, up until the point where buyers and sellers can agree on one combination of price and quantity in the market. At this point, supply and demand curves intersect, the quantity supplied equals the quantity demanded, and the market is said to be in equilibrium.
At equilibrium prices, both buyers and sellers maximize their economic gains relative to the limits of technology and the resources they have available. Not everyone gets everything they want, but all parties in the market balance their wants against unavoidable scarcity of economic goods as best they can. Because of this, competitive equilibrium is considered a kind of ideal goal for economic efficiency.
Benefits of Competitive Equilibrium
The competitive equilibrium serves many purposes, describing how markets might settle on one price for all buyers and sellers, explaining how production and consumption can be brought in to balance without a central planner, and operating as a benchmark for efficiency in economic analysis.
Economists have long observed that in many markets, buyers and sellers tend to settle around one market price for a given good and that businesses tend to be more or less successful at matching the the amounts and types of goods that they bring to market with the things that consumers want. And that all this seems to happen even without a government official or other authority, or any single person, calculating what the official market prices and quantities should be. The theory of competitive equilibrium is the explanation that they devised to explain how this can happen: when buyers and sellers co-cooperatively calculate the appropriate market prices and quantities together through their acts of buying and selling.
Because competitive equilibrium sets a balance between the interests of all market participants, it can be used to analyze the effects of changes to supply and demand and to benchmark the desirability of government policies that alter market conditions. Moreover, it is often used extensively to analyze economic activities dealing with fiscal or tax policy, in finance for analysis of stock markets and commodity markets, as well as to study interest, exchange rates, and other prices.
Special Considerations
The theory relies on the assumptions of competitive markets. Each trader decides upon a quantity that is so small compared to the total quantity traded, such that their individual transactionshave no influence on the prices. All buyers and sellers have the same information, including all information relevant to supply and demand. Buying and selling goods, or shifting goods and resources between markets or lines of production, involve zero transaction costs. Because these assumptions are not very realistic, competitive equilibrium is only an ideal, and a standard by which other market structures are evaluated, rather than a prediction that real world markets will always achieve competitive equilibrium.
Competitive Equilibrium vs. General Equilibrium
Competitive equilibrium is often used to describe just a single market for one good. An extension of competitive equilibrium to all markets in an economy simultaneously is known as general equilibrium. General equilibrium is also called Walrasian equilibrium.
The difference between the two types of equilibria is all about the emphasis; one market or many connected markets considered together. Both types of equilibria can be described as competitive. The analysis of competitive equilibrium in one market, holding conditions in all other markets to be constant, is also known as partial equilibrium, in order to distinguish it from general equilibrium.
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190210 | https://scholarship.claremont.edu/cgi/viewcontent.cgi?article=1741&context=jhm | Published Time: Fri, 09 Dec 2022 19:24:29 GMT
Journal of Humanistic Mathematics Journal of Humanistic Mathematics
Volume 11 | Issue 1 January 2021
Aligning Political Options and Aggregated Personal Opinions on Aligning Political Options and Aggregated Personal Opinions on
the Issues the Issues
Kris H. Green St. John Fisher College
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Kris H. Green, "Aligning Political Options and Aggregated Personal Opinions on the Issues,"
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Humanistic Mathematics , Volume 11 Issue 1 (January 2021), pages 369-389. DOI: 10.5642/
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them. See for more information. Aligning Political Options and Aggregated Personal Opinions on the Issues
Kris H. Green
Department of Mathematics, Computer Science, & Statistics, St. John Fisher College, Rochester, New York, USA
kgreen@sjfc.edu
Synopsis
Much work has been done in studying how to aggregate voter opinions to decide a fair election. These models presuppose that each voter has a solid understanding of their choices and can express that opinion in the election process. We discuss why this is not always the case. Further, we explore some of the issues that arise when considering the multidimensional nature of both voter preference, with respect to the slate of issues in an election, and the platforms of the various candidates, with respect to the same slate of issues. In light of the complications we encounter and with full apologies to Jonathan Swift, a modest proposal is made for conducting future elections in a way that offers all of the voters a true chance to find a voice.
Keywords. decision theory, voter preferences, elections, multidimensional opin-ions
Introduction
The political system in the United States, and in many countries around the world, presents voters with choices so that they can express their indi-vidual opinions regarding policy by electing officials that will represent their perspectives when making decisions. There are two aspects of such votes, each with its own essential mathematical issues. One of these aspects, which I will call the “election resolution problem,” (ERP) is about aggregating the preferences of a population to determine a fair outcome for a given vote.
Journal of Humanistic Mathematics Volume 11 Number 1 (January 2021)
370 Aggregating Political Opinions
This has received a great deal of attention for quite a long time; Arrow’s Theorem, the works of Donald Saari (e.g. ), and recent work by Siegenfeld and Bar-Yam provide some highlights of this. The study of ERP involves voting systems, apportionment, and gerrymandering, and succinctly il-lustrates how these lead to unfairness with near mathematical certainty. But, by and large, there are well-studied methods for resolving ERP. By contrast, the second aggregation problem receives much less attention. Resolving this problem requires aggregating an individual voter’s various opinions and ideas in order to to match them with options in an election. The implicit assump-tion is that if your preferences are not available, you need to expand the candidate pool, possibly by standing for election yourself. But that method of resolving what I will call the “preference matching problem” (PMP) merely sidesteps the issue. In what follows, we will explore PMP and some ways of resolving it. In the end, the simplifications and models I introduce will lead us to an idea for PMP that exploits many aspects of the modern world, but at quite a severe cost. This foray into mathematically modeling voters is meant to give readers a framework for thinking about voting processes without representing a serious view of the situation. I hope you, Dear Reader, will indulge my satire as a personal catharsis in the aftermath of the political campaigning of 2020 in the United States. And the catharsis extends beyond the present work. Later in this issue of Journal of Humanistic Mathematics interested readers will find my first science fiction short story, which takes place in a political system built on the ideas here, where the costs of such a system can be explored freely. We will frame PMP in the following way. Define S to be the slate of can-didates running in an election. An election is about, ultimately, a set of issues. Each issue might relate to a position on a particular piece of legis-lation, or it might be more general opinions about taxes, education, social security, national security, etc. These might even be attitudes or more gen-eral statements, like “prefers to compromise” or “puts certain views first.” Each candidate Ci, i = 1 , . . . , M , will be represented as a vector that en-codes their platform, an expression of their opinions, attitudes, and so forth, on each of the N issues at hand. Thus, S = {Ci|i = 1, . . . , M } and
Ci = {cij |j = 1 , . . . , N }. Likewise, the set of voters is V = {Vi|i = 1 , . . . , K },Kris H. Green 371
and each of the voters will express their preferences on the issues in a vector so that Vi = {vij |j = 1 , . . . , N }. For now, we take the underlying vector space to be binary. That is, each of cij and vij are elements of the set {0, 1} such that cij = 0 indicates that candidate i takes position 0 on issue j and cij = 1 indicates that the candidate takes position 1 on that issue. Voter preferences are defined analogously. It is also useful, at this point, to note that we will assume that all combinations of preference and platform are possible, which is really just a way to say that each issue is independent of the other issues. (If this is not the case, which is entirely likely, one can simply combine the linked issues into a single meta-issue.) Visually, if such can be said in reference to n-dimensional vectors, this means that one can imagine the issue space as a hypercube, where the coordinates of each vertex defines a combination of opinions on the issues. Each voter and each candidate is positioned at one of these vertices, based on their preferences or platform, respectively. In a typical two-party system, then, each party would seek to define its platforms to accomplish two goals. The first is that the platform they endorse be one that is, in some sense, near the preferences of a large number of voters. The second goal is that they can find candidates whose platforms fall within this scope in order that they represent the party effectively. Thus, one expects that party platforms are evolving over time as both the issues (in number and specific content) and the voters’ opinions on those issues change. While modeling the political parties over time on this hypercube might be interesting, we will focus on PMP, formulated as a series of questions: 1. PMP-1. How well does each voter actually understand the values of their preferences, vij ?2. PMP-2. How well does each voter actually understand the platforms of each candidate, cij ?3. PMP-3. What relative importance (weight) does each voter assign to the various issues? 4. PMP-4. How does each voter determine which of the Ci is “closest” to their Vj ?5. PMP-5. How do we ensure that as many voters as possible can match their preferences “close enough” in any election? 372 Aggregating Political Opinions
PMP-4 and PMP-5 will be the primary focus of our analysis and discussion, but along the way some proposals for making progress on the first two will be explored. In fact, a single approach will be proposed that, in principle, resolves all PMP. As a caveat, I would like to acknowledge that many of these ideas are not completely new. At least, they are not new to political theorists or to those who study and develop models of elections and voting. For similar formulations of elections in an issue space, see or . But I hope to frame them in a new light and connect some ideas together in a way that leads to, with apologies to Jonathan Swift, a “modest proposal” for solving PMP that relies on various machine learning tools to help voters. First, we will explore PMP by looking at how a failure to acknowledge the dimensionality of the issue space poses complications for current political practice. We will then investigate successively more complex ways to ac-count for these dimensions in our voting, explicitly involving the electorate as we do. This starts with an extremely simplified representation of elections, and expands to allow more realistic portrayals of the issues. Everything cul-minates in a possible way of resolving all of PMP in one simple system, as the main barriers to resolving it are caused by not having enough candidates participate in an election to represent all distinct sets of voter preferences.
Linear Thinking is Not Enough
The set of options presented to voters in the United States is often pre-sented as a binary choice between individual representatives of two political parties rather than a choice among a larger slate representing a variety of ways of thinking on the issues. This simplification means no matter how many important issues are on the table, we are projecting everything into a one-dimensional space with a dividing line to separate the two parties. This leads to a lot confusing hair-splitting to “define” our politicians. Is a particular candidate on the right or left? Is she a moderate conservative, a liberal conservative, or an extreme conservative? Is the other candidate a democratic socialist? A socialist? A capitalist? All of this confusion comes from attempting to line candidates up along a single axis. One of the most striking examples of this occurred in the first three of the 2019 Democratic Party presidential debates. Although the candidates were ar-ranged on stage (over two nights for some of these debates, due to the sheer number of them) randomly, one could be forgiven for seeing a line of candidate options from “far left” to “moderate left” to just plain “left.” Kris H. Green 373
Indeed, there is evidence that the human brain has co-opted the visual cortex that sees such lines into the processing center that we use for mathematics, so that such a mapping onto a one-dimensional space is inevitable [1, 6]. Most of us realize, hopefully, that the biggest problem with this is that political issues, and the world in general, are multi-dimensional. But we routinely treat them as one-dimensional, with extreme views of democrats on the left and extreme views of republicans on the right, for example. Then we try to force all the candidates to align somewhere on this spectrum, placing dividing line at some point and asking the voters on each side of this point to vote for the party in their “half” of the space. Projecting platforms and preferences this way ignores the richness of the perspectives possible, and it ultimately leaves people confused when a candidate fails to behave consistently with where we think they fall on this single axis. Even worse is that, depending on the projection used, a different ordering of the candidates along the line of projection is possible. Figure 1 illustrates how easy it is to have quite different orderings of candidates by choosing different projections. Using a projection onto a vertical line (labeled Projection 2) places candidate A in between B and C; but projection onto the horizontal line (Projection 1) puts a lot of ideological space between A and C, leaving
B as the middle ground. Using projections in this way to simply a high-dimensional space may add necessary efficiencies to our thinking and communication since treating each politician and each voter as a unique combination of factors and interests could be overwhelming. In many situations, such stereotyping is essential to let us function, as is the case when we approach an intersection in a moving car: We treat this as we would any other intersection, and assume that other drivers will follow the same rules observed and obeyed in the past. If we did not, the number of possible actions each driver could take would lead us to considering a near infinite tree of decisions, paralyzing us instead of allowing us to drive onward to our destination. I argue that in the political arena, however, we have been misusing stereotyping. No longer are the labels acting as a rough-but-convenient shorthand for talking about a candidate’s position. Instead, once they are assigned, we tend to think that these labels should apply to all of a candidate’s actions at all times. Rather than trying to fit a label to the actions and expressed platforms of the candidates, we are trying to force the candidates to fit the labels. 374 Aggregating Political Opinions
-6@@@@@@@@@I1
ABC
rrrrrr
A′ B′ C′
A′′
B′′
C′′
Issue n
Issues 1, . . . , n − 1Projection 1 Projection 2
Figure 1: Schematic illustration of three candidates, A,B, and C, in n-dimensional issue space, showing two different lines of projection that give different orderings for the candi-dates. The dashed line shows the projections of candidate C for clarity. The projection of candidate Conto projection 1 is marked by C′and onto projection 2 by C′′ .Note that for illustrative purposes, the axes for issues 1 , . . . , n −1 are shown as the edge of a hyperplane, and each candidate ranks the each issue on the interval [ −1,1].
Using the hypercube visualization of an election, though, we can frame this projection in a way that is slightly easier to visualize. Start by assuming that each party has articulated a set of platforms that fall within their scope. This basically means that there is a hyperplane passing through the issue-hypercube such that all of the platforms and preferences on one side of the hyperplane are expected to align with party A, while the others are expected to align with party B. In general, we can allow for each party to specify its own hyperplane, which means some voters may be ignored by both parties and some voters are actively courted by both parties. Such a case is illustrated in Figure 2. Each election, then, is a way of asking voters to vote for a candidate that falls on the same side of the hyperplane as the party candidate. Of course, this resolution of PMP-4 presupposes that the voters have solid answers to PMP-1 and PMP-2. Kris H. Green 375
uueeue
B B
(0, 0, 0) (1, 0, 0) (1, 1, 0) (0, 1, 0) (0, 0, 1) (1, 0, 1) (1, 1, 1) (0, 1, 1)
Figure 2: Illustration of issue hypercube with three issues. Preference sets (vertices) are hollow circles if they are aligned with candidate Aand filled circles for those aligned with candidate B. Voter preference sets that are courted by both parties are shown as hollow squares, and hollow triangles mark voters ignored by both parties. It is left as an exercise for the reader to sketch in the two party planes which would lead to these divisions among the preference sets.
Binary Platforms Provide Too Much Freedom
From the start, PMP naturally deals with the dimensionality problem by embracing the messiness inherent to political decision-making and expressing the platforms of candidates and preferences of voters as vectors in some “issue space” of N dimensions. In principle, then, we are not tied to party choices that link opinions about the issues into single platforms about all the issues, tying opinions on education to those about health care. The possible upside is obvious: each voter may be able to find a candidate that better fits their individual preferences, removing the need to oversimplify the decision of which candidate to support. To see some of the downsides to embracing higher dimensions, let us dig deeper into an N -issue space when our underlying platforms and prefer-ences are described as above, with cij , v ij ∈ {0, 1}. This gives 2 N dif-ferent possible platforms for candidates, with the same number of possi-ble voter preferences (equivalently, 2 N vertices on the issue hypercube.) 376 Aggregating Political Opinions
Ideally, each voter would then choose to vote for the candidate whose plat-form is, in some sense, closest to their preference vector. Even assuming that PMP-1 and PMP-2 are resolved, so that each voter is able to make a fully informed choice, there are at least three major problems remaining. First, there is no guarantee a candidate espousing a particular platform during the election process will actually act in a way that is consistent with this plat-form once elected. Apparently, this is not as common as many of the voters think (c.f. and ), and we cannot deal here with such misrepresentation, so we will leave it alone. Second, we are assuming that the issues are inde-pendent of one another, and that it is possible that all 2 N platforms make logical sense. Any correlations among the issues reduces the issues space considerably. But this actually works in our favor, given what follows. The third issue we must address is PMP-4, which is clear from a basic count-ing argument: With only five issues, there are already 32 possible platforms or preferences. But in a two-party system with five issues, at most two of these 32 platforms are available for voters to select. (I say “at most” because it is possible, albeit unlikely, that both candidates have very similar politi-cal stances.) This means that voters are much less likely to find their exact preference expressed by either of the candidates. Voters left out by this then have to decide whether to skip the election or vote for a candidate whose platform does not fully align with their preferences. A popular method, if we are to believe the political theorists and pundits, is for voters to resolve this dilemma by focusing on only a single issue rather than all N of the issues. One of the dangers of this is that after several election cycles voters may find it simpler to change their own opinions to align more fully with the platform of the candidates (or their party). It is this author’s opinion that neither of these is a good outcome, and that it is not what the authors of the United States political system intended, but I will not delve into that here. So how can voters make choices in an N -issue election when fewer than 2N candidates are running for the position while avoiding either single-issue voting or changing their opinions? One method for decision making would be to have each voter to make a list of all the candidates and then com-pare their own preferences to each platform. Each simply writes out new vectors for each candidate computed by checking elementwise equality be-tween the platforms and their preference. Basically, for each candidate-voter pairing, we create a new vector to measure the degree to which they match. Kris H. Green 377
Define the vector comparing Vi to Cj to be Mij = {mijk |k = 1 , . . . , N } with the components of the vector defined as follows: mijk = 1 if vik = cjk and 0 otherwise. Thus, mijk = 1 represents agreement between voter i and candidate j on issue k, and 0 represents disagreement. Then Aij = ∑Nk=1 mijk results in a score that measures the number of issues on which voter i and candidate
j agree. Each voter can then compute these and select the candidate with the highest agreement score. This is exactly what potential voters can try through various online quizzes, such as , where answering the questions truthfully could surprise you! Table 1 shows this comparison for a single voter with preference [0 , 1, 1] in a full election on three issues, in which each possible binary platform is espoused by one of the eight candidates. Candidate, j Platform, Cj Mij Aij
1 0, 0, 0 1, 0, 0 12 0, 0, 1 1, 0, 1 23 0, 1, 0 1, 1, 0 24 0, 1, 1 1, 1, 1 35 1, 0, 0 0, 0, 0 06 1, 0, 1 0, 0, 1 17 1, 1, 0 0, 1, 0 18 1, 1, 1 0, 1, 1 2
Table 1: Comparison of a voter with preference [0, 1, 1] to each of the possible candidates.
In our example, the voter should select candidate 4, whose opinions align perfectly with the voter. But what if only candidates 1, 2, 3, and 7 were running in the election? The voter now has a dilemma: both candidates 2 and 3 have the same agreement score! And what if only candidates 1, 5, and 6 were running? None of these have a score above 1, leaving this voter feeling like none of the options presented is worth supporting. Another clear problem is that we are not guaranteed that each candidate has a unique platform, so that there could be ties for the most matched candidate. 378 Aggregating Political Opinions
Ranking the Issues Isn’t Much Better
We can resolve some of these complications by abandoning one of our un-stated assumptions: that all issues are of equal importance in an election to both the voters and candidates. This is a somewhat unlikely scenario anyway, as many of us weigh the issues differently. This would require each voter Vi to specify a weight vector, Wi = {wij |j = 1 , . . . , N } for the weights — the relative importance — they associate with the issues. Then, instead of a simple sum, the agreement score for voter i against candidate
j is Aij = Wi · Mij = ∑Nk=1 wik mijk and then compare the weighted scores. Such a method is feasible with the right computational support, but with-out it, and without each voter carefully going through the analysis, it is very possible that voters might choose to vote for candidates that are not actually in line with their expressed preferences. Attempting to resolve this is why we need to include PMP-3, although it can lead to some interesting results. For example, suppose a voter’s matching with candidate 1 is the vector M1 =(1 , 1, 1, 0, 0, 1) and their matching with candidate 2 is M2 = (0 , 0, 0, 1, 1, 1). Further, suppose the voter has weighted the issues W = (5 , 1, 1, 4, 4, 2). The voter would then have an unweighted agreement of 4 with the first candi-date and 3 with the second. But the weightings would give a score of 9 to candidate 1 and 10 to candidate 2, thus indicating preference for the second candidate, even though the first candidate matches on more issues, includ-ing the issue that the voter has ranked most highly. This is only one such example, and it does require a particular collection of characteristics in the various weights and matchings. However, it illustrates how voters may need to accept a different way of viewing and ranking candidates in order to get more out of their votes. Furthermore, unless there is an agreed-upon (among voters) system for ranking the issues or some sort of standardized weighting, it would be impossible to compare one voter to another directly. This is not needed, however, if the only purpose is to aid an individual voter in making their decision, so long as each voter remains consistent in their process when applying the weights to each of the candidates. Another, somewhat simpler, solution would be to use what we can call “or-dered platforms.” For this, we assume that a candidate’s opinion about issues is expressed by the order of importance they place on the issue. In order to allow candidates and voters to express their opinions both for and against each issue, one would need to include each issue twice, once as a “support Kris H. Green 379
this issue” option, and once as a “work against this issue” option, so that a candidate strongly in favor of an issue can rank the “pro” side highly and the “con” side with slightly lower importance, or vice versa. At the same time, it opens up the ability for our system to allow candidates and voters to express their willingness to compromise by having very different rankings for the two sides. For example, if a candidate is willing to accept the legislation failing, in order to compromise and let something else pass, then their con position might be more highly ranked than the pro position for an issue. A ranked system, in effect, expresses how the candidates are going to spend their energy and resources, once elected. For a first analysis, we will ignore the candidates’ particular views. We assume that by prioritizing an issue, the candidate will work to their best ability to find a solution to the issue that addresses the common good, so that while the outcome may not be one that a voter would have initially been in favor of, the voter respects the process by which the candidate worked to resolve the issue. Unfortunately, there are far more ordered platforms than binary platforms when N issues are considered, since there are N ! ways to order N issues. For example, a five-issue election results in 120 possible (1x2x3x4x5 = 120) platforms but only 32 (2 5) binary platforms. And placing the dummy issues to represent the opposing sides, we see the situation is much worse: for an N issue election, we would need (2 N )! platforms. So a given voter’s likelihood of finding their opinion perfectly matched by a candidate in a limited election decreases dramatically with the number of issues. Table 2 compares the number of platforms for various values of N ; note that the number of platforms grows so quickly that even with only 8 issues, more candidates would need to run for election than the current population of the entire planet in order to guarantee all possible voter preferences are presented! Issues, N 4 6 8 10 Binary Platforms, 2 N 16 64 256 1024 Ordered Platforms, (2 N )! 40,320 4.79 · 10 8 2.09 · 10 13 2.43 · 10 18
Table 2: Candidates needed to guarantee that all possible voter preferences are expressed in an election based on the number of issues, N, and the method of expressing the platform.
Given the variety of possible platforms in this case, and the low probability that any given voter finds their preferences expressed, it seems that many voters will be ignored. There is some hope here though, since voters can look 380 Aggregating Political Opinions
at partial orderings for agreement, and there are far more partial ordering-matchings possible. For example, if a voter orders the issues in a five-issue election from most to least importance as 12345, a candidate for whom the issues are ranked 13245 is quite similar (only one swap is needed to match perfectly). This gives rise to questions about whether 52341 is a better match than 14325, for example, but there are many ways to compare the similarity of two orderings.
Other Options Just Muddy the Process
This is all getting rather complicated, and seems to be requiring more and more from the voters in order to make a rational selection, so we will take a different path in a search for a simple solution. For example, voters could focus on only matching the rankings of the top m issues. This dramatically reduces the number of candidates required in the field in order to guarantee full representations. Table 3 shows this effect. For example, with a four-issue election, attempting to match all four issues for priority would require 24 candidates, but only four candidates are needed if the first issue is the only one considered important. Note that we could have multiple candidates with similar platforms, which would force voters to consider additional aspects of the candidates, such as personal behavior, family connections, and so forth. However, it might be useful, as we will discuss later, to include those features in issue-space from the start.
k n = 4 n = 6 n = 8 n = 10 n = 12 1 4 6 8 10 12 2 12 30 56 90 132 3 24 120 336 720 1320 4 24 360 1680 5040 11880 5 720 6720 30240 95040
Table 3: Number of candidates needed for full representation in an n-issue election where only the top kissues are considered critical.
One way to implement this is by using the weighted issue system above, and simply assigning weights of 0 to the ( N −m) issues we do not want to consider. Of course, the number of important issues will vary from voter to voter, as will the rankings of those issues. And now we are much more likely to have multiple candidates with equivalent matching scores for a particular voter. Kris H. Green 381
Also, while the equally-ranked candidates may indeed be isomorphic in the views of some voters, one would probably need to make use of uncoded data (past history, for example) to break any ties. We could, of course, combine several of these methods into an“ordered-binary platform” or a “weighted-ordered platform.” A relatively simple way to do this would be to move from an underlying binary vector space of {0, 1} for ranking preferences and platforms. Instead, we can allow vij ∈ [−1, 1]. Negative values would indicate support for the opposing side of the issue, and values near zero would indicate indifference, eliminating the complexity of doubling the dimensions to include separate pro and con sides of each issue. Once the voters and candidates have ranked or weighted the issues, these can be rescaled so that each entry is the original value divided by the square root of the sum of the squares of the of entries: the scaled weights ˜wij = wij /
√∑ w2
ij
then guarantee that each vector has unit length. If we also do this for each candidate’s platform, we can visualize each voter and candidate as a point on a unit hypersphere in the issue space. Once all the platforms are deter-mined, an individual voter can then use any of several methods to select the candidate that is “closest” to them, such as using the standard relationship that the cosine of the angle between vectors is given by their dot product divided by the products of their magnitudes. This then allows each voter to find the the platform of the candidate having the smallest angle with their preference vector by choosing the one having the largest dot product to their weighted preferences. On the other hand, mapping an entire collection of voters V onto this hy-persphere would allow a political party or candidate to use clustering meth-ods to design a platform that would appeal to the largest number of voters while also being clearly distinguished from other platforms. So the weighted issues formulation here would, given sufficient data on the electorate, al-low the parties to divide up the issue-space. Effectively, though, we are now merely in a continuous analogue of the hypercube discussed earlier. So the approach brings with it many the potential difficulties of the original approach. However, if voters could all be mapped onto this hypersphere, it might be possible to employ machine learning to develop policies with-out directly involving representatives in the government to make decisions. 382 Aggregating Political Opinions
For example, a nearest neighbors or clustering algorithm could identify large groupings of opinions and employ evolutionary algorithms to develop policies that would match the preferences of these groups. such policies could be vet-ted by teams of policy workers and made public, so that voters outside the clusters would have a chance to review them and revise their own opinions prior to voting on the policies. A real concern with all of these methods to resolve aspects of PMP is not even stated in our formulation of the problem: Do the candidates actually know their position on each issue? To see why this could be a problem, depending on our underlying vector space, consider using a scale from 0 to 1 indicating “percent of agreement.” Each candidate can clearly determine this, one would hope, but remember that they will, once in office, be asked to make decisions that are new to them and about which they had no previous knowledge. So what actions will they take? One can imagine an experienced candidate filling out a survey on the issues and drawing upon their expe-riences in office to help nuance their answers, while a new candidate with no experience merely uses an educated guess. Thus, each candidate is ac-tually working on a different scale! And while we could focus on objective voting records rather than self-reported agreements to build our scale, new candidates will not have previous voting records. Thus, the system might need to expand to viewing other records for candi-dates in order to determine where they stand on various issues. One could imagine potential candidates being asked to opt in to allow artificial intelli-gence bots to comb their social media and other public statements for data. A.I. could then perform semantic analysis to compute each candidate’s “frac-tion of statements made consistent with an opinion on the issue” or use bank-ing records and other data to produce the scores for each candidate. Again, those new to the scene will have less data available to inform the system, and candidates who have changed their opinions over time would need to have more recent data weighed more heavily. And none of this addresses the ethical issues associated with mining a person’s background. For example, most social data mining requires branching out from the target individual’s data to those most closely associated with them, who might not be aware of this scrutiny nor have been asked to participate. Nor does this address the vast quantity of information on a given person that has virtually no relevance to their political opinions or position. Even more troubling is the need to train the A.I. to do this mapping from personal data into political opinion. Kris H. Green 383
As we are learning all too often, the algorithms may start as unbiased code, but the training data used to prepare them for use introduce bias (see for an analysis of one prominent example). Even more concerning is that the candidates may not even agree on what the list of issues is. So our set of possible issues would need to be the union of the sets of issues as perceived by each candidate and each voter. Depending on the level of detail we use to specify the issues, this could lead to quite a large value of N . For example, three candidates could all be in favor of similar legislation related to climate change. But the first candidate may want changes implemented within three years, a second candidate may want those same changes over five years, and the third might care less about the timeline and more about the particular level of emissions set in the legislation. This would have be encoded as three separate issues, depending on exactly closely we expect to model each candidate’s platform. There are many other approaches to generating the similarity scores needed to help each voter identify their best-matched candidate. And in reality, many issues are not independent. The connectedness results in a smaller set of issues to consider, and a correspondingly smaller slate of candidates. Using cluster analysis would allow us to group similar issues and positions, greatly reducing the dimensionality of the issue space. Then we would see that, perhaps position 0 on issue A and position 1 on issue B are always expressed together by voters. So we no longer need A and B as separate issues, but only as a composite. However, this clustering is essentially how we got into the two party mess in which the United States finds itself currently: we cluster one set of ideas as “Republican” and another set as “Democratic.” The same approaches to those above may be used to analyze the situation, but in this case, each issue is really a meta-issue, expressing a collection of opinions. One potential danger here is that clustering in this way would get you back into a polarized electorate, with a very limited set of options presented to each voter and with little potential for compromise. In fact, recent work shows that this polarized state, once achieved, may be very difficult to break from [12, 16].Once again, we all wind up bundling the issues into a single meta-issue with two sides, Republican or Democrat, rather than looking deeper. And in reality, political parties are always shifting their platforms in response to events and voter opinions, although different parties tend to do this in different ways and for different reasons . 384 Aggregating Political Opinions
All of the complexity above is caused by one common problem: Assuming that there are fewer candidates standing for election than there are voter preferences in the electorate. This could be solved by just having more candidates each election, with platforms more distributed in the issue space, and by having everyone much clearer on all their own preferences. But there is an inherent impracticality to large slates of candidates, aside from the difficulty of getting clear information to the public and the unwieldiness of really long ballots: Once the number of candidates gets too large, the number of members of the electorate who exactly match any one preference will be quite small, assuming that all combinations of preferences are equally likely. Thus, achieving a plurality vote becomes increasingly hard if we insist on perfect matchings, even for small slate of issues. Table 4 illustrates this problem by showing how the number of issues affects the plurality of a vote. Issues Number of Percent of Percent of possible platforms electorate sharing electorate sharing (approximate) any one preference (%) all but one issue (%)
N 2N 2−N N 2−N
5 32 3.125 15.625 10 1024 0.0977 0.97 20 1.05 · 10 6 9.54 · 10 −5 0.0019 40 1.1 · 10 12 9.09 · 10 −11 3.64 · 10 −9
Table 4: Portion of electorate sharing a particular preference set (expressed as percent of population) for different numbers of critical issues, assuming support for each platform is uniformly distributed.
The table assumes that the voter preferences are uniformly spread over the possible options and shows the percentage of the electorate who would share any one particular preference. As you can see, with just ten issues involved, much less than 1% of the voters share a common preference. However, the hypercube visualization reminds us that there would be a large number of preferences adjacent to any particular platform. For each preference, rep-resented by a vertex on the N -dimensional hypercube, there would be N
preferences (vertices), adjacent to it, agreeing on all but one issue. Thus, while only a small portion of the population would be in complete agree-ment, there would be a larger fraction (proportional to to the number of issues, N ) of the population who agree on all but one issue, although the Kris H. Green 385
single disagreement would vary over the entire issue space. Again, that is part of ERP, and there are some ways to resolve this, independent of the political system to be proposed below.
A “Modest Proposal” for Resolving PMP
So what are we to do? We cannot easily increase the candidate pool. And even if we did such a thing, many voters would still struggle to find their preferences matched perfectly by one of the candidates. It seems that we are at an impasse as long as we stick to volunteers entering the political arena and seeking an aggregate vote from the public that selects their platform over others. There is, however, a solution: Stop requiring humans to serve as elected officials and remove the candidates from the system completely. We now have enough computing power to simulate how a decision-maker would implement their platform while in office. And if we can simulate one possible platform, there is nothing to prevent us from simulating all possible platforms. We can thus envision the creation of a “political algorithmic logic” unit (PAL) with artificial intelligence and tunable characteristics so that it could make decisions and vote as if it were a person holding whatever platform of opinions was desired. To implement the system, we poll the electorate to allow each voter to express their preferences. Then we select a Decision Body composed of a multitude of these artificially intelligent PAL systems, each adjusted to represent different voter preferences. Below, we will explore how this resolves PMP, but it is worth briefly noting that this solution also resolves ERP for free! We only need to weight each PAL’s contribution to the actions taken by the Decision Body by weighing their votes in proportion to the number of voters matching with them. Then the aggregate of voter preferences is directly put into office to make decisions. Note that if the platforms/preferences contain only specific issues, then we are effectively talking about a direct democracy where the outcome of any proposal is predetermined by the composition of the Decision Body. To avoid this, instead, the issue space should also include attitudes like “willing to compromise on issue X,” “uncompromising,” or “prioritizes education.” Then the PALs can meaningfully work to convince each other and reach a collective decision that accounts for all the preferences and the interac-tions among the issues and attitudes, without being a direct democracy. 386 Aggregating Political Opinions
This system would also allow the Decision Body to interface and collate the different PAL positions, and factor in information that is unavailable to the public due to security or similar concerns. I would argue that this is critical, since it would be impossible to anticipate all future scenarios up front for matching with voter preferences. Indeed, while issues are critical, it is the way we anticipate a candidate will meet the unknown that often influences our vote or the way we feel about him or her (as explored by thermome-ters in ). In essence, legislative and other decisions would undergo a sort of evolutionary algorithm before emerging from the decision body of PALs. And given the nature of the senators in this model, we would not have to deal with political action committees or lobbyists influencing the system, although there are a host of other dangers to consider. The PAL-based system would automatically resolve PMP-4 and PMP-5. Mining social media, life history information, and so forth, would help de-termine each voter’s preferences on the issues. If necessary, direct surveys of voter awareness, knowledge, and opinion on the issues could be implemented, leading to the resolution — by not directly requiring the voters themselves to even be aware of it — of PMP-1. And since the system provides opportunities for every platform to participate, PMP-2 is also resolved. It is also not hard to envision a method that would simultaneously weight each issue for the voters, resolving PMP-3 as well. Voting computers could quickly determine matchings for each voter with the possible platforms. To be fair, one could also allow multiple points of intervention, where the voter can adjust what the system has determined about their preferences. This would be essential, since people often change their views, leading to different desirable candi-dates and different rankings of the issues. And if one’s entire social media and other history were factored in, recent changes in one’s opinion could be lost.
Conclusion
Clearly, the preceding has ignored much. For example, voting can be treated as an exercise in economic game theory: If there is a candidate that you like, but feel will not win enough votes, are you better off selecting that candidate anyway, or should you vote for a more popular second choice can-didate in order to try and prevent a candidate from your “non-preferred” list from winning? We have also been completely ignoring the publicly available Kris H. Green 387
polling data and the role of popularity in the process. There are several well-studied examples in which election results were influenced by voter decisions made from polling data rather than their preferences, since many chose not to spend time voting when the predictions made public during the election process appeared to already be strongly consistent with their preferences. And the entire formulation here is based on a the idea of the candidates and voters having a common knowledge of the issues. If different voters have different lists of “the issues” then it would be difficult to generate a consensus on even what the PALs should look like for a given election. But electing (or selecting) a mix of artificially intelligent PAL units to make decisions would restore the importance of each individual voter in the system. And it would still allow for a representational system. After all, we cannot have everyone in the country spending all of their time running the country; someone has to teach the children, build the homes, and grow the food. Everyone’s preferences now factor in the decision process. And it turns out that there is already work in using AI to match voter preferences to platforms, as in . Of course, there are bound to be complications in letting a group of AI run the country. Mathematically, it is not hard to envision instances where the combination of the PAL would fail to reach resolution on a particular issue. The system could be manipulated by hackers during the decision-making process. The choices for which issues to include in the platforms could be manipulated; this has deep consequences explored in , for example. I’m sure there are a host of additional complications that I have not raised. But here’s the secret: Most of us believe these problems already exist in the human-centered election process. Compared to the interminable and contentious 2020 election year in the United States, I, for one, welcome our robot overlords... er, decision-makers. And I welcome you, Dear Reader (apologies to the late Isaac Asimov) to explore a fictional future filled with PALs who want the best for you in my short story “The Deterministic Republic.” 1
1Green, K. H. “Deterministic Republic,” Journal of Humanistic Mathematics , Vol-ume 11 Issue 1 (January 2021), pages 445–484. Available at claremont.edu/jhm/vol11/iss1/29 .
388 Aggregating Political Opinions
References
Stanislas Dehaene, Veronique Izard, Elizabeth Spelke, and Pierre Pica, “Log or linear? Distinct intuitions of the number scale in Western and Amazonian indigene cultures,” Science , Volume 320 (30 May 2008), pages 1217-1220. Julia Dressl and Hany Farid. “The accuracy, fairness, and limits of pre-dicting recidivism,” Science Advances , Volume 4: eaao5580 (2018). James Enelow and Melvin Hinich, “The location of American presiden-tial candidates: An empirical test of a new spatial model of elections,”
Mathematical and Computer Modelling , Volume 12 Number 4/5 (1989), pages 461-470. Timothy Hill, “Trust us: Politicians keep most of their promises,”
FiveThirtyEight , April 21, 2016, available at ht.com/features/trust-us-politicians-keep-most-of-their-pro mises/ , last accessed on January 27, 2021. Jasper Lu, David Kai Zhang, Zinovi Rabinovich, Svetlana Obraztsova, and Yevgeniy Vorobeychik, “Manipulating elections by selecting issues,” in Proceedings of the 18th International Conference on Autonomous Agents and MultiAgent Systems (AAMAS ’19) , International Founda-tion for Autonomous Agents and Multiagent Systems, Richland, SC, 2019, pages 529-537. Andre Knopps, Bertrand Thirion, Edward Michael Hubbard, Vincent Michel, and Stanislas Dehaene, “Recruitment of an area involved in eye movements during mental arithmetic,” Science , Volume 324 (19 June 2009), pages 34-39. Brittany Renee Mayes, Kevin Schaul, and Kevin Uhrmacher, “Which of these 2020 Democrats agrees with you most?”, The Washing-ton Post , November 22, 2019, available at tonpost.com/graphics/politics/policy-2020/quiz-which-candid ate-agrees-with-me/ , last accessed on January 27, 2021. Marvin Minsky, Society of Mind , Simon and Schuster, New York, 1986. Donald Saari, “Geometry of chaotic and stable discussions”, American Mathematical Monthly , Volume 111 Issue 5 (2004), pages 377-393. Kris H. Green 389
Gijs Schumacher, Catherine E. de Vries, and Barbara Vis, “Why do par-ties change position? Party organization and environmental incentives”,
Journal of Politics , Volume 75 Number 2 (April 2013), pages 464-477. Avner Seror and Thierry Verdier, “Multi-candidate political competition and the industrial organization of politics”, in CEPR Discussion Papers ,Volume 13121 (2018), Center for Economic Policy Research. Alexander Siegenfeld and Yaneer Bar-Yam, “Negative representation and instability in democratic elections,” Nature Physics , Volume 16
(2020), pages 186-190. doi:10.1038/s41567-019-0739-6 Ian Stewart, “Electoral dysfunction: Why democracy is always un-fair,” New Scientist , Volume 2758 (2010), available at newscientist.com/article/mg20627581-400-electoral-dysfuncti on-why-democracy-is-always-unfair/ , last accessed on January 27, 2021. Luis Ter´ an, Andreas Ladner, Jan Fivaz, and Stefani Gerber, “Us-ing a fuzzy-based cluster algorithm for recommending candidates in e-elections,” pages 115-138 in Fuzzy Methods for Customer Relationship Management and Marketing: Applications and Classifications edited by Andreas Meier and Laurent Donze (IGI Global, 2011). Robert Thomson, Terry Royed, Elin Naurin, Joaqu´ ın Art´ es, Rory Costello, Laurenz Ennser-Jedenastik, Mark Ferguson, Petia Kostadi-nova, Catherine Moury, Fran¸ cois P´ etry, Katrin Praprotnik, “The fulfill-ment of parties’ election pledges: A comparative study on the impact of power sharing,” American Journal of Political Science , Volume 61
Number 3 (July 2017), pages 527–542. Herbert Weisberg, and Jerrold Rusk, “Dimensions of candidate evalu-ation”, The American Political Science Review , Volume 64 Number 4 (1970), pages 1167-1185. Vicky Yang, Daniel Abrams, Georgia Kernell, and Adilson Motter, “Why are U.S. parties so polarized? A ‘satisficing’ dynamical model,”
SIAM Review , Volume 62 Number 3 (2020), pages 646-657. |
190211 | https://www.lessonplanet.com/teachers/khan-academy-convert-to-smaller-units-sec-min-hr | Khan Academy: Convert to Smaller Units (Sec, Min, & Hr) Unknown Type for 3rd - 5th Grade | Lesson Planet
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190212 | https://en.wikipedia.org/wiki/Conformal_map | Jump to content
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Contents
(Top)
1 In two dimensions
1.1 Global conformal maps on the Riemann sphere
1.2 Conformality with respect to three types of angles
2 In three or more dimensions
2.1 Riemannian geometry
2.2 Euclidean space
3 Applications
3.1 Cartography
3.2 Physics and engineering
3.3 Maxwell's equations
3.4 General relativity
4 See also
5 References
6 Further reading
7 External links
Conformal map
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From Wikipedia, the free encyclopedia
Mathematical function that preserves angles
For other uses, see Conformal (disambiguation).
"Conformal projection" redirects here; not to be confused with Conformal map projection.
| |
| Mathematical analysis → Complex analysis |
| Complex analysis |
| Complex numbers |
| Real number Imaginary number Complex plane Complex conjugate Euler's formula |
| Basic theory |
| Argument principle Residue Essential singularity Isolated singularity Removable singularity Zeros and poles |
| Complex functions |
| Complex-valued function Antiderivative Analytic function Entire function Holomorphic function Meromorphic function Cauchy–Riemann equations Formal power series Laurent series |
| Theorems |
| Analyticity of holomorphic functions Cauchy's integral theorem Cauchy's integral formula Residue theorem Liouville's theorem Picard theorem Weierstrass factorization theorem |
| Geometric function theory |
| Complex manifold Conformal map Quasiconformal mapping Riemann surface Schwarz lemma Winding number Analytic continuation Riemann's mapping theorem |
| People |
| Augustin-Louis Cauchy Leonhard Euler Carl Friedrich Gauss Bernhard Riemann Karl Weierstrass |
| Mathematics portal |
| v t e |
In mathematics, a conformal map is a function that locally preserves angles, but not necessarily lengths.
More formally, let and be open subsets of . A function is called conformal (or angle-preserving) at a point if it preserves angles between directed curves through , as well as preserving orientation. Conformal maps preserve both angles and the shapes of infinitesimally small figures, but not necessarily their size or curvature.
The conformal property may be described in terms of the Jacobian derivative matrix of a coordinate transformation. The transformation is conformal whenever the Jacobian at each point is a positive scalar times a rotation matrix (orthogonal with determinant one). Some authors define conformality to include orientation-reversing mappings whose Jacobians can be written as any scalar times any orthogonal matrix.
For mappings in two dimensions, the (orientation-preserving) conformal mappings are precisely the locally invertible complex analytic functions. In three and higher dimensions, Liouville's theorem sharply limits the conformal mappings to a few types.
The notion of conformality generalizes in a natural way to maps between Riemannian or semi-Riemannian manifolds.
In two dimensions
[edit]
If is an open subset of the complex plane , then a function is conformal if and only if it is holomorphic and its derivative is everywhere non-zero on . If is antiholomorphic (conjugate to a holomorphic function), it preserves angles but reverses their orientation.
In the literature, there is another definition of conformal: a mapping which is one-to-one and holomorphic on an open set in the plane. The open mapping theorem forces the inverse function (defined on the image of ) to be holomorphic. Thus, under this definition, a map is conformal if and only if it is biholomorphic. The two definitions for conformal maps are not equivalent. Being one-to-one and holomorphic implies having a non-zero derivative. In fact, we have the following relation, the inverse function theorem:
where . However, the exponential function is a holomorphic function with a nonzero derivative, but is not one-to-one since it is periodic.
The Riemann mapping theorem, one of the profound results of complex analysis, states that any non-empty open simply connected proper subset of admits a bijective conformal map to the open unit disk in . Informally, this means that any blob can be transformed into a perfect circle by some conformal map.
Global conformal maps on the Riemann sphere
[edit]
A map of the Riemann sphere onto itself is conformal if and only if it is a Möbius transformation.
The complex conjugate of a Möbius transformation preserves angles, but reverses the orientation. For example, circle inversions.
Conformality with respect to three types of angles
[edit]
In plane geometry there are three types of angles that may be preserved in a conformal map. Each is hosted by its own real algebra, ordinary complex numbers, split-complex numbers, and dual numbers. The conformal maps are described by linear fractional transformations in each case.
In three or more dimensions
[edit]
Riemannian geometry
[edit]
See also: Conformal geometry
In Riemannian geometry, two Riemannian metrics and on a smooth manifold are called conformally equivalent if for some positive function on . The function is called the conformal factor.
A diffeomorphism between two Riemannian manifolds is called a conformal map if the pulled back metric is conformally equivalent to the original one. For example, stereographic projection of a sphere onto the plane augmented with a point at infinity is a conformal map.
One can also define a conformal structure on a smooth manifold, as a class of conformally equivalent Riemannian metrics.
Euclidean space
[edit]
A classical theorem of Joseph Liouville shows that there are far fewer conformal maps in higher dimensions than in two dimensions. Any conformal map from an open subset of Euclidean space into the same Euclidean space of dimension three or greater can be composed from three types of transformations: a homothety, an isometry, and a special conformal transformation. For linear transformations, a conformal map may only be composed of homothety and isometry, and is called a conformal linear transformation.
Applications
[edit]
Applications of conformal mapping exist in aerospace engineering, in biomedical sciences (including brain mapping and genetic mapping), in applied math (for geodesics and in geometry), in earth sciences (including geophysics, geography, and cartography), in engineering, and in electronics.
Cartography
[edit]
Main article: Conformal map projection
In cartography, several named map projections, including the Mercator projection and the stereographic projection are conformal. The preservation of compass directions makes them useful in marine navigation.
Physics and engineering
[edit]
Conformal mappings are invaluable for solving problems in engineering and physics that can be expressed in terms of functions of a complex variable yet exhibit inconvenient geometries. By choosing an appropriate mapping, the analyst can transform the inconvenient geometry into a much more convenient one. For example, one may wish to calculate the electric field, , arising from a point charge located near the corner of two conducting planes separated by a certain angle (where is the complex coordinate of a point in 2-space). This problem per se is quite clumsy to solve in closed form. However, by employing a very simple conformal mapping, the inconvenient angle is mapped to one of precisely radians, meaning that the corner of two planes is transformed to a straight line. In this new domain, the problem (that of calculating the electric field impressed by a point charge located near a conducting wall) is quite easy to solve. The solution is obtained in this domain, , and then mapped back to the original domain by noting that was obtained as a function (viz., the composition of and ) of , whence can be viewed as , which is a function of , the original coordinate basis. Note that this application is not a contradiction to the fact that conformal mappings preserve angles, they do so only for points in the interior of their domain, and not at the boundary. Another example is the application of conformal mapping technique for solving the boundary value problem of liquid sloshing in tanks.
If a function is harmonic (that is, it satisfies Laplace's equation ) over a plane domain (which is two-dimensional), and is transformed via a conformal map to another plane domain, the transformation is also harmonic. For this reason, any function which is defined by a potential can be transformed by a conformal map and still remain governed by a potential. Examples in physics of equations defined by a potential include the electromagnetic field, the gravitational field, and, in fluid dynamics, potential flow, which is an approximation to fluid flow assuming constant density, zero viscosity, and irrotational flow. One example of a fluid dynamic application of a conformal map is the Joukowsky transform that can be used to examine the field of flow around a Joukowsky airfoil.
Conformal maps are also valuable in solving nonlinear partial differential equations in some specific geometries. Such analytic solutions provide a useful check on the accuracy of numerical simulations of the governing equation. For example, in the case of very viscous free-surface flow around a semi-infinite wall, the domain can be mapped to a half-plane in which the solution is one-dimensional and straightforward to calculate.
For discrete systems, Noury and Yang presented a way to convert discrete systems root locus into continuous root locus through a well-known conformal mapping in geometry (aka inversion mapping).
Maxwell's equations
[edit]
Maxwell's equations are preserved by Lorentz transformations which form a group including circular and hyperbolic rotations. The latter are sometimes called Lorentz boosts to distinguish them from circular rotations. All these transformations are conformal since hyperbolic rotations preserve hyperbolic angle, (called rapidity) and the other rotations preserve circular angle. The introduction of translations in the Poincaré group again preserves angles.
A larger group of conformal maps for relating solutions of Maxwell's equations was identified by Ebenezer Cunningham (1908) and Harry Bateman (1910). Their training at Cambridge University had given them facility with the method of image charges and associated methods of images for spheres and inversion. As recounted by Andrew Warwick (2003) Masters of Theory:
: Each four-dimensional solution could be inverted in a four-dimensional hyper-sphere of pseudo-radius in order to produce a new solution.
Warwick highlights this "new theorem of relativity" as a Cambridge response to Einstein, and as founded on exercises using the method of inversion, such as found in James Hopwood Jeans textbook Mathematical Theory of Electricity and Magnetism.
General relativity
[edit]
In general relativity, conformal maps are the simplest and thus most common type of causal transformations. Physically, these describe different universes in which all the same events and interactions are still (causally) possible, but a new additional force is necessary to affect this (that is, replication of all the same trajectories would necessitate departures from geodesic motion because the metric tensor is different). It is often used to try to make models amenable to extension beyond curvature singularities, for example to permit description of the universe even before the Big Bang.
See also
[edit]
Biholomorphic map
Carathéodory's theorem – A conformal map extends continuously to the boundary
Penrose diagram
Schwarz–Christoffel mapping – a conformal transformation of the upper half-plane onto the interior of a simple polygon
Special linear group – transformations that preserve volume (as opposed to angles) and orientation
References
[edit]
^ Blair, David (2000-08-17). Inversion Theory and Conformal Mapping. The Student Mathematical Library. Vol. 9. Providence, Rhode Island: American Mathematical Society. doi:10.1090/stml/009. ISBN 978-0-8218-2636-2. S2CID 118752074.
^ Richard M. Timoney (2004), Riemann mapping theorem from Trinity College Dublin
^ Geometry/Unified Angles at Wikibooks
^ Tsurusaburo Takasu (1941) Gemeinsame Behandlungsweise der elliptischen konformen, hyperbolischen konformen und parabolischen konformen Differentialgeometrie, 2, Proceedings of the Imperial Academy 17(8): 330–8, link from Project Euclid, MR 0014282
^ Selig, Michael S.; Maughmer, Mark D. (1992-05-01). "Multipoint inverse airfoil design method based on conformal mapping". AIAA Journal. 30 (5): 1162–1170. Bibcode:1992AIAAJ..30.1162S. doi:10.2514/3.11046. ISSN 0001-1452.
^ Cortijo, Vanessa; Alonso, Elena R.; Mata, Santiago; Alonso, José L. (2018-01-18). "Conformational Map of Phenolic Acids". The Journal of Physical Chemistry A. 122 (2): 646–651. Bibcode:2018JPCA..122..646C. doi:10.1021/acs.jpca.7b08882. ISSN 1520-5215. PMID 29215883.
^ "Properties of Conformal Mapping".
^ "7.1 GENETIC MAPS COME IN VARIOUS FORMS". www.informatics.jax.org. Retrieved 2022-08-22.
^ Alim, Karen; Armon, Shahaf; Shraiman, Boris I.; Boudaoud, Arezki (2016). "Leaf growth is conformal". Physical Biology. 13 (5) 05LT01. arXiv:1611.07032. Bibcode:2016PhBio..13eLT01A. doi:10.1088/1478-3975/13/5/05lt01. PMID 27597439. S2CID 9351765. Retrieved 2022-08-22.
^ González-Matesanz, F. J.; Malpica, J. A. (2006-11-01). "Quasi-conformal mapping with genetic algorithms applied to coordinate transformations". Computers & Geosciences. 32 (9): 1432–1441. Bibcode:2006CG.....32.1432G. doi:10.1016/j.cageo.2006.01.002. ISSN 0098-3004.
^ Berezovski, Volodymyr; Cherevko, Yevhen; Rýparová, Lenka (August 2019). "Conformal and Geodesic Mappings onto Some Special Spaces". Mathematics. 7 (8): 664. doi:10.3390/math7080664. hdl:11012/188984. ISSN 2227-7390.
^ Gronwall, T. H. (June 1920). "Conformal Mapping of a Family of Real Conics on Another". Proceedings of the National Academy of Sciences. 6 (6): 312–315. Bibcode:1920PNAS....6..312G. doi:10.1073/pnas.6.6.312. ISSN 0027-8424. PMC 1084530. PMID 16576504.
^ "Mapping in a sentence (esp. good sentence like quote, proverb...)". sentencedict.com. Retrieved 2022-08-22.
^ "EAP - Proceedings of the Estonian Academy of Sciences – Publications". Retrieved 2022-08-22.
^ López-Vázquez, Carlos (2012-01-01). "Positional Accuracy Improvement Using Empirical Analytical Functions". Cartography and Geographic Information Science. 39 (3): 133–139. Bibcode:2012CGISc..39..133L. doi:10.1559/15230406393133. ISSN 1523-0406. S2CID 123894885.
^ Calixto, Wesley Pacheco; Alvarenga, Bernardo; da Mota, Jesus Carlos; Brito, Leonardo da Cunha; Wu, Marcel; Alves, Aylton José; Neto, Luciano Martins; Antunes, Carlos F. R. Lemos (2011-02-15). "Electromagnetic Problems Solving by Conformal Mapping: A Mathematical Operator for Optimization". Mathematical Problems in Engineering. 2010 e742039. doi:10.1155/2010/742039. hdl:10316/110197. ISSN 1024-123X.
^ Leonhardt, Ulf (2006-06-23). "Optical Conformal Mapping". Science. 312 (5781): 1777–1780. Bibcode:2006Sci...312.1777L. doi:10.1126/science.1126493. ISSN 0036-8075. PMID 16728596. S2CID 8334444.
^ Singh, Arun K.; Auton, Gregory; Hill, Ernie; Song, Aimin (2018-07-01). "Estimation of intrinsic and extrinsic capacitances of graphene self-switching diode using conformal mapping technique". 2D Materials. 5 (3): 035023. Bibcode:2018TDM.....5c5023S. doi:10.1088/2053-1583/aac133. ISSN 2053-1583. S2CID 117531045.
^ Kolaei, Amir; Rakheja, Subhash; Richard, Marc J. (2014-01-06). "Range of applicability of the linear fluid slosh theory for predicting transient lateral slosh and roll stability of tank vehicles". Journal of Sound and Vibration. 333 (1): 263–282. Bibcode:2014JSV...333..263K. doi:10.1016/j.jsv.2013.09.002.
^ Hinton, Edward; Hogg, Andrew; Huppert, Herbert (2020). "Shallow free-surface Stokes flow around a corner". Philosophical Transactions of the Royal Society A. 378 (2174). Bibcode:2020RSPTA.37890515H. doi:10.1098/rsta.2019.0515. PMC 7287310. PMID 32507085.
^ Noury, Keyvan; Yang, Bingen (2020). "A Pseudo S-plane Mapping of Z-plane Root Locus". ASME 2020 International Mechanical Engineering Congress and Exposition. American Society of Mechanical Engineers. doi:10.1115/IMECE2020-23096. ISBN 978-0-7918-8454-6. S2CID 234582511.
^ Warwick, Andrew (2003). Masters of theory: Cambridge and the rise of mathematical physics. University of Chicago Press. pp. 404–424. ISBN 978-0-226-87375-6.
Further reading
[edit]
Ahlfors, Lars V. (1973), Conformal invariants: topics in geometric function theory, New York: McGraw–Hill Book Co., MR 0357743
Constantin Carathéodory (1932) Conformal Representation, Cambridge Tracts in Mathematics and Physics
Chanson, H. (2009), Applied Hydrodynamics: An Introduction to Ideal and Real Fluid Flows, CRC Press, Taylor & Francis Group, Leiden, The Netherlands, 478 pages, ISBN 978-0-415-49271-3
Churchill, Ruel V. (1974), Complex Variables and Applications, New York: McGraw–Hill Book Co., ISBN 978-0-07-010855-4
E.P. Dolzhenko (2001) , "Conformal mapping", Encyclopedia of Mathematics, EMS Press
Rudin, Walter (1987), Real and complex analysis (3rd ed.), New York: McGraw–Hill Book Co., ISBN 978-0-07-054234-1, MR 0924157
Weisstein, Eric W. "Conformal Mapping". MathWorld.
External links
[edit]
Wikimedia Commons has media related to Conformal mapping.
Interactive visualizations of many conformal maps
Conformal Maps by Michael Trott, Wolfram Demonstrations Project.
Conformal Mapping images of current flow in different geometries without and with magnetic field by Gerhard Brunthaler.
Conformal Transformation: from Circle to Square.
Online Conformal Map Grapher.
Joukowski Transform Interactive WebApp
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190214 | https://www.youtube.com/watch?v=umWiEZ-VpRM | Calculations using Avogadro's number (part 1) | Chemistry | Khan Academy
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Atomic mass units (a.m.u.) are much smaller than grams and are used to measure atomic-scale masses. Avogadro's number (6.022 x 10²³) represents the quantity of particles in one mole, linking atomic masses to measurable amounts in grams. Molar mass, expressed in grams per mole, bridges these concepts, enabling the conversion between the mass of a substance and the amount in moles.
Sections:
00:00 - Introduction
00:25 - What is an Atomic mass unit (amu or u)
04:20 - Avogadro's number and the mole
08:12 - Molar mass
10:28 - Calculating the number of atoms in a sample
11:29 - Calculating the molar mass of CO2
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Introduction I have about 3.21 gram of Sulfur powder over here my question to you is how many atoms of sulfur are there at first this question sounds ridiculous I mean there's going to be lots and lots of atoms how in the world are we going to count that that's what we're going to find out in this video we're going to do that by introducing the idea of mole so let's begin to come up with the idea of moles we first need a new unit of mass What is an Atomic mass unit (amu or u) to deal with the masses of atoms see atoms are very tiny their masses are going to be incredibly tiny so kilograms or gram is going to be very inconvenient to use so we come up with a new unit called the atomic mass unit AMU or U it's a very tiny unit of mass just like grams or kilogram it's a unit of mass but of course whenever we learn a new whenever we learn about a new unit we need to ask ourselves how big is that unit what is the definition of that unit how big is one U well here's how we Define what a u is okay you take a single atom of carbon 12 now its mass by definition is 12 U this is not something that we have measured this is something that we fixed we fixed the mass of a carbon 12 atom to be 12 U exactly okay now what is 1 U well if the mass of a carbon 12 atom is 12 u 1 U is 12th of its mass right so we Define one atomic mass unit one U as one 12th of the mass of a single atom of the carbon 12 isotope does that make sense well I'm sure at this point you may be having some questions like why did we decide to use carbon as a reference and not any other elements well it turns out that we actually started with hydrogen because it's one of the lightest elements then we ran into some problems and then we switched to oxygen because again it's extremely abundant then again we ran into some other problems and then finally we decided to go with carbon which is also abundant we'll not Del into the histories and details of what really happened but yeah we have to choose some element as a reference and we ended up choosing carbon as a reference another question you could be having is why do we fix the mass of a single atom of this carbon 12 to be 12 U why not any other number why 12 well for that you can see that over here carbon has how many protons and neutrons in it well it has a total of I mean it has six protons and six neutrons so it has a total of 12 protons and neutrons 12 particles I think of protons and neutrons together over here because they have pretty much similar mass I mean a neutron is actually slightly heavier than a proton but for our purposes to get an intuition over here their masses we can pretty much think of them to be almost equal to each other so it has a total of 12 particles right now by fixing the mass of those 12 particles to be 12 U look at what we're doing we are basically saying hey let's fix the mass of a single proton or a NE neon to be about one U that was the whole intention okay so you can also think one U is kind of a representation of a mass of a single proton or a neutron but again this is not exact because masses of protons and neutrons are not exactly equal to each other so a proton and neutron will have a mass very close to one U but it's not exactly Vu but it's a good way to think about what a what a u represents it represents sort of the mass of a proton or Neutron anyways now that we understand this here's a question what do you think is the mass of the a single atom of Oxygen 16 isotope a single atom of this what would be its mass in U atomic mass unit well it has a total of 16 particles 16 protons and neutrons together and since each particle each Proton or Neutron has a mass of one U and there are total 16 oxygen mass will be about 16 U again you can see it's not going to be exactly 16 U because mass of each Proton or Neutron is not exactly one U but it's going to be very close to that similarly if you take an isotope of say chlorine a particular isotope the most abundant isotope of Florine which has 35 protons and neutrons together in it well then its mass is will be close to 35 U makes sense right okay now here's a Avogadro's number and the mole question we're going to ask ourselves let's go back to carbon each carbon has a mass of 12 U right by definition now how many carbon atoms do I need to take take together such that the total mass of all of those carbon atoms together becomes 12 G you can imagine it's going to be lots and lots of atoms right because each atom has a very tiny mass and we want together 12 G so we probably need to take billions and billions and billions of atoms but the big question is how many atoms do I need to take so that they all add up to give me 12 G of mass Well turns out we figured it out again we'll not get into the details of how we figured it out okay the history is actually pretty interesting but again we will not talk about that over here but we figured it out and it turns out to be this number you need to take about 6.022 and there are some other decimals over here some numbers here times 10^ 23 which is a huge number okay if you take these many carbon atoms together carbon 12 atoms together they will together have a mass of 12 G this number is what we call the avagadro number named after the scientist AO avagadro who worked a lot on this idea but anyways you can now see the significance of this number I can now count the number of atoms in a carbon isotope if you give me 12 gram of carbon I know it has these many number of carbon atoms in it carbon 12 okay these many number of carbon 12 atoms in it if you give me 24 G of carbon it must there must be twice the amount if you give me 6 G of car carbon then there must be half the amount you tell me the mass of the carbon 12 isotope that I'm holding in my hand and I can now use this number to tell you how many atoms there are beautiful isn't it in other words this becomes the conversion factor for our tiny unit of mass from our tiny unit of mass U to our more familiar big unit of mass grams if you take U and you multiply with this number you get grams and whenever you have an Ardo number of things with you we call it a mole just like how when you have 12 things with you we call it a dozen these many things if you have together it could be things of it could be anything it could be these many atoms then we'll call it a mole of atoms or it could be these many babies then we'll say we have a mole of babies it's a ridiculous number but you get the idea and this word mole actually comes from the Latin molecule which translates to a very tiny amount of something but anyways what is a mole a mole represents aard R number these many number of things it could be number of atoms molecules particles anything and what's so special about this number it's a conversion factor from the tiny unit of mass U to grams you take this number multiplied by this number and you will now get it in grams okay now let's see if you understand this what do you think would be the mass of one mole of Oxygen 16 atoms if I had an avagard number of oxygen 16 atoms together what do you think collectively would its mass be well and a a guardo number of two values will give me a mass of 12 G so an avocado number of 16 use will give me a mass of 16 G that's what we mean by a conversion factor okay it works for any atom which has any Mass you just multiplied by this and now you will get the mass in grams similarly if I had an avagard number of chlorine 35 if I had one mole of chlorine 35 atoms with me then it will have 35 G of mass make sense and so another way to talk about all of these things whatever I just said now another way to talk about this is we Molar mass say the marolar mass of carbon 12 is 12 g carbon 12 has a mass of 12 G per mole makes sense right we would say Oxygen 16 will have 16 G per mole I mention Oxygen 16 because remember there are other Isotopes as well different isotopes will have different masses so their molar mass would be different so Oxygen 16 isotope has a molar mass of 16 G per mole and chlorine 35 has a molar mass of 35 G per mole okay same thing whatever I just said it's technical way of stating the same thing over here all right the last thing we need to do to make this more practical is to remember that over here we considered pure cases I took a pure carbon 12 Isotopes where every single atom was carbon 12 or I took a pure chlorine isotope where every single atom was chlorine 35 but in reality that's not the case if I take a chunk of chlorine a lot of it will be chlorine 35 but there will be some other Isotopes as well like another abundant isotope next to chlorine 35 is chlorine 37 and that sounds really complicated but what's important and Powerful is that that doesn't matter to us this whole idea still works okay here's what I mean let me take an example if we look at our if you look at our periodic table then you can see that the atomic mass of chlorine is given to be not 35 it's 35.45 so significant deviation from 35 why because this also accounts for the fact that if you take a chunk of chlorine it'll also contain a lot of chlorine 37 in it so what we do is sort of like take an average this is an vited average we say so this is the average atomic mass of chlorine so since I know the average atomic mass of chlorine is 35.45 if I now take one mole of chlorine not pure as it exists as a mixture in nature then it's one mole will have a mass of 35.45 G that's it similarly if I take 1 mole of carbon which you know it's not exactly 12 G because there are other Isotopes it'll be 12.01 G you see what I mean a mole is a conversion factor take one mole of anything it'll be this number in Gs and Calculating the number of atoms in a sample so now now we can try and answer original question we asked ourselves if you have 3.21 G of sulfur how many atoms there are why don't you pause the video and see if you can now answer this question yourself if I take one mole of sulfur if I take a guardo number of sulfur atoms it will have a mass of 32.1 G roughly 32.1 G so 32.1 G represents 1 mole of sulfur but how much sulfur do I have I have not 32.1 I have 3.21 G which is just oneth of a mole that's why I took 3.21 to just keep the calculation simpler we can do it in our head this is 1/10th of a mole so how many atoms we must be having one10 of a mo so one10 of the avagard number so the answer would be the aardra number which is 6.022 10 ^ 23 divide by 10 1/10 of it so it'll be 6.022 10 ^ 22 okay here's our final question if I take one mole of carbon dioxide what do you think will its mass be what is the Calculating the molar mass of CO2 molar mass of one mole of carbon dioxide can you pause the video and try to think about this okay let's do this step by step I know if I have one mole of carbon dioxide then it must be having an avagadro number of molecules of carbon dioxide right remember if I had half a mole of carbon dioxide it means that I would have half the aagard or number of carbon dioxide Mak sense right okay anyways now comes the question how many carbon atoms must be there there and how many oxygen atoms must be there what do you think well a single carbon dioxide molecule has one atom of carbon if I have five molecules of carbon dioxide I have five carbon atoms which means if I have these many molecules of carbon dioxide I should have exactly that many amount of carbon atoms meaning I have one mole of carbon atoms with me okay what about the number of oxygen atoms well each carbon dioxide molecule has two atoms of carbon oxygen and so if I had five for example molecules of carbon dioxide I would have twice the amount 10 atoms of oxygen and therefore if I have these many molecules of carbon dioxide I would have twice the amount okay so which means I have 2 moles of oxygen atoms with me and I can now look at the periodic table to find the mass of one mole of carbon it's 12.01 07 G and for oxygen it would be 15.9994 G that will be the mass of 1 mole of oxygen but then we have to multiply it by two because over here we have 2 moles simplifying this will give me the molar mass of carbon dioxide so one mole of carbon dioxide will have this much mass or we can also say that carbon dioxide has a mass of 44.95 G per mole same thing it's the same thing okay they all mean the same thing of course we can round it off and we are actually doing a numerical |
190215 | https://www.ck12.org/assessment/ui/?test/detail/practice/chemistry/law-of-multiple-proportions-practice&collectionHandle=chemistry&collectionCreatorID=3&conceptCollectionHandle=chemistry-::-law-of-multiple-proportions&isPageView=true | Law of Multiple Proportions | Practice | CK-12 Foundation
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190216 | https://www.chegg.com/homework-help/questions-and-answers/portion-plane-x-y-z-1-lies-first-octant-oriented-counter-clockwise-viewed--wish-compute-wo-q123629868 | Solved portion of the plane x+y+z=1 that lies in the first | Chegg.com
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portion of the plane x+y+z=1 that lies in the first octant, oriented counter-clockwise when viewed from above. We wish to compute the work. done by the vector fleld F=(x−y2,x4,z2) on an object moving along the path c. This problem walks you through the computation using Stokes' Theorem. a.) Compute the unit normal vector n~ to the surface x+y+z=1. Make sure
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Question: portion of the plane x+y+z=1 that lies in the first octant, oriented counter-clockwise when viewed from above. We wish to compute the work. done by the vector fleld F=(x−y2,x4,z2) on an object moving along the path c. This problem walks you through the computation using Stokes' Theorem. a.) Compute the unit normal vector n~ to the surface x+y+z=1. Make sure
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portion of the plane x+y+z=1 that lies in the first octant, oriented counter-clockwise when viewed from above. We wish to compute the work. done by the vector fleld F=(x−y 2,x 4,z 2) on an object moving along the path c. This problem walks you through the computation using Stokes' Theorem. a.) Compute the unit normal vector n~ to the surface x+y+z=1. Make sure you make it a unit vector and that the z-component is positive so that the normal points upward. n=(F)ते०(3,51,31)∥∇N=mationon 3So 2x=1 n˜(−31,−51,−11) b.) Compute the curl of F~=(x−y 2,x 4,z 2) and find (∇×F~)⋅n~. (v∼)n oxt⇒(0−(4 x y′,0−2 x 3,4 x)(0 x+)an−(−4 x y,−2 x 3,−4 x)×(−31,−31,−31)=34 x 2 y−2 x 3 c.) Compute the Jacoblan 1+f x 2+f y of the surface x+y+z=1. Hint: Rewrite os z=1−x−y. #3 continued... d.) Sketch the shadow region R in the x y-plane formed when we restrict x+y+z=1 to the z=0 plane. e.) Find ∫CF⋅T d s by computing the surface integral ∬Q(∇×F)⋅n d S.
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190217 | https://study.com/learn/lesson/square-root-property.html | Square Root Property | Overview, Formula & Examples - Lesson | Study.com
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Math Courses / SAT Mathematics Level 2: Help and Review Course
Square Root Property | Overview, Formula & Examples
Lesson
Transcript
Christian Killian, Laura Pennington
Author Christian Killian
Christian has a bachelor's degree in business administration, a master's degree in media communication and psychology, and credits toward a PhD in social psychology. They excel at math and science and enjoy explaining things to others. They are also OSHA 30 certified.
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Instructor Laura Pennington
Laura received her Master's degree in Pure Mathematics from Michigan State University, and her Bachelor's degree in Mathematics from Grand Valley State University. She has 20 years of experience teaching collegiate mathematics at various institutions.
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Learn about the Square Root Property, its formula, and how to use it to solve quadratics. Updated: 11/21/2023
Table of Contents
What is the Square Root Property?
Square Root Formula
Properties of Roots
Solving Quadratic Equations Using the Square Root Property
Solving Area Problems Using the Square Root Property
Lesson Summary
Show
Frequently Asked Questions
What is the square root formula?
The Square Root Formula is x = +/- sqrt c, which is simplified from x^2 = c. Both positive and negative numbers have the same answer when multiplied by themselves, so all quadratic values with exponent 2 will have two possible answers.
What are the properties of square roots?
There are three properties of square roots. They are the Multiplication Property of Roots, the Square Root Property of Equality, and the Quotient Property of Roots.
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Table of Contents
What is the Square Root Property?
Square Root Formula
Properties of Roots
Solving Quadratic Equations Using the Square Root Property
Solving Area Problems Using the Square Root Property
Lesson Summary
Show
What is the Square Root Property? ---------------------------------
The Square Root Property is used to calculate the number that, when multiplied by itself, equals a sought-after variable. The symbol used for square roots is x, where x is any number that is the product of two identical numbers. 4 is the product of 2 and 2, or 2 2. 32, while not a perfect square (called an imperfect square), is the approximate product of 5.66 x 5.66, or 5.66 2.
The Square Root Property states that if x has exponent of 2, then we can solve for it by taking the square root of both sides and adding ± to the solution.
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0:03 Square Root Property
1:44 Using the Square Root Property
4:22 Another Example
6:10 Lesson Summary
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Square Root Formula -------------------
The Square Root Formula is x 2 = c, which when solved for x becomes x = ±c. This formula represents the Square Root Property by solving as accurately for x as is possible without another perfect square on the other side.
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Properties of Roots -------------------
There are twelve main properties of mathematical roots depending what is being done with them.
If there is more than one number being multiplied with square roots, we can combine them together under one square root, connected by the multiplication symbol. a x b = a⋅b
When dividing numbers that both have square roots, we can combine them similarly to #1. c / d = c/d
If there are two of the same number inside the radicand(the parentheses inside the square root), they can be reduced down to one of that number. (e⋅e) = e
Another way to write the square root of a number is using the exponent 1/2. f = f 1 2
If there are two or more numbers where the radicands match, addition and subtraction can be performed. g h + i h = (g+i)h
Moving the square root to the other side of the equal sign makes it a square integer. j = k, where j x j = k
Likewise, moving a square integer to the other side of the equal side will make it a square root. L = m, where m x m = L
The numbers 2, 3, 7, and 8 in the ones place of a number cannot be a perfect square.
Numbers with an odd number of zeros at the end will also not be a perfect square.
Rational numbers, or numbers that are whole integers, are the square roots of a perfect square. 81 = 9
Odd numbers result in odd squares and even numbers result in even squares. 144 = 12
Negative numbers that are square rooted result in imaginary numbers. −16 = 4 i, where i is the indicator of an imaginary number
Multiplication Property of Roots
The Multiplication Property of Roots states that if there are two or more numbers with matching radicands, the whole numbers can be multiplied together to simplify the final product. This is the exact same as #5 in the last section, just with multiplication. Let's look at an example.
Simplify 5 2 x 8 2.
Since the radicands are both 2, we can just multiply 5 x 8 and get out answer: 40 2
Square Root: Property of Equality
Properties of Equality are used to isolate variables on either side of the equal sign by performing the same operation to both sides. With square roots, both sides of the equation are raised to the second power for variables involving square roots or square rooted if the variable has an exponent of 2.
Simplify x=3.
The first thing we want to do is isolate x on the left side of the equation. To do this, we raise both sides to the second power, or 2. This gives us (x)2 = 3 2.
X cancels itself out and we can calculate the right side of the equation to get the answer: x = 9.
Simplify x 2 = 49.
For this problem, we will want to square root both sides to isolate x. x 2 = 49
Since the square root of 49 is 7, our answer is x = 7.
Quotient Property of Roots
The Quotient Property or Roots states that the square root of a fraction is equal to the square root of the numerator over the square root of the denominator. (x/y) = x y
Solve 25/16
Since the numerator and the denominator are both perfect squares, we know that the answer will not contain any square roots. Using the quotient property of roots, we can rewrite the fraction with two square roots instead of one. 25 / 16.
From here, we can solve the square roots and end up with the answer 5/4, or 1.25.
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Solving Quadratic Equations Using the Square Root Property ----------------------------------------------------------
The Square Root Property is used in solving quadratic equations by eliminating the square exponents to isolate the variable being solved. The formula x=±c gives us two possible answers, the positive and negative of the number that can be multiplied by itself to equal x.
Solve x 2 = 17.
In order to isolate x, we have to square root both sides of the equation. x 2 = 17
Since 17 is not a perfect square, we will leave it in its square root form. It may look like this is the answer, but we must not forget the last crucial step: adding the +/- before it. Thus, x = ±17.
Solve for x: x 2 - 27 = 0
First, let's combine the whole numbers on the right side of the equation. 0 minus negative 27 is the same as 0 + 27, so our new equation is x 2 = 23. Then we can square root both sides just like in the last example.
x 2 = 23
x = ±23
Solve for x: (x - 5)2 = 36.
Since 36 is a perfect square, we know we will have no square roots in our solution. The parentheses add a step to our calculation, but it is simply the reverse of what we did in the last example.
(x−5)2 = 36
x - 5 = ±6
x = 5 ± 6
x = 11 or x = -1
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Solving Area Problems Using the Square Root Property ----------------------------------------------------
The Square Root Property can also be used to calculate the dimensions of a geometric figure whose area is a perfect square, such as a square and a circle.
Solve for the side length of a square with an area of 16 square units.
The length of the sides can be calculated from the area
Using the formula x=±c, we can plug 16 in for c and solve for x.
The square root of 16 is 4, so our answer is ±4. Since dimensions cannot be negative, the side lengths on all four sides of this square are 4 units.
Solve for the radius of a circle with an area of 39 units squared.
The Square Root Property applies to circles too
Recall from geometry that the formula for an area of a circle is A = π r 2. If our A is 39, then to solve for r we need to divide both sides by π, then use the Square Root Property to isolate the r from its exponent. Using 3.14 for π, we end up with the following solution.
39 = 3.14 x r 2
39 / 3.14 = r 2
12.42 = r 2
r = 3.52 units
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Lesson Summary --------------
In this lesson we covered the Square Root Property and its many uses. The Square Root Property solves for x when x is either a square root or an exponent, multiplying both sides by the inverse to isolate x as the only variable on one side of an equation. The Square Root Formula is x = ±c, which is derived from x 2 = c. C will either be a perfect square, or a number that is the product of two of the same number, or an imperfect square, in which case the square root will remain in the answer. With imperfect squares, only an approximate answer can be calculated by performing the final calculation, so this is considered its simplified form.
There are twelve properties of roots, three of which we looked at in-depth in this lesson. The Multiplication Property of Roots states that if two or more numbers contain the same number inside the square root (radicand), then the whole number portions can be multiplied together. The Square Root Property of Equality says that we can perform the same operation on both sides of an equation to simplify or solve for a particular variable. Finally, the Quotient Property of Roots copies the square root onto both the numerator and the denominator when they are originally part of the same radicand.
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Video Transcript
Square Root Property
Have you ever moved into an unfurnished apartment or home? If so, then you probably know that one of the best parts of moving is decorating your new living area! Imagine that you've just moved into a new home that has a large square dining room with hardwood floors. You decide that you want to get a nice area rug for this room.
You take the measurements of the room and find that you will want a square area rug with sides 12 feet in length to cover the floor adequately.
You go online to look at area rugs and find a few that you really like. However, the rugs are listed by area, not by dimensions, so you're not sure which ones will fit the space appropriately. You narrow it down to three different rugs with the areas of 100 square feet, 121 square feet, and 144 square feet.
You know that to find the area of a square rug, you would use the formula A = s 2, where A is the area of the square, and s is the length of a side of the square. You realize that if you plug each of the areas into this formula for A, you can solve for s. If you get 12 for s, then you know it will fit the space perfectly! In other words, we want to solve three different equations.
s 2 = 100
s 2 = 121
s 2 = 144
Okay, we know what you need to do. Now, we just need to figure out how to do it! It just so happens that there is a property we can use to solve these specific types of equations, and that property is called the square root property.
The square root property can be used to solve certain quadratic equations, and it states that if x 2 = c, then x = √c or x = -√c, where c is a number.
Using the Square Root Property
In words, the square root property states that if we have an equation with a perfect square on one side and a number on the other side, then we can take the square root of both sides and add a plus or minus sign to the side with the number and solve the equation.
Let's use this to see which of the area rugs you've found will fit your new dining room! First, let's consider the rug with area 100 ft 2. We plug A = 100 into the area formula and use the square root property to solve for s.
We write:
A = s 2
We plug in 100 for A. Now our equation reads:
100 = s 2
We use the square root property and have two equations:
s = √100, or
s = - √100
When we simplify, we have:
s = 10 or s = -10
Since we're talking about the length, our answer will be a positive number. So, our answer is:
s = 10 feet
We see that this rug with area 100 ft.2 has sides of length 10 feet. This rug is too small! Let's look at the rug with area 121 ft 2. We plug A = 121 into our formula and solve for s again.
We write:
A = s 2
We plug in 121 for A. Now our equation reads:
121 = s 2
We use the square root property and have two equations:
s = √121, or
s = - √121
When we simplify, we have:
s = 11 or s = -11
Since we're talking about the length, our answer will be a positive number. So, our answer is:
s = 11 feet
We see that this rug with area 121 ft 2 has sides of length 11 feet. This one is also too small. Let's hope the third one will be a good fit! To check the third one, we plug A = 144 into the formula and solve for s.
We write:
A = s 2
We plug in 144 for A. Now our equation reads:
144 = s 2
We use the square root property and have two equations:
s = √144, or
s = - √144
When we simplify, we have:
s = 12 or s = -12
Since we're talking about the length, our answer will be a positive number. So, our answer is:
s = 12 feet
We get that this rug with area 144 ft 2 has side length 12 ft. Awesome! It's a perfect fit! That sure makes the decision of which rug to get a whole lot easier!
Another Example
Let's take note that the square root property can be used to solve quadratic equations in which we can isolate a perfect square on one side of the equation and a number on the other side of the equation. That is, we can use this property to solve quadratic equations that can be put in one of two forms:
ax 2 = c
or
(ax + b)2 = c
For example, suppose you want to hang a large decorative clock that is in the shape of a circle in your new place, and you need to know the distance from the center of the clock to the edge of the clock to know if it will fit on your wall the way you want it to. In other words, you need to know the radius of the clock.
You know the area of the clock is 1,018 square feet, and the formula for the area of a circle is A = π r 2, where A is the area of the circle, and r is the radius of the circle. You can plug the area in for A in the formula, rearrange the equation to get it in a form that we can use the square root property to solve, and then solve for r.
We first write:
1018 = π r 2
We divide both sides by π, so we now have:
1018 / π = r 2
We simplify 1018 / π ≈ 324
Now, our equation reads:
324 = r 2
We interchange sides, and it now reads:
r 2 = 324
We use the square root property and have the two equations:
r = √324 or r = -√324
Simplifying these equations gives us:
r = 18 or r = -18
Since this is a distance we're talking about, we can discard the negative answer, and we see that the radius of the clock is approximately 18 inches long.
Lesson Summary
The square root property is a property that can be used to solve quadratic equations. It states that if x 2 = c, then x = √c or x = -√c, where c is a number. This property can be used to solve an equation that can be put into one of the following two forms:
ax 2 = c
or
(a x + b)2 = c
As we've seen, the square root property can come in very handy in real life applications. Thanks to this property, you've found a perfectly fitting rug for your dining room!
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SAT Mathematics Level 2: Help and Review
22 chapters | 225 lessons
Ch 1. Basic Math Review: Help and...
Ch 2. Square Roots: Help and Review
Evaluating Square Roots of Perfect Squares 5:12
Estimating Square Roots | Overview & Examples 5:10
Simplifying Square Roots When not a Perfect Square 4:45
Simplifying Square Root Expressions | Steps & Examples 7:03
Discriminant Formula, Rules & Solutions 4:12
Principal Square Root | Definition, Calculation & Examples 5:02
Square Root | Definition, Sign & Problems 10:15
Square Root Property | Overview, Formula & Examples 6:49
Next Lesson Square Root of a Number | Adding, Multiplying & Simplifying
Subtracting Square Roots 3:44
Ch 3. Basic Algebraic Expressions in Math:...
Ch 4. Exponents in Math: Help and...
Ch 5. Linear Equations & Inequalities in...
Ch 6. Absolute Value Equations &...
Ch 7. Polynomials in Algebra: Help and...
Ch 8. Functions in Math: Help and...
Ch 9. Rational Expressions in Algebra: Help...
Ch 10. Matrices: Help and Review
Ch 11. Sequences and Series in Math: Help and...
Ch 12. Exponentials and Logarithms in Math:...
Ch 13. Complex Numbers in Math: Help and...
Ch 14. Coordinate Geometry Basics: Help and...
Ch 15. Graphing Linear Equations &...
Ch 16. Advanced Coordinate Geometry: Help and...
Ch 17. Three-Dimensional Geometry: Help and...
Ch 18. Overview of Trigonometry: Help and...
Ch 19. Probability, Combinations &...
Ch 20. Working with Data & Statistics: Help...
Ch 21. Logic: Help and Review
Ch 22. Limits in Math: Help and Review
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190218 | https://math.stackexchange.com/questions/1050184/difference-between-poisson-and-binomial-distributions | statistics - Difference between Poisson and Binomial distributions. - Mathematics Stack Exchange
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Difference between Poisson and Binomial distributions. [closed]
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If both the Poisson and Binomial distribution are discrete, then why do we need two different distributions?
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asked Dec 3, 2014 at 17:23
Bareerah AnsariBareerah Ansari
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3 There are many more than 2 discrete distributions. Poisson counts the number of occurrences in an interval given a certain average occurence rate per interval. It can have values 0, 1, 2, 3, ... Binomial counts the number of occurrences out of a fixed number N of possibilities, where any one occurrence happens with probability p. It can take values 0, 1, 2...N. Quite different situations.Paul –Paul 2014-12-03 17:28:26 +00:00 Commented Dec 3, 2014 at 17:28
3 If both apples and oranges are fruits, why do we need both?user147263 –user147263 2014-12-03 20:14:24 +00:00 Commented Dec 3, 2014 at 20:14
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The Binomial and Poisson distributions are similar, but they are different. Also, the fact that they are both discrete does not mean that they are the same. The Geometric distribution and one form of the Uniform distribution are also discrete, but they are very different from both the Binomial and Poisson distributions.
The difference between the two is that while both measure the number of certain random events (or "successes") within a certain frame, the Binomial is based on discrete events, while the Poisson is based on continuous events. That is, with a binomial distribution you have a certain number, n n, of "attempts," each of which has probability of success p p. With a Poisson distribution, you essentially have infinite attempts, with infinitesimal chance of success. That is, given a Binomial distribution with some n,p n,p, if you let n→∞n→∞ and p→0 p→0 in such a way that n p→λ n p→λ, then that distribution approaches a Poisson distribution with parameter λ λ.
Because of this limiting effect, Poisson distributions are used to model occurences of events that could happen a very large number of times, but happen rarely. That is, they are used in situations that would be more properly represented by a Binomial distribution with a very large n n and small p p, especially when the exact values of n n and p p are unknown. (Historically, the number of wrongful criminal convictions in a country)
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answered Dec 3, 2014 at 17:56
KSmartsKSmarts
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2 Is the Poisson Distribution (en.wikipedia.org/wiki/Poisson_distribution) different from Poisson Binomial Distribution (en.wikipedia.org/wiki/Poisson_binomial_distribution)? If yes, how?Minu –Minu 2016-10-02 16:49:36 +00:00 Commented Oct 2, 2016 at 16:49
1 A Bernoulli trial is an experiment with two outcomes with fixed probabilities p p and 1−p 1−p. The Binomial distribution provides the probability of getting some number of successes amongst a number of Bernoulli trials that have the same p p value. The Poisson Binomial distribution, on the other hand, allows for different values of p p for each of the individual Bernoulli trials. The Poisson distribution is something different; in this context, it is primarily relevant as a limiting case of the Binomial distribution.Vyas –Vyas 2017-06-12 22:32:20 +00:00 Commented Jun 12, 2017 at 22:32
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The binomial distribution counts discrete occurrences among discrete trials.
The poisson distribution counts discrete occurrences among a continuous domain.
Ideally speaking, the poisson should only be used when success could occur at any point in a domain. Such as, for example, cars on a road over a period of time, or random knots in a string over a length, etc. We are talking about infinitely many infinitesimally small trials, each having at most one success.
In practice, though, the poisson can be used to approximate the binomial under certain conditions, but it is only a rough approximation. Such as using the Normal curve in place of a Binomial under the right conditions.
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edited Dec 6, 2014 at 23:23
answered Dec 3, 2014 at 17:59
CogitoErgoCogitoSumCogitoErgoCogitoSum
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Why is this down voted?Gabriel Fair –Gabriel Fair 2017-09-12 22:21:25 +00:00 Commented Sep 12, 2017 at 22:21
4 Why do any of my answers get down-voted? Happens all the time.CogitoErgoCogitoSum –CogitoErgoCogitoSum 2017-11-23 21:42:10 +00:00 Commented Nov 23, 2017 at 21:42
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190219 | https://www.vocabulary.com/dictionary/embezzle | SKIP TO CONTENT
/ɪmËbÉzÉl/
IPA guide
Other forms: embezzled; embezzling; embezzles
When a person embezzles, it usually means that he is stealing money from his employer. If he is caught embezzling, it probably also means that he will soon be unemployed.
The word embezzle implies more than simply "to steal." When a person embezzles, he or she takes advantage of an employer's trust for personal gain. Embezzling is a so-called "white-collar crime" which often involves some sort of cover-up, like falsifying financial records or stealing small amounts of money over a long period of time. The word embezzle comes from an Old French word meaning "maltreat or ravage," besillier, and an embezzler can be said to ravage someone else's money.
Definitions of embezzle
verb
appropriate (as property entrusted to one's care) fraudulently to one's own use
“The accountant embezzled thousands of dollars while working for the wealthy family”
synonyms: defalcate, malversate, misappropriate, peculate
see moresee less
types:
fiddle
commit fraud and steal from one's employer
type of:
steal
take without the owner's consent
Cite this entry
Style:
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DISCLAIMER: These example sentences appear in various news sources and books to reflect the usage of the word embezzle'. Views expressed in the examples do not represent the opinion of Vocabulary.com or its editors. Send us feedback
Word Family
Vocabulary lists containing embezzle
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190220 | https://www.ibchem.com/IB25/s3.23.php | Colourful Solutions - IB Chemistry
Colourful Solutions > Functional groups: Classification of organic compounds > Homologous series
Standard level
The sheer scale of organic chemistry means that it is important to categorise
compounds with similarities in order to make sense of the vast number
of organic compounds possible. Categorisation relies on similarities between
and within structures, giving rise to the concept of homologous series.
Structure 3.2.3 - A homologous series is a family of compounds in which successive members differ by a common structural unit, typically CH2.
Guidance
Tools and links
Definition
An homologous series is a group of compounds with similar structural features
that can be represented by a general formula whose members differ by one -CH2-
unit.
You can think of an homologous series as a 'family' or group of compounds,
whose chemical behaviour is very similar and whose physical properties show
trends when ascending the members of the group (in terms of number of carbon
atoms).
Each homologous series has a class name that must be distinguished
from the name of the functional group that it contains.
For example: The homologous series of alcohols (class) contains the hydroxyl
functional group.
Note: The IBO no longer requires use of the term "class" for the 2025 exams. The homologous series represents the family of organic compounds.
| |
| |
^ top
General formula
A general formula is one in which the numbers of carbons are represented by
the letter 'n' and the numbers of other elements as a function of 'n'
So if there are two carbons and four hydrogens, n is equal to 2 (the number
of carbons) and the value for the hydrogens = 2n. The general formula becomes
CnH2n.
General
Formula data
| Series | General formula |
--- |
| Alkanes | CnH(2n+2) |
| Alkenes | CnH2n |
| Alkynes | CnH(2n-2) |
| Alcohols | CnH(2n+1)OH |
| Halogenoalkanes | CnH(2n+1)X |
| Aldehydes | CnH(2n+1)CHO |
| Ketones | CnH(2n+2)CO |
| Carboxylic acids | CnH(2n+1)COOH |
| Amines | CnH(2n+1)NH2 |
| Amides | CnH(2n+1)CONH2 |
| Esters | CnH(2n+1)COOCnH(2n+1) |
| Ethers | CnH(2n+1)OCnH(2n+1) |
| Nitriles | CnH(2n+1)CN |
| Series | General formula |
--- |
| Alkanes | CnH(2n+2) |
| Alkenes | CnH2n |
| Alkynes | CnH(2n-2) |
| Alcohols | CnH(2n+1)OH |
| Halogenoalkanes | CnH(2n+1)X |
| Aldehydes | CnH(2n+1)CHO |
| Ketones | CnH(2n+2)CO |
| Carboxylic acids | CnH(2n+1)COOH |
| Amines | CnH(2n+1)NH2 |
| Amides | CnH(2n+1)CONH2 |
| Esters | CnH(2n+1)COOCnH(2n+1) |
| Ethers | CnH(2n+1)OCnH(2n+1) |
| Nitriles | CnH(2n+1)CN |
Note: Learning specific general formulae is not required, just be aware that they exist and that all homologous series can be represented in this way.
^ top
Trends in properties
The chemical properties of the members of an homologous series are due to the
presence (or lack of) functional groups. As all members of the same series have
the same functional groups, then the chemical properties are similar. They are
not identical because chemical properties are also modified by other factors,
such as the position of the functional group, or branching in the carbon chain.
A case in point is the reactivity of the halogenoalkanes. This is known to
depend on the position of the halogen atom on the carbon chain, as well as
the shape of the carbon chain itself.
The physical properties of an homologous series show a trend as the series
is ascended. This is due to the increased relative mass, as each member
of the series differs from the previous member by one -CH2- unit.
Physical properties include:
Both density and boiling point depend on the intermolecular forces in the compound.
These may be due to London dispersion forces only, or a combination of London dispersion force and dipole-dipole attractive forces. In either case, the London dispersion force increases as the relative molecular mass increases, causing a corresponding increase in density and boiling point.
Boiling points of the
alkanes
Solubility in water depends on the ability of the water molecules to attract
dipoles in the organic compound. Functional groups may contain dipoles that
make the organic molecule soluble to a lesser, or greater extent. These dipoles
are said to be hydrophilic (water loving). However, the carbon chain itself
has no dipoles and is not attractive to water. It is said to be hydrophobic
(water hating).
The difference between each successive member of an homologous series is one -CH2- unit. The CH2 group is non-polar (hydrophobic) and this increases
the percentage of the molecule that is unattractive to water.
Organic compounds consequently tend to be less soluble as an homologous series is ascended.
^ top
Quick
check time
| Quick check 1 - Homologous series. Enter your answer in the space provided. | |
--- |
| | |
| | | | --- | | | | | Tried | Correct | | | | | --- | | | | |
| | |
--- |
| | |
| Tried | Correct |
| | |
--- |
| | |
Worked examples
| |
| Homologous series have the same degree of saturation and the same functional groups. They differ only by one CH2 group. The molecule 1-chloropropane has one chlorine atom and no unsaturation, so 1-chlorobutane is the same homologous series. |
Homologous series have the same degree of saturation and the same functional
groups. They differ only by one CH2 group.
The molecule 1-chloropropane has one chlorine atom and no unsaturation,
so 1-chlorobutane is the same homologous
series.
Q1013-02 Which homologous series
has ethanol for a member?
| |
| Ethanol ends with the letters -ol and therefore is a member of the alcohols |
Ethanol ends with the letters -ol and therefore is a member of the
alcohols
| |
| The ending associated with carboxylic acids is the -COOH group, hence the molecule CH3COOH is an acid |
The ending associated with carboxylic acids is the -COOH group, hence
the molecule CH3COOH is
an acid
| |
| Homologous series differ by one CH2 unit and have all of the same functional groups. The molecule CH3CH2CH3 belongs to the same homologous series - the alkanes. |
Homologous series differ by one CH2 unit and have all of
the same functional groups.
The molecule CH3CH2CH3
belongs to the same homologous series - the alkanes.
| |
| The ketones (alkanones) must have carbon chains either side of the carbonyl group. |
The ketones (alkanones) must have
carbon chains either side of the carbonyl group.
| |
| Aldehydes contain the -CHO terminal group, CH3CH2CHO. |
Aldehydes contain the -CHO terminal group, CH3CH2CHO.
| |
| Amides contain the -CONH2, or -CONH- groups. Therefore CH3CONH2. |
Amides contain the -CONH2, or -CONH- groups. Therefore CH3CONH2.
| |
| The alkyne series contains CC bonds and the nitrile series has CN bonds. Therefore the answer is alkyne and nitrile. |
The alkyne series contains CC
bonds and the nitrile series has CN
bonds. Therefore the answer is alkyne and
nitrile.
| |
| Neighbouring members of an homologous series differ by one CH2 group. |
Neighbouring members of an homologous series differ
by one CH2 group.
| |
| The empirical formula gives the simplest ratio of atoms within a molecular fomula. There must be an integral number of empirical formula units in a molecular formula. The relative mass of this empirical formula = (2 x 12) + (4 x 1) + (1 x 16) = 44. If the relative formula mass = 88 there are 88/44 empirical formula units in a molecule. The molecular formula is C4H8O2 |
The empirical formula gives the simplest ratio of atoms within a molecular
fomula.
There must be an integral number of empirical formula units in a molecular
formula.
The relative mass of this empirical formula = (2 x 12) + (4 x 1) +
(1 x 16) = 44.
If the relative formula mass = 88 there are 88/44 empirical formula
units in a molecule.
The molecular formula is C4H8O2
^ top |
190221 | https://www.investopedia.com/ask/answers/010815/what-difference-between-book-value-and-salvage-value.asp | Skip to content
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Book Value vs. Salvage Value: An Overview
Book value and salvage value are two different measures of value that have important differences. Book value attempts to approximate the fair market value of a company, while salvage value is an accounting tool used to estimate depreciation amounts of tangible assets and to arrive at deductions for tax purposes.
Key Takeaways
When valuing a company, there are several useful ways to estimate the worth of its actual assets.
Book value refers to a company's net proceeds to shareholders if all of its assets were sold at market value.
Salvage value is the value of assets sold after accounting for depreciation over its useful life.
Book Value
Book value (also known as net book value) is the total estimated value that would be received by shareholders in a company if it were to be sold or liquidated at a given moment in time. It calculates total company assets minus intangible assets and liabilities. Book value is a metric that helps analysts and investors evaluate whether a stock is overpriced or underpriced when compared to the company's actual fair market value, an estimate of the price for which the company could be sold. Net book value can be very helpful in evaluating a company's profits or losses over a given time period.
Salvage Value
Salvage value is a tool used in accounting to estimate the value that a tangible asset can be sold for when it has reached the end of its useful life—in short, what the asset can be salvaged for when a company can no longer make viable use of it. The salvage value is used to determine annual depreciation in the accounting records, and the salvage value is used to calculate depreciation expense on the tax return.
Salvage value can sometimes be merely a best-guess estimate, or it may be specifically determined by a tax or regulatory agency, such as the Internal Revenue Service (IRS).1 The salvage value is used to calculate year-to-year depreciation amounts on tangible assets and the corresponding tax deductions that a company is allowed to take for the depreciation of such assets.
Special Considerations: Liquidation Value
A third consideration when valuing a firm's assets is the liquidation value. Liquidation value is the total worth of a company's physical assets if it were to go out of business and the assets sold. The liquidation value is the value of a company's real estate, fixtures, equipment, and inventory. Intangible assets are excluded from a company's liquidation value.
Liquidation value is usually lower than book value but greater than salvage value. The assets continue to have value, but they are sold at a loss because they must be sold quickly.
Liquidation value does not include intangible assets such as a company's intellectual property, goodwill, and brand recognition. However, if a company is sold rather than liquidated, both the liquidation value and intangible assets determine the company's going-concern value. Value investors look at the difference between a company's market capitalization and its going-concern value to determine whether the company's stock is currently a good buy.
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IRS. "Publication 946 (2023), How To Depreciate Property."
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Elimination Half-Life
In subject area:Immunology and Microbiology
Elimination half-life is defined as the time it takes for the plasma concentration of a drug to decrease by half, which is influenced by the drug's systemic clearance rate and volume of distribution. A drug with a short elimination half-life is removed quickly from circulation, indicating a lower potential for side effects.
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Pharmacokinetic and Toxicokinetic Modeling
2014, Encyclopedia of Toxicology (Third Edition)R.W. Seabury, C.M. Stork
Elimination Half-Life (t1/2)
Elimination half-life is the time required to produce a 50% reduction in blood or plasma concentration. This value is estimated using the following equation:
Since the first-order elimination rate constants ke and β can be calculated by dividing VD by Cl, the half-life of a xenobiotic that follows a one- or two-compartment model can be calculated as follows: (1) one-compartment model – t1/2 = 0.693/ke and (2) two-compartment model – t1/2 = 0.693/β. These values should remain relatively consistent in xenobiotics following these models. Conversely, the half-lives of xenobiotics undergoing zero-order elimination, such as overdoses with aspirin and acetaminophen, cannot be calculated using eqn because the time required to produce a 50% reduction in blood/plasma concentration is variable (i.e., elimination is not proportionally related to concentration).
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Reference work2014, Encyclopedia of Toxicology (Third Edition)R.W. Seabury, C.M. Stork
Chapter
Pharmacokinetic Modeling
2009, PharmacologyJames P. Byers, Jeffrey G. Sarver
10.5.3.3Elimination Half-Life
The elimination half-life (t½,elim) can be defined in the same way as was done previously for absorption and distribution by the equation
(10.79)
This equation can then be combined with Equation (10.78) to give the relationship between clearance (CL) and the elimination half-life as
(10.80)
As with previous half-lives, this indicates that elimination is 50% complete after one elimination half-life, 75% complete after two elimination half-lives, and about 99% complete after seven elimination half-lives. Since elimination represents the final removal of drug molecules from the body, the elimination half-life also serves as the determining factor of how much of the originally absorbed drug remains in the body, so that 50% of the absorbed drug remains after one elimination half-life, 25% remains after two elimination half-lives, and only about 1% remains after seven elimination half-lives. The elimination half-life is the half-life value reported in drug handbooks as an indication of how long a drug remains active in the body.
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Book2009, PharmacologyJames P. Byers, Jeffrey G. Sarver
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Pharmacologic Principles and Pain: Pharmacokinetics and Pharmacodynamics
2009, Handbook of Veterinary Pain Management (Second Edition)William W. MuirIII, Richard A. Sams
HALF-LIFE
The elimination half-life (t1/2) is the time required for the Cp of the drug to decrease to 50% of an earlier value (Box 7-1). The units for half-life are expressed as time (minutes, hours).
♦
: The half-life of a drug can also be used to determine the time required for an infused drug to reach steady state (Box 7-2).
♦
: The t1/2, total clearance, and Vd are related by the following equation:
♦
: Changes in total clearance or Vd alter the t1/2. For example, reduced renal clearance resulting from renal disease or toxicity decreases total clearance and increases t1/2.
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Book2009, Handbook of Veterinary Pain Management (Second Edition)William W. MuirIII, Richard A. Sams
Chapter
Profiles of Drug Substances, Excipients, and Related Methodology
2013, Profiles of Drug Substances, Excipients and Related MethodologyGennady Ananchenko, ... Anna Tikhomirova
6.4Elimination
Upon administration, Alendronate is cleared from plasma by deposition into bone and excretion via urine. The renal clearance was found to average 4.26 L/h. Terminal elimination half-life was estimated to be on average 10.5 years . A very long elimination half-life is a reflection of the rate of bone turnover and as such related to Alendronate’s mechanism of action.
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Book series2013, Profiles of Drug Substances, Excipients and Related MethodologyGennady Ananchenko, ... Anna Tikhomirova
Chapter
Clinical Pharmacokinetics
2012, Principles of Clinical Pharmacology (Third Edition)Arthur J. AtkinsonJr.
Concept of Elimination Half-Life
If the rate of drug distribution is rapid compared with the rate of drug elimination, the terminal exponential phase of a semilogarithmic plot of drug concentrations vs time can be used to estimate the elimination half-life of a drug, as shown in Figure 2.10. Because Equation 2.10 can be used to estimate k from any two concentrations that are separated by an interval t, it can be seen from this equation that when C2 ={1/2}C1:
so:
(2.13)
For digoxin, t1/2 is usually 1.6 days for patients with normal renal function and k = 0.43 day−1 (0.693/1.6 = 0.43). As a practical point, it is easier to estimate t1/2 from a graph such as Figure 2.10 and to then calculate k from Equation 2.13 than it is to estimate k directly from the slope of the elimination-phase line.
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Book2012, Principles of Clinical Pharmacology (Third Edition)Arthur J. AtkinsonJr.
Chapter
An introduction to pharmacokinetics
2014, Veterinary Anaesthesia (Eleventh Edition)
Total elimination clearance (ClE) and elimination half-life (t1/2β)
ClE is an independent variable relating the rate of irreversible drug removal from the body to the plasma or blood concentration:
Thus, ClE is the sum of the elimination clearances of all the organs and tissues of the body, principally the liver and the kidneys. The elimination half-life, t1/2β, is a dependent variable and is the time required for the amount of drug in the body to decrease by one-half:
These pharmacokinetic variables can be useful for drugs with a rapid onset of action, such as IV administered propofol, but they do not provide a complete description of their pharmacokinetics. The total volume of distribution is not realized until after extensive drug distribution and redistribution has occurred. Thus, predicted early drug concentrations based on the dose and Vdss will be very low. Although elimination clearance begins from the time the drug arrives at the clearing organs, the relatively slow decline in drug concentrations due to elimination clearance becomes a significant factor in the relationship of plasma (or blood) concentration with time only after the initial rapid decline due to the distribution and redistribution phase is over (Fig. 3.6).
It is now appreciated that the offset of clinical effect is not simply a function of half-life. It may be affected by the rate of equilibration between plasma and effector site and duration of infusion. Hughes et al. (1992) proposed the use of context-sensitive half-time (t1/2 context) and defined this as the time for the plasma concentration to decrease by 50% after termination of an IV infusion designed to maintain a constant plasma concentration. Context refers to the duration of infusion. They demonstrated that context-sensitive half-lives of commonly used IV anaesthetic agents and opioids could differ markedly from elimination half-lives and were dependent on duration of infusion.
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Book2014, Veterinary Anaesthesia (Eleventh Edition)
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Pharmacokinetics and Pharmacology of Drugs Used in Children
2019, A Practice of Anesthesia for Infants and Children (Sixth Edition)Brian J. Anderson, ... Charles J. Coté
Half-Life
Half-life, the time for a drug concentration to decrease by one-half, is a familiar exponential term used to describe the kinetics of many drugs. Half-life is a first-order kinetic process because the same proportion or fraction of the drug is removed during equal periods of time. As described earlier, the greater the starting concentration, the greater the amount of drug removed during each half-life.
Half-life can be determined by several methods. If concentration is converted to the natural logarithm of concentration and graphed versus time, as described in Eq. 7.2, the slope of this graph is the elimination rate constant, k. For both accuracy and precision, at least three concentration-time points should be used to determine the slope, and they should be obtained over an interval during which the concentration decreases at least by half. In clinical practice, for infants and small children, however, k is often estimated from just two concentrations obtained during the terminal elimination phase. With multiple data points, the slope of ln C versus time may be calculated easily by least squares linear regression analysis. Half-life (T1/2) may be calculated from the elimination rate constant, k (time−1), as follows:
Eq. 7.3
Graphic techniques may be used to determine half-life from a series of timed measurements of drug concentration. The concentration-time points should be graphed on semilogarithmic axes and used to determine the best-fitting line either visually or by linear regression analysis. This approach is illustrated in Fig. 7.1, in which the least squares regression line has been fitted to the concentration-time points and crosses a concentration of 20 µg/mL at 100 minutes and a concentration of 10 µg/mL at 200 minutes. The concentration decreased by one-half in 100 minutes, so the half-life is 100 minutes. The elimination rate constant is 0.693/100 per minute or 0.00693 per minute.
Elimination half-life is of no value for characterizing the disposition of many intravenous (IV) anesthetic drugs during dosing periods relevant to anesthesia. A more useful concept is that of the context-sensitive half-time (CSHT) where “context” refers to the duration of the infusion. This is the time required for the plasma drug concentration to decrease by 50% after terminating the infusion.23 The CSHT is the same as the elimination half-life for a one-compartment model and does not change with the duration of the infusion. However, most drugs in anesthesia conform to multiple compartment models and the CSHTs are markedly different from their respective elimination half-lives.
CSHT may be independent of the duration of the infusion (e.g., remifentanil, 2.5 minutes); be moderately affected (propofol, 12 minutes at 1 hour, 38 minutes at 8 hours); or display marked prolongation (e.g., fentanyl, 1 hour at 24 minutes, 8 hours at 280 minutes). This is a result of the return of drug from peripheral compartments to plasma after stopping the infusion. Peripheral compartment sizes and clearances in children differ from adults and at termination of the infusion, more or less drug remains in the body in children for any given plasma concentration compared with adults. The CSHT for propofol in children, for example, is greater than that in adults.24 The CSHT gives insight into the PK of a drug, but the parameter may not be clinically relevant; the percentage decrease in concentration required for recovery from the drug effect is not necessarily 50%.
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Book2019, A Practice of Anesthesia for Infants and Children (Sixth Edition)Brian J. Anderson, ... Charles J. Coté
Chapter
β-Lactam antibiotics
2010, Antibiotic and Chemotherapy (Ninth Edition)David Greenwood
Metabolism and excretion
No metabolites have been detected. Elimination is almost exclusively renal, predominantly via the glomerular filtrate, with 80–90% of the dose appearing in the urine in the first 24 h. Elimination half-life is inversely correlated with creatinine clearance: as the values fall to 2–12 mL/min, the mean plasma half-life rises to 16 h. In patients maintained on hemodialysis the half-life fell to 2.8 h on dialysis. No accumulation occurred over 10 days in severe renal impairment on a daily dose of 0.5–1 g.
Concentrations of 6.6–58 mg/L have been found in bile 25–160 min after the dose at times when the mean serum concentration was 77.4 mg/L. In T-tube bile there was considerable interpatient variation, with mean concentrations of 34 mg/L at 1–2 h after the dose. No accumulation occurs in patients with impaired hepatic function, but the presence of ascites, low plasma albumin and accumulation of protein-binding inhibitors may increase the volume of distribution.
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Book2010, Antibiotic and Chemotherapy (Ninth Edition)David Greenwood
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Pharmacokinetics and Toxicokinetics
2013, A Comprehensive Guide to Toxicology in Preclinical Drug DevelopmentMichael Schrag, Kelly Regal
Half-Life
Half-life (t1/2) is defined as the amount of time required for the drug concentration measured in plasma (or other biological matrices) to be reduced to exactly half of its starting concentration or amount. After IV dosing, the drug concentrations in plasma decline due to both elimination and distribution . In Figure 3.6b, a plasma profile is shown that is characterized by an initial rapid decline, followed by a slower decline at later time points. The first phase or rapid decline is assumed to be primarily due to distribution, while the later phase of decline is slower and assumed to be primarily due to elimination. It is important to note that both processes are occurring in both phases, but the dominant process differs between the two phases. Eventually, after sufficient time has passed, distribution is assumed to be complete. Consequently, the elimination half-life is generally determined from the terminal or elimination (dominant) phase of the plasma concentration versus time curve (Figure 3.6b). While the most accurate half-life is determined after IV administration, it is not unusual to evaluate oral half lives in a multi-dose toxicology study. Changes in oral t1/2 (across time and/or increasing doses) can be an indication of an alteration in the CL mechanism(s) or an indication of the type of toxicology, e.g., reduced renal or liver function.
The following equation is used to calculate the half-life:
(3.20)
where λz is the elimination rate constant, presented in Eq. (3.10) (also see Figure 3.6). Because t1/2 and λz are inversely proportional, a compound with a long half-life will have a long, flat terminal phase, corresponding to a small λz or terminal slope value, similar to the profile show in Figure 3.6b. In contrast, a compound with a shorter half-life will have a steeper slope and a larger λz value. See Practical Considerations #9 in the section ‘Data Evaluation: Look at a Visual Representation of the Data (e.g. graphs)!’ for further discussions on determining half-life.
The parameters of CL and Vd are closely related to half-life, i.e., half-life is a ‘dependent’ parameter [10,19]. After a simple rearrangement of Eq. (3.10), Eq. (3.21) shows the relationship between clearance, volume of distribution, and the elimination rate constant, .
(3.21)
Considering Eq. (3.20), solving for λz and substituting into Eq. (3.21) yields the following equation, which reflects the dependence of half-life on clearance and volume of distribution:
(3.22)
Thus, t1/2 will increase as CL decreases or Vd increases (Vz in this case). Equation (3.22) is written specifically to highlight that CL and Vd, two independent parameters, both control t1/2 which is a dependent parameter. Inulin, which is used to assess kidney function, has a clearance of 7.2 L/hr . Because the volume of distribution is limited to extracellular water (~16 L), the resulting and observed half-life is 1.5 hr. The example of Sensipar® is not as clear [12,13]. Estimates of Vss range from 1000 to 1200 L (for a 70 kg human) while the CL estimate after a single 0.29 mg/kg IV dose is 1.1 L/hr/kg (~87% of hepatic blood flow; see the section ‘Practical Considerations’, section ‘Normalization of Clearance Values to Liver Blood Flow’). These data suggest that the t1/2 of Sensipar® should be 11.1 hr. The t1/2 reported in the IV study was 19.9 hr while a PO study suggested a t1/2 of 10.3 hr. Independent of the variability in the reported t1/2 for Sensipar®, it is interesting to note that despite high CL, the large Vss facilitates a long t1/2 (consistent with Eq. (3.22)) as well as desirable efficacy in patients. Thus, it is not always true that a drug with high CL will have a short t1/2.
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Book2013, A Comprehensive Guide to Toxicology in Preclinical Drug DevelopmentMichael Schrag, Kelly Regal
Chapter
Therapeutic Drug Monitoring
2012, Small Animal Clinical Diagnosis by Laboratory Methods (Fifth Edition)Dawn Merton Boothe
Sample Collection
The elimination half-life of the benzodiazepines is short, and PDCs fluctuate significantly during an 8-hour or 12-hour dosing interval. Two samples are recommended (see Table 18-1).
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Book2012, Small Animal Clinical Diagnosis by Laboratory Methods (Fifth Edition)Dawn Merton Boothe
Related terms:
Monoclonal Antibody
Blood Plasma
Minimum Inhibitory Concentration
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Steady State
Low Drug Dose
Kidney Function
Blood Level
Absorption
Metabolite
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190223 | https://www.youtube.com/watch?v=fXveOlxKtng | Power Series: Combining Series (Reindexing) Examples
James Wenson
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Posted: 9 Jul 2021
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hello everyone in this video i'm going to be taking a look at some power series examples specifically in which i'm going to be a a look at some expressions involving multiple series and we're going to be combining those into a single series an expression with a single infinite series through the use of re-indexing hopefully you'll see what i mean by that let me get into the three examples for you now i have to preface this video by saying that uh this is a video that's made for an introductory differential equations course so i'm assuming that if you're watching this that you have taken some sort of calculus series you know calculus 2 calculus 3 course where you were introduced to power series and and all that so i'm not this isn't going to go that deep into power series as you would see in a calculus 2 course okay so my instructions for this first example and every example for the rest of the video is going to be rewrite the given expression using a single power series whose general term involves x to the k power right so so for right now no i have the sum from n equals 1 to infinity of n times you know some constant c sub n times x to the n minus first power all right so here the general term involves x to the n minus one minus and then the infinite series you know from n equals zero to infinity of c sub n or that's a constant c sub n times x to the nth power all right so the general term for this series involves x to the nth so what i'm going to be doing and this is really in practice for you know some some skill this is a skill that's going to be necessary when attempting to solve a differential equation using power series to find a power series solution for some differential equation so right now my in my index variables the indices are in right and the thing that's changing right that's the index and what i'm going to do is i'm going to re-index uh to the point where eventually i have you know my index will be i'll call it k and i'll have the expression rewritten so there's a single series involving x to the k all right so the important thing here before before re-indexing is to possibly pull some terms out of the sums so that every series starts with the same power of x so i'll make that note off to the side here so before re-indexing going to manipulate the expression so each series in the expression starts with the same power of x so i'll just say get each series to start with the same power of x or the same power of whatever the variable is the independent variable right it doesn't have to be x all the time so if you look at these two series this series starts you know when n equals one so the first term of this series would be 1 times you know c1 times x to the 0 power all right when n equals 1 you have x to the 0. so this this series starts with in its first term you know x to the 0 power or just a constant this other series also starts with x to the zero so this is a very actually easy example because i have it's you know it's already set up where the series are starting with the same power of x so now that they have you know now that all my series start with the same power of x i'm going to re-index each series uh meaning the following you know again just to have a generic um i have each series have instead of depending on x to the n minus 1 or x to the nth i i want them to depend on x to the k so in this series i'm going to replace n minus 1 this exponent on the variable replace it with k or think about this as replacing all the ends with k plus one and let's see what that series will look like so i'd have the sum from now if n equals one then and we're re-indexing to be have an index of k when n equals one k would be 0 right so it start it would start off with k equals 0 to infinity because as n approaches infinity k would approach infinity as well of and then n again is the quantity k plus one times c sub n right so times c sub k plus one and then times x to the n minus one or x to the k okay and these really are again just because they have a different letter now k instead of n they are the same series and you can see that by just looking at the first several terms they have this they have the same exact terms right when k equals zero you get one times c sub one times x to the zero exactly the same term that starts off this series when n equals one and so on you could you could check other terms but these are the exact same series just re-indexed right now instead of the index of n you have an index of k okay and then basically like you're starting a whole new problem even though it is part of the same expression this series here do the same thing just to get it to have the same index variable as the other series instead of x to the nth i want x to the k so i'm going to let k replace n right in this second series and let's see what we get we get the sum now basically it's just the same series but k replaces n so from k equals 0 to infinity of c sub k times x to the k all right now it's still not a single power series ever i have two power series but now that i have two series with the same index and they both have the same initial the starting value same initial index value same starting value if you want to call it that then i can combine them all right then i can combine them to make a single sum or an expression with a single power series so in combining these we'd have this is equal to the the sum you know from that k equals 0 to infinity of and then this minus this and again notice how both of these have x to the k in common i'll factor that out so we'd have a k plus 1 times c sub k plus 1 minus c sub k times x to the k and there we have it right here is our original series our original expression with with the two series written as a sing an expression with a single power series right ones one sigma notation and they really do represent the same expression they really these are the same and you can you can test this out by writing out you know the first five terms of this write out the first several terms of these and you'll see that and combine some like terms and you will see that they are indeed the same okay so again this is going to be an important skill for when we are finding power series solutions of differential equations okay so let's do this again so my next example same same instructions rewrite the given expression using a single power series whose general term involves you know x to some power x to the k power again it doesn't have to be k you can make it whatever variable you index variable you wish i'm just using k all right all right so again i have this expression with you know the sum of two sums right just a large sum that i would like to rewrite so that there's a single sigma a single summation notation single power series all right so as before you know first off my indices you know they're the same variable they're both this n but they don't start at the same value and so i want to change that up and also you know before i re-index with the k i'd like to you know manipulate this expression so that all the series in the expression start with the same power of x all right so if you look at this first series when n equals one you have you know two times one times some constant c one times x to the zero so this this series starts with x to the zero again all right now here's a now throughout hopefully more and more you know the three examples i've chosen get you know slowly more complicated this series when n equals zero you know starts with six times some constant c zero times x to the zero plus one x to the first so this series if you were to write out you know the first term starts with x to the first power all right so these these are not starting with the same power of x but in order to do the you know proper re-indexing and get a single series in here a single summation in this expression i want to make sure the series start with the same power of x all right so just take the lower ones right this this series starts with a lower power of x and just basically write out terms take out terms from that series until you get to a series in this expression that starts with the same power as x of x as the higher one here's what i mean so if i pull out say the first term when n equals one so this series here so i start right writing out terms when n equals 1 i'd have 2 times c 1 times you know x to the 0 but there's no reason to write x to the 0 right that's a as long as x is not the number zero um that would be one anything to the zero any non-zero number to the zero power be one all right plus and then you know now the summation would look like this because i pulled out the term when n equals one the summation is now starting at n equals two all right so from n equals two to infinity of you know the same look right 2 times n times c sub n times x to the n minus 1. and now this series we are starting at n equals 2 this series starts with x to the first and that's exactly what i want all right i want all known so then this series already starts with x to the first so i'm going to leave it the way it is the sum from n equals zero to infinity up six times c sub n times x to the n plus one right now all now all the series now i have this 2c1 out here that's going to stay there any terms i pull away from a series is going to stay in the expression alright so now comes the re-indexing all right because both series in my expression now start with the same power of x now comes the re-indexing where i get instead of x to the n minus 1 and instead of x to the n plus 1 i'm going to have x to the to the k power in both of these so now i'm going to re-index all right so in this first series instead of you know i don't want x to the n minus 1 i want x to the k so i'm going to let k replace the n minus 1. or if you'd like you know n be replaced by k plus 1 right just adding 1 to both sides here well now the expression looks like this sorry i should be putting equal signs here so i still have this again i have this term i pulled out earlier this 2c1 that's not going anywhere plus and now this sum after reindexing with k looks like the following we have the sum now when n equals two k would be equal to one right so the new lower limit for the index is you know k equals one to infinity right because again as n as n goes to infinity k would go to infinity of 2 times n so 2 times the quantity k plus 1 times c sub n or c sub k plus 1 times x to the n minus 1 which is you know times x to the k okay plus and then again same thing over here again imagine you're starting over it's a totally different series i want to re-index so that you know instead of having terms involving x to the n plus 1 i have terms involving x to the k power i use the same indexes over here so i'm going to again replace the power here n plus 1 i'm going to replace that with k so i'm going to have k be replace replacing n plus 1 or n equals k minus 1 same thing so now this sum becomes the following the sum now it should have if you did everything correctly you know starting the series off with the same power of x and then do this re-indexing the lower limit on each sum the lower index value should be the same if you're doing everything right and that is the case here right when n equals 0 k equals 1 is the lower limit same lower limit as the other sum so that means i can put them together and we can put them together as a single sum because they have that same lower limit k equals 1 to infinity of and then 6 times c sub n but n here is k minus 1 times x to the n plus 1 or x to the k all right and now finally i can put both these sums together so this is equivalent to 2 times that c1 plus and then putting these sums together we have the sum from k equals 1 to infinity right and then both these expressions get added together and they both have you know x to the k in common so i'll factor that out right so we have 2 times k plus 1 times c sub k plus 1 then plus 6 times some constant c sub k minus 1 times x to the k and there you go here's how that expression uh there is rewritten as a as an expression with a single sum in it all right yeah there's this extra term that i pulled out the 2c1 but still there is only a single power series involved here and that's what we wanted to get and again you can verify this by going back to the original sum and now just take the original sum the the original expression write out a bunch of terms combine some like terms then write out a bunch of terms you know of this expression here this this 2c1 plus this sum and you'll see that indeed they are the same series they represent the same stuff they have the same exact terms but again i'll leave that to you to verify all right i just got one more example of this same same instructions okay but hold on it's a little more complicated this one because uh my last example has three sums involved but the setup is the same all right the setup is the same now all three of these sums you know have a different lower limit for the index n equals two n equals one and equals zero so obviously obviously i cannot combine them right now and they're also dependent they're also involving different powers of x which i don't like if i want to write this as a single an expression with a single sum involved a single power series i'd like the same power of x in every sum and this in the same lower value for the index so just like earlier i want to make sure every single one of these sums starts with the same power of x all right so this first sum starts with x to the 0 right when n equals 2 your first term would have x to the 0. this second sum starts with x to the first right when n equals one you got x to the nth here and the fi this final sum starts with x to the zero all right so the lower the highest of these is x to the first so that's the one i'm going to leave alone it's these other two now if i start with x to the zero in order to get to x to the first i just need to pull one term out of each of these right so i'm going to rewrite all this here by pulling one term away from this sum one term away from this third sum and leaving the second sum alone then all the sums that are left over will start with x to the first all right so if i say when n equals 2 this is going to be 2 times 2 minus 1 so two times one times c two and then times x to the zero all right so that's the first term pulled out of that and i'll also pull out the first term of this third sum over here when n equals zero over here we just have c zero x to the zero so plus c zero alright so that's the first term pulled out of this first sum the first term pulled out of this third sum and then i'll rewrite the rest all right so i've already pulled out when n equals 2 here so that the leftover sum will be when n equals 3 to infinity of this you know n times n minus 1 times constant c n times x to the n minus 2. this sum i didn't pull anything from so i'm going to leave it the way it is minus now just because i'm going to be putting all these together i'm going to put the 2 inside the sum right you can pull constants out put them in so i make that the sum from n equals 1 to infinity of you know 2 times n times c sub n times x to the nth all right and then finally plus and again i pulled the first term that was this c sub zero here i pulled the first term away when n was equal to zero so now the sum starts with when n equals one so n equals 1 to infinity of you know c sub n times x to the nth all right now every single one of these series starts with x to the first in their first term right so now that i've got all the series in my expression to start with the same power of x let's re-index you know make it make it so that they all involve x to the k power so again for this for this first sum i'm going to have k replace n minus 2 or have n be replaced by k plus 2 same thing in this second sum i'm simply going to have k replace n and the same thing for this third sum i'm simply going to have k replace n so pretty simple those last two all right so finally we have this is equal to now i'll rewrite this so we have c sub zero plus two c sub two you know c 0 and c 2 are constants plus then this sum you know rewritten with the new index of k is you know k equals 1 right because when n equals three k equals one the new lower limits one to infinity of and then n is k plus two then n minus one would just be k plus one so it's k plus two times k plus one times c sub k plus two and then times x to the k and then the second sum we have minus the sum from k equals 1 to infinity of 2k all right n equals k here times c sub k and then times x to the k and then final sum sum again just ends replacing k again here so from k equals 1 to infinity of c sub k x to the k and now every single one of these has the same lower limit on this index of k same lower limit for the summation they all involve x to the k power i can pull that out and in fact these last two also involve c to the k right so you're going to see me pull out c to the k from those last two as well so here's what we end up getting when we write this involving a single summation so we have that c0 plus 2c2 plus and then sum from k you know k equals 1 to infinity of the following so i've got you know k plus 2 and i'll put it in the other order k plus 1 times k plus 2 times you know some constant c sub k plus 2. then over here i'm going to do this you know both of these coefficients of x to the k have c to the k in it and negative 2 k plus 1 of those so i'll have the quantity 1 minus 2 k times c sub k and then all that all this is a coefficient of x to the k wonderful there it is this is a the original sum the original expression now written as an expression with one a single power series in it right a single summation notation and if again if you'd like and i'll i'll leave it to you you know if to see that this truly is equivalent to the original series the original expression i had you know take that original expression write out the first several terms of each sum combine like terms if you need and then take this one write out the first several terms of of all this and and you're going to see the more and more terms you go out that they're all the same you get the same exact terms in this series in this expression as you would in this expression right so that they are indeed equivalent okay and that's it all right so you notice i went through the same proceed now the same procedure in all three examples you know try to get all your series to start with the same power of x if you need to pull out some terms from certain series to do that then do that then re-index right just basically go through each series replace all the powers on the x all the exponents with you know some other letter k or j or whatever you want and if you did all that properly you should be able to then express the same expression with you know with a single power series wonderful alright so hopefully watching me go through these three examples help you out when you're doing problems similar to this and thank you very much for watching |
190224 | https://projecteuclid.org/journals/journal-of-differential-geometry/volume-46/issue-1/Harmonic-functions-with-polynomial-growth/10.4310/jdg/1214459897.pdf | j . d i f f e r e n t i a l g e o m e t r y 45 (1997) 1-77 H A R M O N I C F U N C T I O N S W I T H P O L Y N O M I A L G R O W T H TOBIAS H. COLDING k WILLIAM P. MINICOZZI II 0. Introduction Twenty years ago Yau, , generalized the classical Liouville theo-rem of complex analysis to open manifolds with nonnegative Ricci curva-ture. Specifically, he proved that a positive harmonic function on such a manifold must be constant. This theorem of Yau was considerably generalized by Cheng-Yau (see ) by means of a gradient estimate which implies the Harnack inequality. As a consequence of this gradient estimate (see ), one has that on such a manifold even a harmonic function of sublinear growth must be constant. In order to study further the analytic properties of these manifolds one would like to restrict the class of functions to be considered as much as possible while minimizing loss of information (cf. , ). From the results of Cheng and Yau, it follows that a natural candidate is the class of harmonic functions of polynomial growth (note that they must be of at least linear growth). In fact, in his study of these functions, Yau was motivated to make the following conjecture (see , , and ; see also the excellent survey article by Peter Li, ): Conjecture 0.1. (Yau). For an open manifold with nonnegative Ricci curvature the space of harmonic functions with polynomial growth of a fixed rate is finite dimensional. We recall the definition of polynomial growth. Definition 0.2. For an open (complete noncompact) manifold, M n, given a point p G M let r be the distance from p. Define H d(M) to be Received January 9, 1996. The first author was partially supported by NSF Grant DMS 9504994 and the second author by an NSF Postdoctoral Fellowship. 1 2 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii the linear space of harmonic functions with order of growth at most d. This means that u G H d if u is harmonic and there exists some C > 0 so that \u\ < C(l + r d). The main result of this paper is the following. Theorem 0.3. Conjecture 0.1 is true if M has Euclidean volume growth. M n is said to have Euclidean volume growth if there exists p £ M and a positive constant V such that Vol(B r(p)) > Vr n for all r > 0. Note that by the Bishop volume comparison theorem (see ) we have that Vol(B r(p)) < Vo(l)r n for r > 0. Here as in the rest of this paper V ^(r) denotes the volume of the geodesic ball of radius r in the n-dimensional space form of constant sectional curvature A. We show Theorem 0.3 by giving an explicit bound on the dimension of H diM) depending only on n and d. From the new results given by the investigation initiated by the first author in , , and , and later on further developed by the first author jointly with Cheeger in , , and , and finally the joint work of the first author with Cheeger and Tian in , we have a good understanding of the geometry of spaces with Ricci curvature bounded from below. For the present paper, it is particularly important that it was shown in that every tangent cone at infinity of a manifold satisfying the assumptions of Theorem 0.3 is a metric cone. For an open manifold M n with nonnegative Ricci curvature, we say that a metric space, Mx,, is a tangent cone at infinity of M if it is a Gromov-Hausdorff limit of a sequence of rescaled manifolds (M,p,r~ g), where r j —> oo. Recall that by Gromov's compactness theorem, , any sequence, r i — > oo, has a subsequence, r j —> oo, such that the rescaled manifolds (M,p,r~ g) converge in the pointed Gromov-Hausdorff topology to a length space, Moo. Examples of Perelman (; see also for further examples) show that MQO is not unique in general even if M has Euclidean volume growth and quadratic curvature decay (cf. and ). We also note that examples of Perelman (see ) most likely can be modified to give examples of manifolds with nonnegative Ricci cur-vature, Euclidean volume growth and infinite topological type. It is a classical result that the space of harmonic functions of polyno-mial growth on Euclidean space is spanned by the spherical harmonics. h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 3 Recall that the spherical harmonics are the homogeneous polynomials whose restriction to every sphere centered at the origin is an eigenfunc-tion of the spherical Laplacian. We will observe in Section 1 that this is a general property of metric cones. That is, the harmonic functions of polynomial growth on a metric cone with smooth cross-section can be written as a linear combination of harmonic functions which separate variables (into the radial and cross-sectional directions). Further, they are homogeneous in the radial direction; it follows that the restriction to the cross-section gives an eigenfunction, where the eigenvalue depends on the dimension and the order of growth. We will show that asymp-totically this picture still holds in the general case of nonnegative Ricci curvature and Euclidean volume growth (cf. Theorem 4.60). That is, on many sufficiently large annuli, harmonic functions of polynomial growth will almost separate variables and be approximately homogeneous in the radial direction. It seems worth pointing out some of the difficulties that arise in the general case of nonnegative Ricci curvature and Euclidean volume growth compared with the model case of a cone. Here we will only indicate three such. The first is the low regularity of the cross-section of tangent cones at infinity (cf. ). The second is that the frequency function (see Section 2 for the definition of the frequency function) is no longer monotone in the general case; see Section 11 and . Thirdly, the frequency function is not known to be bounded; see for further discussion of this. Simple examples show (see Section 11) that there exist manifolds with nonnegative Ricci curvature which admit no nontrivial harmonic function with polynomial growth; in fact, we can take such a manifold to have positive sectional curvature. However, to our knowledge no such example exists with nonnegative Ricci curvature and Euclidean volume growth; see for further discussion of this. Important contributions on this Conjecture of Yau and related prob-lems have been made by Donnelly-Fefferman, Kasue, Li, Li-Tam, Wang, and Wu (see , , , , , , , , , and ). In related work, F.-H. Lin has studied asymptotically conical elliptic oper-ators. The organization of this paper is as follows: Section 1 is concerned with the description of harmonic functions with polynomial growth on cones and serves to illustrate the methods 4 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii that we will employ in the general case. In Section 2 for later use we introduce an important tool which is a generalization of Almgren's frequency function. A lower bound for the frequency of a harmonic function on good annuli is given in Section 3. We study in Section 4 the monotonicity properties of the frequency function for harmonic functions on manifolds with nonnegative Ricci curvature and Euclidean volume growth. We also study the asymptotic homogeneity properties of harmonic functions with polynomial growth on these manifolds. Section 5 deals with orthogonality properties of harmonic functions on these manifolds. We get in Section 6 an explicit upper bound for the number of or-thonormal functions with bounded gradient on a compact manifold with Ricci curvature bounded from below and diameter bounded from above. Section 7 contains the proofs of some elementary results for functions of one variable with bounded growth, that will be used later on. Given a set of independent harmonic functions with polynomial growth on a manifold with nonnegative Ricci curvature and Euclidean volume growth, we show in Section 8, how to produce large annuli and a set of independent harmonic functions with good properties. This together with the results of Section 7 allows us to convert the (global) polynomial growth condition to information on a definite scale (local). With the aid of the results of Sections 7 and 8 we obtain, in Section 9, a technical result that will be needed in the inductive step of the proof of Theorem 0.3. Using the results of the previous sections, Theorem 0.3 is proved in Section 10, by giving a bound on the dimension of the space of harmonic functions with bounded growth (and suitable independence properties) on any sufficiently large annulus in an open manifold with nonnegative Ricci curvature and Euclidean volume growth. Section 11 furnishes various examples that illustrate the difficulties in the Euclidean volume growth setting compared with the model case of a cone; see for further discussion of this. In the appendix, we will collect some consequences of the first vari-ation of energy that we need for this paper. Finally, we point out that in a joint paper with Cheeger (see ) we study the case of linear growth harmonic functions. Throughout this paper, if N is a closed manifold, we take the conven-h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 5 tion that a function g is an eigenfunction with eigenvalue A if Ag + \g = 0. With this convention, A is a negative operator but we will say that the eigenvalues are positive. Acknowledgment. We wish to thank Jeff Cheeger and Gang Tian for numerous helpful discussions, and Chris Croke, Peter Li, Fang-Hua Lin, Grisha Perelman, and Richard Schoen for their interest. The results of this paper were announced in . Some time after the submission of this paper, in part by further developing the ideas presented here, we solved the general case of the conjecture of Yau, . We wish to point out that in this paper in addition to showing the finite dimensionality of H d, we also describe the asymptotic structure of the harmonic functions with polynomial growth (like the almost separation of variables). This finer description is in part due to the asymptotic cone structure of the manifolds considered here. It should also be pointed out that the bounds for the dimension of the space H d given in this paper depend exponentially on d. In a subsequent paper to , we gave polynomial bounds sharp in the order of growth, see . 1. Harmonic functions with polynomial growth on cones In this section N n _ 1 will be a closed smooth (n — l)-dimensional manifold. The study of function theory on the Euclidean cone on N is meant to illustrate the methods that we will employ in the proof of Theorem 0.3. Note however that the results of this section will not be used in the proof of Theorem 0.3. We will often further assume that Ric N > (n — 2). This condition is equivalent to the Euclidean cone C(N) = (0, oo) xr N n _ 1 having nonnegative Ricci curvature. In this section u(r, 9) is a smooth function on the Euclidean cone C(N) which may be extended continuously to the vertex. On such a cone, the Laplacian can be written as (1.1) AC{N)u=—u+———u+-AN u{r,-). In general, we will say that a function is homogeneous of degree p if it is of the form u(r, 9) = r p g(9). First, we claim that if u(r, 9) = f(r)g(9) is harmonic, then f(r) = r p for some p > 0 and g is an eigenfunction of N. To see this note that 6 t o b i a s h . c o l d i n g & w i l l i a m p . m i n i c o z z i ii f 1.1) becomes :i.2ì AC(N)u = f'g + (n - 1)— g + —AN g. From this we have that (1.3) AN g + Xg = 0. Note also that if Ric N n-i > (n — 2) then A = 0 or A > (n — 1). Substi-tuting (1.3) in (1.2) gives 0 (1.4) Therefore (1.5) f"g+ (n- l)—g —^Xg f " + ( n - 1 ) r - A f ) g . f"+{n-l)f--\^ 0. We easily see that if p(p — 1) + (n — l)p — A = p2 + (n — 2)p — A = 0, i.e., (1.6) p = 2 n - 2 ) + ( n - 2 ) 2 + 4A then f(r) = r p is a solution of (1.5). Note that we take only the non-negative solution p because the negative solution has a pole singular-ity at the vertex. Further, we see that if we require that p < d then A < d(d+ n — 2), and therefore (1.7) A = p(p + n — 2) . Collecting the previous calculations, we have the following elemen-tary lemma. Lemma 1.8. A function u(r,9) = f(r)g(9) on C(N) is harmonic if and only if (1.9) and (1.10) where A = p(p + n — 2). AN g + \g = 0 f(r h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 7 Let E(N) denote the linear space spanned by the eigenfunctions of N with eigenvalues less than or equal to A. Further, we let 0 = Ao < Ai < A2 < • • • denote the distinct eigenvalues of N and let p j > 0 be determined by Xj = p j (p j + n — 2). The following theorem is well known (see ). Theorem 1.11. (Harmonic functions on a cone). If u is a har-monic function on C(N), then (1.12) u(r,9) = ^2a j r p g j(9), j where the a j are constants. Furthermore, u has polynomial growth if and only if this is a finite sum. Proof. We may assume that u(0) = 0. By the spectral theorem applied to N, we may write (1.13) u(l,0) = Y^a j g j(0). j Consider the harmonic function (1.14) v(r,0) = u(r,0) - ^ a j r p j g j(0) . j Note that v vanishes on dB\ and at the vertex; by the maximum prin-ciple, v vanishes identically. The second claim follows easily from the first. q.e.d. We will now obtain a second proof of Theorem 1.10 that is closer to the proof of Theorem 0.3. If u is a Lipschitz function on C(N) then we set (1.15) D{r) = r2~n Z \Vu\2, (1.16) I(r) = r1-n Z u2, JdB r(p) du dr ;i.i7) F(r) dBAp) 8 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii and finally the frequency (cf. and Remark 2.16) r R B r(p)\Vu\2 D(r) 1.18 U(r dB r p) u I ( ;i.2o) Lemma 1.19. If u is harmonic then U is monotone nondecreasing. Proof. To show this note that D'(r) I'r (logU)'(r) D(r) I(r) Further, from the first variation of energy, i.e., Proposition A.23, we have that D'{r)=r2-n Z \Vu\2 + ^—n D(r) JdB r(p) r :i.2D 2r 2-n dB r(p) du dr F(r) and, since Au2 = 2|Vu|2, I'(r)=r1-n {Vu2,Vr) + rl~n u2Ar + - -I(r) dB r(p) JdB r(p) r 1 n n.22) Therefore 1-n Au2 = 2rl~n |Vu|2 = 2 D r B r(p) B r(p) r ;i.23) (logUy{r) = D r 2 D ( r ;i.24) ;i.25) D{r) rI{r) 2-n R \du\2 , , D(r) rI(r) 2r 3-2n D(r)I(r) 8B r(p) du dr d 8B r(p) 2 u dB r{p) dr h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 9 From the Cauchy-Schwarz inequality we get that (1.26) ( l o g U ) ' > 0 . q.e.d. We will later see (Theorem 1.66 and Lemma 1.26) that in the case of a cone many U are in fact constant. Next we have the following: Lemma 1.27. If u is harmonic and U is constant, then u(r,9) = f(r)g(9). Conversely, if u(r,9) = f(r)g(9) is harmonic, then U = p, f(r) = r p and ^g + p(p + n — 2)g = 0. Proof. Since U is constant, then by the equality in the Cauchy-Schwarz inequality, see (1.23), we have du . . 1.28 — = h(r)u. or Integrating (1.28) shows that u(r, 9) = f(r)g(9). The lemma now follows from Lemma 1.7 and an easy computation. q.e.d. Lemma 1.29. If u e H d(C(N)) then U < d. Proof. Equation (1.22) is equivalent to (1.30) ( l o g I ( r ) ) ' = H U r . r Integrating equation (1.30) yields (1.31) I(r) = exp( r ^^-dt)I(s). Since U is monotone nondecreasing, we see that U must be bounded by d. q.e.d. Definition 1.32. (Order at infinity). If u is harmonic, then we define the order at infinity of u, ord00(u), by (1.33) ord00(u) = lim U(r). r—>oo Note that this limit exists since U is monotone nondecreasing by Lemma 1.18. When u has polynomial growth, then Lemma 1.28 shows that ord00(u) is finite. Likewise, the monotonicity of U allows us to make the following definition. 10 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Definition 1.34. (Order at the vertex). If u is harmonic, then we define the order at the vertex of u, ordo(u), by (1.35) ordo(u) = lim U(r). r—»0 Lemma 1.36. If u and v are harmonic functions, then (1.37) ordoo^u + v) < maxford00(u), ord ^ ^v)} • Proof. By the Cauchy-Schwarz inequality, we have that ( 1 3 8 ) log(I u +v) maxford 0 0(u), ord 0 0(v)}, then there exist an R > 0 and an e > 0 such that for any r > R, (1.40) (logI u+v(r))' > max f(logI u(r))', (log v(r))'} + - . Since - is not integrable, this would contradict the inequality in (1.38); therefore, (1.37) follows. q.e.d. Lemma 1.41. Suppose that u and v are harmonic functions on C(N). If in addition v(r,9) = r p g(9), then (1.42) r1'n Z uv = r2p Z uv. JdB r JdBi Proof. Using Green's formula, we get that d- (rx-n Z uv) =rx~n Z ^-(uv) dr JdB r JdB r or l-n J u 1-n Z dv : i - 4 3 ) =r v ^ + r udr '1.44) =2r1~n Z u^-. dB r dr h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 11 From the homogeneity of v, we have that -v = p v. Substituting this into (1.44), Integrating (1-45) yields (1-42) and the lemma follows. q.e.d. Definition 1.46. We say that two harmonic functions, u and v, on C(N) are orthogonal if (1.47) / uv = 0. Note that by Lemma 1.40, if v is homogeneous, and u and v are orthogonal in the sense of Definition 1.45, then (1.48) uv = 0 dB r for all r > 0. Also note that from the maximum principle it follows that the left side of (1.47) defines an inner product on the space of harmonic functions on C(N). Lemma 1.49. Suppose that u is harmonic on C(N) with ord ^ ^u) = d < oo and that u is orthogonal to the homogeneous harmonic func-tions whose growth is less than d. Then for r > s > 0, we have (1.50) D(r) > (-) D(s). Proof. Let A be given by (1.7), that is, (1.51) \ = d(d + n-2). By the orthogonality assumption and Lemma 1.40, we get the following scale-invariant Poincare inequality for the cross-section dB r: (1.52) dB R j j > 4 . h dB uc where r u is the tangential gradient. Note that (1.53) j r u j 2 = j r T uj2 + du dr 12 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Using (1.53), we can rewrite (1.52) as (1.54) rD'(r) - (2 - n)D(r) - F(r) > 2\I{r) . From the first variation of energy (see equation (1.21)) it follows that (1.55) D'(r) = 2Fr-. r Eliminating F(r) in (1.54) and using (1.55), we have ,, ^ 2(2-n)D(r) 2\I(r) (1.56) D'(r) - — '—-t > — . Dividing (1.56) through by D(r) and noting that I = U 1 > d 1, give D'(r) 2(2-n) 2A 2A (1.57) — - - > > . D(r) r ~ rU{r) ~ dr Substituting (1.51) for A in (1.57), combining the - terms, and rewriting the first term as a logarithmic derivative, we obtain 2d (1.58) ( l o g D ( r ) ) ' > — . r Integrating (1.58) yields (1.50). q.e.d. Lemma 1.59. If u is harmonic, u(0) = 0, and ordo(u) = 0, then u is identically zero. Proof. We may assume that u is not constant; this implies that I(r) is positive for every r > 0. By (1.31) we get (1.60) I{r) < 2 d I (r where d = U(l) and we take r < 1. From the scale-invariant Poincare inequality, we have that 2 2 T 2 2 2 (1.61) A i / u2<r2 i r u ^Kr2 | r u , JdB r JdB r JdB r where Ai is the first eigenvalue of the Laplacian on N, and r u is the tangential gradient of u. Using (1.60) and the monotonicity of I, and integrating (1.61) from r to r we are led to r 2-2dX1rI(r) <\xrI (-) < 2Xt I(t)dt 2r This shows that U(r) > Ai2 - 2 and the lemma follows. q.e.d. ;i.62) 2rD(r h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 13 Corollary 1.63. Suppose that u is harmonic on C(N) with ordoo^u) = d < oo and that u is orthogonal to the homogeneous har-monic functions whose growth is less than d. Then u is homogeneous. Proof. By Lemma 1.58, we may assume that ordo(u) > 0; let c = ordo(u). By the definition and monotonicity of U, (1.64) cD{r) < I{r) < dD{r) . Therefore, by Lemma 1.48, for r > s > 0, we have c /r\2d (1.65) I r ^ d s ) I s -Setting r = 1 and taking s < 1, we see that (1.66) I(s) < cjs2d I(l) . By equation (1.31), (1.66) implies that ordo(u) = d. Since U is mono-tone, we conclude that U is constant. The corollary now follows from Lemma 1.26. q.e.d. We are now ready to give a second proof of Theorem 1.10. Theorem 1.67. (Harmonic functions with polynomial growth on cones; second version). If N n~l is a closed (n — 1)-manifold, then (1.68) dim(H d(C(N))) = dim(E d{d+n_2)(N)). In fact, if u G H d(M) then (1.69) u(r,0)= X rVj g t f ) , p j-Cd where g j is an eigenfunction with eigenvalue Xj. Proof. The inequality ">" in (1.68) follows from Lemma 1.7. We will show the reverse inequality, i.e., "<", by induction on j . For j = 0 we have by Lemma 1.28 that ord00(u) = 0; by the monotonicity of U, and Lemma 1.18, u must be constant. Assume now that the theorem is true for p j and will show that it is true for p j+i- Given u G H p + 1, by the inductive hypothesis and Lemma 1.35 we may assume that u = u' + u" where u' G 'HPj+1, u" = P k (n — 2) then the case d = 0 in Theo-rem 1.10 or Theorem 1.66 is essentially a special case of the Liouville theorem of Yau, ; the cases 0 < d < 1 follow from the gradient esti-mate of Cheng-Yau, , and is in this case equivalent to Ai > (n — 1) (Lichnerowicz's theorem, ). Example 1.71. In many cases where Ric N > (n — 2) and N is diffeomorphic to S n _ 1, it is possible to round off the metric on a cone while preserving the condition that the Ricci curvature is nonnegative. In fact the change in the metric can often be done by a compactly supported change in the warping function. As a consequence of Theorem 1.66 and Proposition 11.5 we see that for such a perturbation dim(Td) = dim(E d (d +n_ 2 )(N)). 2. Tools to study the growth of harmonic functions on manifolds From now on, unless explicitly stated otherwise, let M n be an n-dimensional open manifold with nonnegative Ricci curvature. Set Vol{B r{p)) (2.1) V M = lim oo note that by the volume comparison theorem this limit exist (in fact the quantity in (2.1) is nonincreasing) and is independent of the point p. We will also assume that M has Euclidean volume growth, that is, V M > 0. Fix a point p G M and let G denote the global Green's function on M with singularity at p. It is well known that G exists in this setting (see for instance ). For ease of exposition, we will henceforth restrict our attention to the case of n > 3. The case n = 2 was done earlier by Li-Tam, (in fact, for surfaces with finite total curvature). For another proof in the case n = 2 using nodal sets see Donnelly-Fefferman, . Set When M is R n, the function b defined in (2.2) is just the distance function to p. When studying the global analytic properties of M, the function b is the proper replacement for the distance function (cf. Proposition 2.21; see also -, ). h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 15 (2.3) With this choice of b we have V M Vb ( 2 - n ) V n(i b n _ 1VG IVbI (2.4) Ab=(n-l)i—-t and (2.5) Ab2 = 2n\Vb\2. We define the following quantities (2.6) I(r) 1-n u 2|Vb|, b=r (2.7) D(r) 2 - n IVuI b<r (2.8) F(r) 3 — n du dn |Vb|, and finally the frequency function (cf. and Remark 2.16) by D(r) (2.9) U(r Ir) Observe that if r < s, then (2.10) 2 —n D r < r D(s) Remark 2.11. S.Y. Cheng, , showed that locally the critical sets (sets where the function is constant and its gradient vanishes) of any harmonic function are of codimension two on any smooth manifold (see also Hardt-Simon, ; their results are valid for low regularity elliptic equations). Since the critical sets of b coincide with those of G (which is harmonic), it is easy to see that these calculations are valid on all level sets of b (and not just at regular values). 16 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Differentiating (2.6) gives D(r) (2.12) I'{r) = 2 and therefore (2.13) (logI(r))' = H U r r From (2.13) we have for s > r > 0 (2.14) I(s) = exp (2 Z s U t - d t I{r) . The quantity I u(r) is a weighted average of u2, and Ii(r) is the weighted volume of the level set b = r. By (2.12), Ii(r) is constant. From the definition of b it is easy to see that (2.15) I ( r ) = n V M 3 and M n is an n-dimensional manifold with Ric M > 0 and Euclidean volume growth, then there exists a constant C > 1 such that (2.18) r2~n <G(x,y) < Cr2~n . It follows from (2.18) that there exist positive constants (depending on M) C\ and Cì such that (2.19) C\r 3 and M n is an n-dimensional manifold with Ric M > 0 and Euclidean volume growth, then for each fixed x Ci M (2.22) lim Gxy V° ( 1 ) r(y)^oo r2 n V M Observe that (2.22) implies the strengthening of (2.19): b (2.23) lim - = 1. r—>oo r Furthermore, from Section 4 of and (2.23) given any S > 0, there exists R = R(p, S) > 0 such that for all r > R, we have (2.24) Z ||Vb| 2- l|2 < 5Vol(b< r) b<r and (2.25) b |Hess(b2) - 2g\2 < 5Vol(b < r) , where g is the metric tensor on M. 18 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii In fact, all that we essentially require of G (for this section and Sections 3 and 4) is that it is harmonic on an annulus and C° close to a multiple of r2~n, where r is the distance to the center of the annulus. It then follows from that b has the properties similar to (2.24) and (2.25) (see Section 4 of and cf. ). We will also use the following meanvalue inequality of Li-Schoen, which for convenience we state only for the case of nonnegative Ricci curvature. Proposition 2.26. (Li-Schoen, ). Suppose that M n is an n-dimensional manifold with Ric M > 0 and v is a nonnegative subhar-monic function on M. Then (2-27) s u p v < C Z v, B r(p) Vol(B,(p)) B {p) l 2 where C = C{n). Often, we will get natural integral bounds for harmonic functions and their gradients; the meanvalue inequality, Proposition 2.26, will allow us to get supremum bounds on a subset. Finally, we will use that for each r, I(r) defines a quadratic form on the linear space of harmonic functions. The associated bilinear form is given by (2.28) rx~n uvjVb for harmonic functions u and v. Note that from the maximum principle we have that for the regular values, s, of b, (2.28) defines an inner product on the space of harmonic functions on {x j b(x) < s}. Clearly, this also follows from the monotonicity of I. 3. Lower bound of the frequency In this section, we will give several versions of a lower bound for the frequency of a harmonic function. In a future paper we plan on undertaking a more careful study of this and some of its consequences. We now define quantities analogous to those of Section 2 which are technically easier to work with. Let (3.1) E(r) = r2-n Z jVuj 2jVbj 2 b 0, i7o > 1, and y > 1, there exists R = R(p, j , e, QQ) > 0 such that if r > R, 1 < Ci < QQ, and u is any harmonic function on M with (3.4) then for all r < s < Çlr D{2Qr) < jD(r) (3.5) log D(s) E(s) e Proof. Note that |1 — s| < -^ implies that (3.6) |log(s)| < i_ t < e. From and the asymptotics of the Green's function, Proposition 2.21 (see also the remarks following that proposition), given any S > 0, there exists R = R(p, S) > 0 such that for all r > R, we have (3.7) and (3.8) b l o g -r S. IVbI b<r l\ < 52Vol(b< r) Note that (3.8) implies by the Cauchy-Schwarz inequality that (3.9) |Vb| 2- il < 5Vol(b< b<r 20 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii We shall assume that S is small enough to arrange that exp(25) < | . By definition, we have for s > R, jD(s)-E(s) (3.10) 2-n 2 2 < s sup jVuj s b<s V u j 2 ( l - jVbj2) 1 1 - jVbj2| b<s < s2 sup jVuj2 ÄV n(l) exp(nS) , b<s where the last inequality follows from (3.7), (3.9), and the Bishop vol-ume comparison theorem. From the Bochner formula, jVuj2 is a subharmonic function. Since exp(25) < | (3.7) and Proposition 2.26 yields that for r > R, (3.11) C sup jVuj2 < —-Q- 2r- 2D(2Qr) ; b<îîr V M where C\ = C(n) > 0; in (4.22) we do this again in more detail. Using (3.4), (3.10), and (3.11) we obtain, for r < s < Çlr, jD(s)-E(s)j< (3.12) < V M V M V M D(2Çir)SV n(l) exp(nS) yD(r)5V n(l) exp(n5) jD(r)5V n(l) Q V Finally, to finish the proof, we use the trivial bound, for s between r and Çlr, (3.13) and set (3.14) D(r) 2 - n j u j 2 r 2-n b 0 . o q.e.d. If the frequency is locally bounded from above, we will get a lower bound for the frequency. We will have two versions of this lower bound. First, Lemma 3.22 will give a crude lower bound for the frequency func-tion and later, in Section 4, Corollary 4.40 a more refined version. 22 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Lemma 3.22. (Lower bound of the frequency; crude version). Let M be an open manifold with nonnegative Ricci curvature and Euclidean volume growth and let p G M be fixed. Given Ci > 2, there exist C = C(n) > 0 and R = R(p) > 0 such that for any harmonic function u with u(p) = 0, we have for r > R, (3.23) I(r) < CQ-2D(Qr) . Furthermore, if U(s) < d for r < s < Çtr, we get a lower bound for U{Çlr); that is, r)2-2d (3.24) —C- < U(Qr) . Proof. By (2.23), we can choose R = R(p) > 0 such that for r > R 4 (3.25) b 2 l o g -r < log 3 which implies that the set fb < r} is contained in a ball of radius p r. As in (3.11), by Proposition 2.26, we have (since Ci > 2) C (3.26) sup|Vu| 2 < — - Q - 2 r - 2 D ( Q r ) , b 0. By integrating equation (3.26) along geodesics starting at p and using the fact that u(p) = 0, we get 4 C (3.27) s u p u 2 < - — - Q - 2 D ( Q r ) . b<r 3 V M The claim (3.23) now follows from the weighted volume bound for the level set, (2.15), with C = ±nC. If U < d, then by (2.14), (3.28) I(Qr) < Q2d I(r) < CQ2d-2D(Qr) , and the second claim follows. q.e.d. We are now prepared to give a uniform lower bound for the maximum of the frequency on arbitrary annuli outside a compact set. In contrast to Lemma 3.22 the importance of this result is that it does not require any control on the function. h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 23 Corollary 3.29. (Uniform lower bound of the maximum of the fre-quency). Let M be an open manifold with nonnegative Ricci curva-ture and Euclidean volume growth and let p G M be fixed. There exist C L = C L{n) > 0 and R = R(p) > 0 such that for any harmonic function u with u(p) = 0, we have for r > R, (3.30) max U(s) > C L . r<s<2r Moreover, given e > 0, there exists QL = f2L(n, e) > 2 such that for any harmonic function u with u(p) = 0, we have for r > R, (3.31) max U(s)>(l-e). Proof. Suppose that d is a uniform upper bound for the frequency; we will show that d cannot be too small. We apply Lemma 3.22 with Q = 2 to get an R = R(p) > 0 and a C = C{n) > 0 such that for r > R, if U(s) < d for s between r and 2r, then 92-2d (3.32) U ( 2 r ) > _ _ . Hence, we have 22-2d (3.33) d>——>0. This implies that d > C L = C L(n) > 0. Moreover, with the same R = R(p), for any Ci > 2 and the same C = C(n) > 0, if r > R and U(s) < (1 — e) for s between r and Çlr, then (3.34) 1 > U(Qr) > - C 2 e . This is not possible for Ci > QL = f2L(n, e) = max{C~, 2}. q.e.d. Remark 3.35. We will see in Section 4 that there are many large annuli on which the frequency function is almost monotone. As a result, Corollary 3.29 will imply a lower bound for the frequency function on the outer parts of these annuli. See Section 4 for a precise version, and for other results in this direction. We will now show how to get control of the growth of D just from a bound on the growth of I (cf. Theorem 4.60). In later sections, this will allow us to work with bounds only on the growth of I. 24 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Proposition 3.36. (Bounding the growth of D by the growth of I). Let M be an n-dimensional manifold with nonnegative Ricci curvature and Euclidean volume growth, and let p G M be fixed. There exists R = R(p) > 0 such that for r > R and any Ci > 1, if u is any harmonic function on M with u(p) = 0 such that (3.37) I(2Qr) < CI (r) , then (3.38) D(Qr) < C 2 C i D ( r ) , where Cì = Ci{n). Proof. By Lemma 3.22 we get an R = R(p) > 0 and a K = K(n) > 0 such that for r > R, (3.39) I(r <KD{r). From (2.12), we have 2ür D( \ (3.40) 2 Z —-^ = I(2fìr) - I(Or) < I(2îîr) , J Sir s and hence 2îîr (3.41) 2 Z s n"2 D(s) < {2Qr)n-1I{2Qr) . Jür By definition, (2.7), s n~2D(s) is monotone nondecreasing, and therefore (3.41) yields (3.42) 2 (Qr)n-1 D(ttr) < 2n~l (fir)n"1I(2ttr) . Dividing through by 2(Qr)n~1 gives (3.43) D(Qr) < 2n" 2I(20r) . Combining (3.37), (3.39) and (3.43), we obtain (3.38) with C2 = 2n~2K. q.e.d. h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 25 4. Almost monotonicity of the frequency and almost separation of variables In this section we will show that when M has Euclidean volume growth the frequency function behaves much like it did in the cone case. In particular, we will first show that the frequency function is almost monotone (Proposition 4.11). This will allow us to show that harmonic functions with polynomial growth come close to separating variables on infinitely many large annuli (Theorem 4.60). Differentiating log W, we get (4.1) W r = E r - I r ' which together with (2.13), that is, with (logI)' = 2r~l U, implies E'(r) 2D(r) (4.2) (logW(r)y Er) rI(r) From the first variation of energy (Proposition A.23) it follows that (3.15) is equivalent to E'(r) =2r 2-n b=r du dn A-n IVbI + \Vu\2Ab2 b<r (4.3) 1 - n Hess(b2) (Vu, Vu) + (2 - n) b<r E(r) F(r) 2 , , 2—K—t + -E(r) 1 - n b<r Hess(b2)(Vu, Vu) where the second equality follows from (2.5) and (2.8). Substituting (4.3) for E'(r) into (4.2), we get (4.4) (log W(r))'=^ + F r \ r rE(r) 1 - n R b 0. (logW(r)y = 2 r2-n RlnriVbl" 1 2D(r) D(r) rI(r) (4.9) + + 2 r 2F rE rl~n R b\ 1, there exists R = R(p, 7, e, QQ) > 0 such that if 1 < Q < QQ, r > R, and u is any harmonic function on M with (4.12) D(2Qr) < jD(r) , then (4.13) Z min{(logW0'(t),0}dt> - e . r In fact we will show that (4.14) ZÜr(logWy(t)- [(logWy(t)]ess\dt 0, there exists Ri = Ri(p, 5) > 0 such that for all r > R\, we have (4.15) b l o g -r S. (4.16) \Vb\2 -l\2 <52Vol(b<r) b<r and (4.17) Z |Hess(b2) - 2g\2 < 52 Vol(b < r) , b 0 such that (4.18) exp(25) < 1 , 28 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii then (4.15) implies that for s > R\, fb<sgcB rjip), B s p-3(p) Cfb< 2sg . (4.19) From Proposition 3.3, we get an R2 = R2(p,y,S,Qo) > 0 such that D and E are equivalent; that is, for r > R2 and s between r and Qr, we have (4.20) log D(s) E(s) < S. We set R = maxfRi, R2g. Note also that Lemma 3.16 together with (4.20) implies that D is almost monotone; that is, for s between r and Qr, D(r) <E(r)exp(S) < E(s)exp(S) R\, j r u j 2 is bounded by sup jruj < sup j u j b<îîr (4.22) r(p) jruj b<2îîr < Ci V M " 1 ttr p ) = CQ-2r-2D(2Qr), where C\ = C(n) > 0 comes from Proposition 2.26 and C = C V M ( p ) - n 2 n - 2 . Bounding the normal derivative by the full gradient, and using the weighted area bound for the level sets, (2.15), we see that (4.22) gives a bound for F. For s between r and Çlr, (4.23) F(s) 3 — n b=s du dn jrbj <CnV M D(2Qr). h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 29 We now bound the second line of (4.9) by ^~n R b<s Hess(b2)(ru,ru) 2 s s (4.24) 1 2 + -s E(s) s 2 s u p b<s | r u | 2 E~{s) E{s)-D{sY E(s) |Hess(b2) - 2g\ b R a bound on the above second term. We will now bound the first term in the second line of (4.24). From the monotonicity of E (see (4.21)) and (4.22), we have for s between r and (4.25) s2sup b<s\ru\2 D{mr) < C— < C7exp(ò) , E(s) E(r) where the second inequality follows from (4.20) and the hypothesis (4.12). We use the estimate (4.17) together with (4.15), the Cauchy-Schwarz inequality, and the Bishop volume comparison theorem to bound (4.26) b<s |Hess(b2) - 2g\ < £V n(l) exp(n^) . Putting it all together, we get a bound on the second line of (4.9), for r > R and s between r and Qr, 2 s s l-n R b<s Hess(b2)(ru,r E(s) (4.27) <-[C7exp(Ä)][ÄV n(l)exp(nÄ)] + - [exp(Ä) - 1] . Integrating (4.27) yields Sir 2 s s l-n R b<s Hess(b2)(ru,r Es) (4.28) < [Cyexp(5)][8V n(l) exp(nS)] log^ 0 + 2 [ e x p ( 5 ) - l ] l o g O 0 -C(î) 2 V o ( l ) 7 ^ o g ^ o + 2 [ e x p ( 5 ) - l ] l o g 0 0 . 30 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii It remains to bound the third line of (4.9); this must be done in an integral sense. We have (4.29) Fs) is 2-n b=s u i v b 1 sE(s) D(s) 2s2"n R I f u d V b l - IVbI-1) b=s 1 on 1 VI I I I / + D(s) D{s)-E{s) E(s) F(s) D(s) Using (4.23) and (4.20), and then (4.21) and (4.12), we get D(s) - E(s) E(s) F s D(s) (4.30) <(exp(5) - l)Cjexp(2S) < ( e x p ( 5 ) - l ) C 7 g . A similar application of (4.22), and then (4.21) and (4.12) yields 2-n R I du I b=s IdnI IVbI - IVbI" 1! Dis) (4.31) < Cjs~n ||VbVbl"11 ds exp(25) b=s IIVbI - | V b l _ 1 | d s . 4 < j C T s b=s By the co-area formula, we see that ür s \Vb\ - |Vb| _ 1| ds b=s I V b I 2 - 1 {r<b<Qr} (4.32) r - n / | | V b | 2 - l| {b<îîr} Vol({b < Qr}) < SV ^(l)Qn exp(nS) n < 5vn(i)nn ' 4 where the third to last inequality follows from (4.16), and the second to last from (4.15) and the Bishop volume comparison theorem. h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 31 Combining (4.30), (4.31), and (4.32), we get an integral bound for (4.29), "2F(s) 2s2-n R f e =s|u| 2|Vb|- 1" Sir sE(s) D(s) (4.33) < - C 7 ( e x p ( 5 ) - l ) l o g ^ o + G0 2 C V U n . To control the four terms from (4.28) and (4.33), we first choose l Si = min { - l o g - , e 4 C ^ ) 2 V n ( l ) 7 l o g ^ o (4.34) re + 2 3 \ 2 e[C7V n ( l ) n _ 1 Next, notice that for 0 < s < \ log | , (4.35) e x p s - l = therefore, choose 82 by taking ( l 4 (4.36) £2 = min - l o g - , e 4 2 exp tdt < — s 32 log^o 128 C 7 l o g ^ c Taking S = min{5i, 62}, each of the four terms from (4.28) and (4.33) is bounded by | , and the proposition now follows. q.e.d. Corollary 4.37. Let M n be as in Proposition Jhll and let p G M be fixed. Given a positive constant e and 1 < Çio, there exists R = R(p, 7, e, S7o) > 0 such that if 1 < Ci < fio, r > R, and u is any harmonic function on M satisfying (4.38) D(2Qr) < jD(r) , then for all r < s < t < £ir we have that (4.39) where d = W(Qr). tx 2(l + e)d 32 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Proof. This follows from Proposition 4.11 together with (2.13) and the equivalence of E(r) and D(r) (Proposition 3.3). q.e.d. In light of Proposition 4.11, the lower bounds for the maximum of U from Section 3 can now be used to derive pointwise lower bounds for the frequency on many annuli. Corollary 4.40. (Uniform lower bound of the frequency). Let M be a manifold with nonnegative Ricci curvature and Euclidean volume growth, and let p G M be fixed. Given | > e > 0, we let QL = ^ L(n, e) be given by Corollary 3.29. Given y > 0 and S7o > &L> there exists R = R(p, 7, e, QQ) > 0 such that if r > R, QL < Q < QQ, and u is any harmonic function on M with u(p) = 0 and (4.41) D{2Çl2r) < yD{r) , then for s between Çlr and Çl2r, (4.42) (1 - 3e) < U(s) . Proof. From Corollary 3.29, there exists a R\ = R{p) > 0 such that for r > Ri, (4.43) ( ! " ) < max U(s) . Given S > 0, Proposition 4.11 and Proposition 3.3 yield an Ri = Rzip-i 7> ài ^o) > 0 such that for r > R2, Q2r (4.44) / min{(logW)'(t),0}dt> -8, and for s between r and Q2r, U(s) (4.45) log-6. W{s) Combining (4.44) and (4.45) we get for r < s\ < s2 < Q2r (4.46) U{sl) < esW(sl) < e2SW(s2) < e3SU(s2) . The corollary now follows from (4.43) and (4.46) by choosing 1 1 e (4.47) 5 = - log 3 1 - 3e and R = max{Ri, R2J- q.e.d. h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 33 Definition 4.48. (Almost separation of variables). Suppose that M n is an open manifold, p G M, e > 0, fr < b < Çlr} is an annulus, and u is a function on fr < b < Çlr}. We say that u e-almost separates variables on the annulus fr < b < Çlr} if there exists a function h : R — 7 - R such that for any r < s\ < s2 < £lr, (4.49) fsl<b<s2g b- h(b)u\Vb\ on < eI u(s2) The next goal is to show that harmonic functions with polynomial growth in fact almost separate variables on many large annuli. As in the conical case, we will show that a harmonic function almost separates variables by analyzing almost equality in the Cauchy-Schwarz inequality which implied the positivity of [(logW)']ess. This term is small because [(logW)'] is small and almost equal to [(log W)']ess by Proposition 4.11. We need some preliminary results which now follow. Proposition 4.50. (U almost constant implies u almost separates variables). Let M n have nonnegative Ricci curvature and Euclidean volume growth, and let p G M be fixed. Given e, do, y > 0, and S7o > 1; there exists S = S (do, e) > 0 such that we have the following: there exists R = R(p, 7, do, e, S7o) > 0 such that if 1 < Ci < Çio, r > R, and u is any harmonic function on M satisfying (4.51) D(2Qr) < jD(r (4.52) and (4.53) max U(s) < do , rKsKÇlr U (Sir log U r ) S < 2 then u e-almost separates variables on the annulus r < b < Çlr} in the sense of Definition Jh48. In fact, we can take h(s) f U(s) and choose S > 0 such that (4.54) S exp (6) < — . do Proof. Under the hypotheses, for each S > 0 there exists R$ > 0 such that for R$ < r < s < Çlr, U(s) and W(s) are almost equal and 34 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii W(s) is almost monotone. That is, given S > 0, Proposition 3.3 and Proposition 4.11 guarantee the existence of an R$ = Rs(p, J, S, ^o) > 0 such that for R$ < r < s < Qr, (4.55) log D(s) E(s) < 2 ' and (4.56) |(logW0'(t)- [(logW)' s)]ess|dt< -Note that (4.55) is equivalent to having for r < s < Çlr (4.57) log U(s) W{s) < Further observe that (4.55) implies by Lemma 3.16 that if s < s2 (4.58) D(s) < exp Ç E (s) < exp i E{s2) < exp (5) D{s2) Therefore, by the co-area formula, we have for r < s\ < s2 < fir, fsi<b<s2g f)u b~n b— -U{b)u\Vb\ dn s si s si 2-n b=s du dn du i s s — — IVbl 2 - U(s) u |Vb b=s dn 2 dn +U2(s)s-\Vb-1-2s1-n U(s) u u b=s dn u2\Vb\ s D(s)[(logW)'( sl <^exp(6)D(s2)Z r [(logWy h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 35 <-exp(S) D(s2) (logWy(s) + \exp(5) D(s2) Z r (logW)'(s)- [(logW)'{ (4.59) < 2exp(5) D^) < 2exp(5) D^) W(nr)5 log W(r) 2 U(Qr) 3 / < exp (S) U(s2) I(s2) S < S exp (S) do I(s2 ). The claim now follows by choosing S by (4.54) and then taking R = R$. q.e.d. As an application of the techniques developed, we give now an asymptotic description of harmonic functions with polynomial growth on manifolds with nonnegative Ricci curvature and Euclidean volume growth. Namely, we show that a harmonic function with polynomial growth on such a manifold almost separates variables on an infinite se-quence of large annuli. By improving the proof, we will in Section 10 give a generalization of this for a set of independent harmonic functions. This generalization will be a key step in the proof of Theorem 0.3. Since it is the generalization of Theorem 4.60 given in Section 10, and not Theorem 4.60 itself, that we need in the proof of Theorem 0.3 the reader can choose to skip Theorem 4.60. Theorem 4.60. (Asymptotic description of harmonic functions with polynomial growth). Let M n be as in Proposition 4.11, and u G H d(M). Given Ci > 2 and e > 0, there exists a sequence r j —> 00 such that u e-almost separates variables on the annulus A r^r. Proof. We can assume that u(p) = 0. By the Cheng-Yau gradient estimate, |Vu| grows polynomially of order at most d — 1, and d > 1 if u is nonconstant. Combining this with the Bishop volume comparison theorem, we see that D(r) grows polynomially of order at most 2d. Choose an eo > 0 such that (4.61) e0 exp e0 < 12d Let nL = QL(n, ±) > 2 and Rx = Ri(p) > 0 be given by Corollary 3.29. Then for any r > R\, the maximum of U on the interval [r, ÇÎL r] 36 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii is at least | . Set fin = max{S7, QL}- Choose m > 2n and S > 0 so that (4.62) I exp 25 < - , and log 8d en 4-63 - 2 — < -%-. y ' m - 2 4 Since D grows polynomially of order at most 2d and hmd 4.64 2d < , V ' 2m+l, there is a sequence r i — > oo such that (4.65) D(nlm+1r) < n5 0dm D(r), (cf. Corollary 7.9). By Proposition 4.11 and Proposition 3.3, we get an R2 = R2(p,n5 0dm,8,nlm+1)>o such that for r i > R2, (4.66) Z ° ' min{(logW)'(t),0}dt> - 5 , and for s between r i and ^ " r , (4.67) l D(s) < 6. Set R = max{Ri, R ^ } -From (2.10), we have (4.68) D(nlm r i) < Qldm+n~2 D{r i) . By the definition of U and (2.13), s Dir 2m < (5dm + n - 2) log O0 + 35 <6dmlog(i7o) , h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 37 where the inequality uses (4.62), (4.63), and (4.68). Since U is nonnega-tive, the bound (4.69) implies that for some s i between fiJJr and Çt^m r i we have (4.70) U(s i)<3d. By (4.67) and the 5-almost monotonicity of W on the interval [r i, Çt^m r i\, (4.66), the bound (4.70) implies that for r i > R2, and s between r i and O m r i (4.71) W{s) < 3dexp(25) < d, where the last inequality follows from (4.62). By the 5-almost mono-tonicity of W on the interval [r i, ^ " r ] , (4.66), the choice of fio, and (4.67), we have for r > R and s between f2or i and ^cm r i' (4.72) i<lexp(-25)<W(s), where again the last inequality follows from (4.62). Combining (4.71) and (4.72) yields that for s between f2or i and ^cm r i' (4.73) - <W(s) <4d. Again by the 5-almost monotonicity of W on the interval [r i, Çt^m r i\, (4.66), together with (4.73), we see that for r i > R, there exists an integer 2 < k i < m such that ,4.74) log(W^)<2S+logd<^. ' \W(Sk-lr i ) ' m-'2 2 By (4.67) and (4.74), we conclude that (4.75) l o g f U«k r ) « . . From (4.67) and (4.73) it is seen that for s between f2or i and ^ r i, (4.76) l<U(s)<5d, 8 38 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii 2 U s ds s 0 such that for r i > m a x i ? , R3}, u e-almost separates variables on the annulus f^0 Î_ r i < b < fl0r i}. Finally, we note that when u e-almost separates variables it also does so on all subannuli, and the claim follows since Ci < S7o- q.e.d. 5. Preserving almost orthogonality In this section, we will use the previous work on almost separation of variables to show how to preserve the almost orthogonality condition for harmonic functions on an annulus. The importance of the results of this section is that it will allow us to show that two harmonic functions u and v have a definite separation at b = r provided that: (1) they have a definite separation at b = Çlr, (2) the growth of u and v from b = r to b = Qr has a definite bound, (3) we have good control of v between b = r and b = Çtr. (In fact v needs to be very close to separating variables on the annulus fr < b < Qr}.) We continue to take M n to be an n-dimensional manifold with non-negative Ricci curvature and Euclidean volume growth. Proposition 5.1. (Almost preserving orthogonality). Fix p G M, Ci > 1, and suppose that u and v are harmonic functions on the an-nulus r < b < Çlr}, v S-almost separates variables on fr < b < Sir}, r < s2 f Qr, and (5.2) s2 1~n Z uv\Vb\ = Q. b=s2 h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 39 Then for r < s\ < s2, (5.3) si l-n uvjVbj b=sl < " s 4d+2 I u{s2)I v{s2) where d = max{fev(s) j si < s < s2}, and h v is as in Definition 4.48. Proof. By differentiation, we get d ds A-n uvjVbj b=s (5.4) {l-n)s-n / uvjVbj b=s i_n Z ( du dv Z b=s Os Os l-n Ab + s1 n uv-b=s using equation (2.4), we have d ds A-n uvjVbj A-n (5.5) 2s l-n Vbj du dv v— + u— jVbj us us dv on where the second equality follows from Green's formula together with the assumption that u and v are harmonic. Define err(s) by (5.6) s d ds A-n uvjVbj b=s 2hjs) A-n uvjVbj b=s + err(s) . By (5.5) and the definition (5.6), we get (5.7) jerr(s)j < 2s1"n s u- h v(s) b=s on uvjVbj b=s It follows from the Cauchy-Schwarz inequality that s u- h v(s) b=s on uvjVbj b=s juj b=s dv s— h v(s) vjVbj on (5.8) u 2 j V b j s p - - h v(s) vjVb\ - ^ j b=s b=s dn j jVb ^ u s) Z (s^-h v(s)vjVbj - J - . b=s dn Vb 40 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Combining equations (5.7) and (5.8) gives (5.9) s-'lerrisil2 < 4I u(s) s~n b fs^--h v(s)v Vb l Vbl Integrating equation (5.9), by the co-area formula and the monotonicity of I (specifically I u(s) < I u(t) for s < t), for r < s\ < t < Qr, we have s-1\err(s)\2ds<4I u(t) f b~n b— - h v(b) v\Vb si<b<t} (5.10) < A8I u(t)I v(t) . dv Here the second inequality follows since v 5-almost separates variables on the annulus fr < b < Çlr}. If we now write (5.11) g(s) = s i _n uv\Vb\, b=s we see that by assumption (5.12) g(s2) = 01 and that (5.6) implies (5.13) s\g'(s)\ < 2d\g(s)+ |err(s)|. It remains to get an upper bound for |g(si)|. For ease of exposition, we set (5.14) a2 = 45I u(s2)I v(s2). If g(si)l ^ a) then we are done. Suppose therefore that |g(si)| > a, and let s be the smallest s > s\ such that \g(s)\ = a (such an s < s2 must exist since g(s2) = 0). Replacing g with — g if necessary, we have for si < s < s , (5.15) g(s)>a. From (5.13) and (5.15) it follows that for s\ < s < s , (5.16) s Ulogg)'\< 2d + J y-^ . h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 41 Now integrating (5.16) leads to / x / x s Id |err(s) (5.17) l o g g ( s l ) - l o g g ( s ) < s — + as ds. By absorbing inequality 2xy < X x + A y we get 2 (5.18) err s < 1 |err(s) - 2 a + 2. Using (5.18) and substituting g (s) = a, from (5.17) we obtain si log g(s1) < log a + (2d + 1) log (5.19) si 1 H— a 4 s 1|err(s)| 2ds. Combining (5.19) with the estimate (5.10) gives s logg(si) < l o g a + (2d+ 1) log • s i 1 + ^ a 2 4 5 u ( s 2 ) I 4 s 2 ) log a + (2d + 1) log — + -si 4 Exponentiating (5.19) yields (5.20) 2d+l a exp - , and the result follows since s < s2 and exp 5 < 2. q.e.d. For the applications it is crucial that the S in Proposition 5.1 is chosen small compared with Ci and the growth of u and I v from b = r to b = Qr. Namely given this then Proposition 5.1 implies that u and v are almost orthogonal at b = r in the following sense. Definition 5.22. (Almost orthogonality). Let u and v be harmonic functions defined in a neighborhood of b = s. Given e > 0, we say that u and v are e-almost orthogonal at s if (5.23) A-n uv\Vb\ b=s <eI(s)I(s). 42 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Noting that any linear functional on a Hilbert space is determined by its kernel and its action on any element orthogonal to the kernel, we see that Proposition 5.1 has the following simple corollary. Corollary 5.24. (Almost preserving the inner product). If u and v are harmonic functions and v 8-almost separates variables on fr < b < Sir}, then for r < s\ < s2 < Çlr s2 1~n Z uv\Vb\ - exp 2 Z s h s ds s!1"n Z uv\Vb\ b=s2 si s b=s\ / \ 6d+2 (5.25) < 3 2 < n s I u(s2) I v(s2) , where (5.26) d= max h v(s) . Proof. Orthogonally decompose u into (5.27) u = u\ + av , where (5.28) s2 l-n Z ulv\Vb\ = Q. b=s2 We apply Proposition 5.1 to u\, and use the fact that v S-almost sepa-rates variables to control the remainder. q.e.d. We note that in the applications we will use Proposition 5.1 and not Corollary 5.24. 6. Bounding the number of almost orthonormal Lipschitz functions In this section we will bound the dimension of the space of almost L2-orthonormal functions with a given Lipschitz bound under very general conditions. Definition 6.1. Given (X,d) a compact metric space, define C(X) to be the set of Lipschitz functions on X. We set Ck(X) = fue C{X) | Lip(u) < k}. h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 43 Definition 6.2. (^-almost orthonormal functions). Let (X,fj,) be a measure space with a probability measure, fj,, and suppose that fi, • • -, f m are L2 functions on X. We say that the f i are ^-almost orthonormal if (6.3) and for i / j (6.4) f X f i f j X < 1]. In the next proposition, we think of r as the scaling factor and Do and k as the constants. Proposition 6.5. Let (X,d,ß) be a compact metric space with a probability measure, fj,, and diam(X) < Dor. Given k > 0, there exist at most N — 1 ^-almost orthonormal functions in L kr-i(X), where N = N(Do, k, v) and v is the maximal number of disjoint balls of radius -^. Proof. Let fi, • • -, f m be such functions. We let B\,..., Bv be a maximal disjoint covering of X by balls of radius -r\ x\,..., xv denote the centers of the balls. It follows from maximality that double the balls covers X. We now partition X into v (disjoint) subsets Si,.. .,SU, where B i C S i and S i is contained in twice B i. Let (P, fj,') denote the set of points fXj g with probability measure //, where fj,''(Xj) = [J,(S j). We can therefore identify functions on P with functions on X which are constant on each S j . Since the average of each f i is one and we have bounds on the Lipschitz constant and diameter, (6.6) sup jfj < kDo + 1 X Let A denote the set f s j s G Z , jsj < 10(kD + 1)g- We will now construct an injective map M from the orthonormal set of functions to the set of maps from P (the points fxg) to A: let M (f i) (Xj) G A be any closest point of A to f i ( construction, for all y £ S j , (there are at most two possibilities). By jfy) - M(f i)(x j)j < j f y ) - f ( x j)j + jf i(x j) - M(f i)(x j)j 44 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii and hence (6.8) X jf i-M(f i)j i 2 1 < - io By the triangle inequality together with (6.8), we get for i / j , X jf i-f j j X jM(f i) - M(f j)j (6.9) 1 + X f j-Mf j 5 Furthermore, since the f i are ^-almost orthonormal, we have (6.10) and for i / j , (6.11) Consequently, for i / j , (6.12) X jf i f i f j X 1 < 2 -1 < jf i - f j X Combining (6.9) and (6.12) yields that for i / j , (6.13) 0 < ^ < (X jM(f i) - M(f j)j Hence, M is injective. The proposition follows by counting the cardi-nality of the set of maps between two finite point sets (in fact, N < ( 2 0 ( k D ) + l ) + !)")• q.e.d. Remark 6.14. (Divergence of eigenvalues). Given a gradient bound C(A) for all eigenfunctions with eigenvalues at most A, that is (6.15) sup jruj < C(A) sup juj X X h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 45 as is the case in the Cheng-Li-Yau gradient estimate (see , , and ), Proposition 6.5 gives a definite rate of divergence of the eigen-values (compare and Weyl's asymptotic formula). In applications X = fb = r} and the functions will be the restriction of harmonic functions on M with bounded growth. Moreover the restriction of the harmonic functions will be approximately eigenfunctions with eigenval-ues given in terms of the frequency. Due to the fact that these functions are restrictions of harmonic functions on M we have a gradient bound already on M. This bound is given in terms of the frequency. For this reason we need not deal with spectral properties of X. 7. Growth properties of functions of one variable In this section, we will prove some elementary results for functions of a single variable with polynomial growth. The first two results (Lemma 7.1 and Corollary 7.9) show the exis-tence of infinitely many annuli with bounded growth. The basic idea is that for any set of 2k functions with polynomial growth of degree at most d, we can find a subset of k functions and infinitely many annuli for which the degree of growth from the inner radius to the outer radius of each of the functions in the subset is at most 2d. We will think of this elementary fact as a weak version of a uniform Harnack inequality for a set of functions with polynomial growth. This simple idea of restricting attention to a large subset in order to make the constants independent of the number of functions in the set will be used over and over again. In the next section, we will produce functions of one variable with the properties of the functions of this section. The main results of this section are Corollary 7.9 and Corollary 7.21. Whereas Corollary 7.9 will be used to start the proof of Theorem 0.3 (see Corollary 8.14), and Corollary 7.21 in the inductive step in the proof of Theorem 0.3; see Section 9. Lemma 7.1. Suppose that fi,..., f; are positive nondecreasingfunc-tions on (0, oo) such that for some d, K > 0 and all i, (7.2) f i(r) 1, k < l, and any C > Ql~k+1, there exist k of these functions fai,..., fak and infinitely many integers, m > 1, such that 46 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii for i=l,...,k, (7.3) fai(Qm+1)<Cfai(Qm). Proof. We will show that there are infinitely many m such that there is some rank k subset of ff i}, where the subset could vary with m, satisfying (7.3). This will suffice to prove the lemma; since there are only finitely many rank k subsets of the l functions, one of these rank k subsets must have been repeated infinitely often. Set l (7.4) g(x) = Y f i(x); i=l note that (7.5) g{r) < K l(r d + l)l and g is a positive nondecreasing function. Assume that there are only finitely many such m and let m — 1 be the largest. Then for all j > 1 we have (7.6) g(ttmo+j) > C l-^ginm^-1) . Iterating this gives (7.7) g(ttmo+j) > C j(l-k+1ïg{Qmo) . From the upper bound on g, equation (7.5), we have for all j > m that (7.8) c (Qj)dl > C j(l-k+1)g(Qmo) , ld where c = c(l,m0,Q,K). Since C > Ql-k+i and g(Qm°) > 0 this is impossible, yielding a contradiction. q.e.d. Corollary 7.9. (Weak version of a uniform Harnack inequality for a set of functions with polynomial growth). Suppose that fi,..., fik are positive nondecreasing functions on (0, oo) such that for some d , K > 0 and all i, (7.10) f i(r) 1, there exist k functions f a i , . . . , fak and infinitely many integers, m > 1, such that for i = 1 , . . . , k, (7.11) fai(ttm + 1) 1, suppose that f is a positive nonde-creasing function on [r, Çl'm r] such that for some do > 0, (7.14) f(fìm r) < Qd°m f{r) . Then for d = d m m1; there exists some j with 0 < j < m — 2 such that (7.15) f(j+1r) < nd f(üj r) and (7.16) f(j+2r) < Q2d f(Qj r) . Proof. Suppose that the lemma is false; then for every j , we have either (7.17) f(j+1r) > Qd f(Qj r) or (7.18) f(j+2r) > n2d f(nj r) . In particular, either for k = 1 or for k = 2, we must have that (7.19) f(Qk r) > Qkd f(r) . Continuing inductively, by (7.17) and (7.18) we get (7.19) for k = m or for k = m — 1. Hence the monotonicity of f, together with (7.19) implies that (7.20) f(ttm r) > ^m-^d f(r) . 48 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii By the assumption (7.14), and the definition of d, (7.20) yields the desired contradiction. q.e.d. Lemma 7.13 has the following easy corollary. Corollary 7.21. (Double growth condition). Given Ci > 1, suppose that fi,..., f km are positive nondecreasing functions on [r, Çl'm r] such that for some do > 0, and all i = 1 , . . . , km, (7.22) i ( ^ m r ) < d m i(r). Then for d = d m m 1, there exist k functions f a i , . . . , fak and some j with 0 < j < m — 2 such that for i = 1 , . . . , k, (7.23) fai(j+1r) 1. The reason for this is that in the inductive step of Theorem 0.3 we will need to find an annulus and a subset f f a g of ff i g such that these fai have almost the same degree of growth on this annulus as on a certain larger annulus, and uai has bounded frequency (the bound must be uniform in terms of the polynomial rate of growth of uai). To achieve this, we apply Corollary 7.21 to find a pair of annuli one contained in the other and so that we have controlled growth on both annuli for fai. The bounded growth on the larger annulus and the almost monotonicity of the frequency then imply the desired frequency bound for uai on the interior annulus; see Section 9 for further details. 8. Constructing independent harmonic functions with good properties from given ones In this section, given a linearly independent set of functions in H d we will construct functions of one variable which reflect the growth and h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 49 independence properties of this set. In particular, here we shall establish that these functions of one variable satisfy the conditions of Section 7. The results of this section rely heavily on the properties of harmonic functions on manifolds with nonnegative Ricci curvature; we use in par-ticular that u is monotone nondecreasing for all harmonic functions. In Section 10, we will use these results to show that given linearly in-dependent harmonic functions with polynomial growth we can produce annuli and harmonic functions on these annuli which are separated and have controlled growth. We begin with two definitions. In the first definition we construct the functions whose growth properties will be studied. Definition 8.1. (w ijr and f i). Suppose that u\,..., u k are linearly independent harmonic functions. For each r > 0 we will now define an orthogonal basis w i<r with respect to the inner product (8.2) rx~n Z uv\Vb\, b=r and functions f i . Set wi<r = w\ = u\ and fi(r) = I ul(r). Define w i<r by requiring it to be orthogonal to u j for j < i with respect to the inner product (8.2) and so that on fb = r} we have i-l (8.3) u i = X Xji(r)u j + w ir . j = 1 Set (8.4) i(r) = r 1-n Z w ir|Vb| . b=r Definition 8.5. (Barrier). We will say that a function f is a (left) barrier for a function g at r if f(r) = g(r) and for s < r, f(s) < g(s). We will use the barrier property to conclude that the growth of g from s to r is not larger than the growth of f from s to r (cf. Remark 8.16). In the next proposition, we will establish some key properties of the functions f i from Definition 8.1. Proposition 8.6. (Properties of f i). If u\,.. .,u k G H d(M) are linearly independent, then the f i from Definition 8.1 have the following 50 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii three properties: There exists a constant K > 0 (depending on the set fu i}) such that (8.7) 2d f i{r)<K{r d + l), f i is a positive nondecreasing function, (8.8) and (8.9) f i is a barrier for I w ir at r. Proof. Note first that (8.10) f i(r) < I u i(r) . Furthermore, for s < r i-l i(s) A-n (8.11) s r 1-n 1-n b=r E^ j = 1 i-1 EAj j = 1 i-l E^ j = 1 i(s)u j i(r)u j i(r)Uj IVbI IVbI IVbI = f r ) , where the first inequality of (8.11) follows from the orthogonality of w i<r to u j for j < i, and the second inequality from the monotonicity of I for harmonic functions (see (2.12)). Since u i are linearly independent, by (8.11) we get (8.8). Using (8.11), we also see that f i is a barrier for I w i r at r; this shows (8.9). Finally, we shall verify (8.7). It follows from the asymptotics of the Green's function that r is bounded and therefore Ui G H d implies that there exists a constant K such that |u(x)| < K(b(x)d + 1). Using the C° bound on u i and the weighted volume bound for the level set b = r, (2.15), we get (8.12) f r ) <I u i(r) < K 2(r d + l)2I u^(r) <2K2(r2d+l)nV n(l). h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 51 If we set K = 2K2nV n(l) then we obtain (8.7). q.e.d. Although we will not use it, we note that since log f i is a barrier for log I w i, we also get that (8.13) (log i ) r > (log w > ) = 2 U r . The following corollary of Corollary 7.9 and the properties of the f i will be used to get initial control of the growth in the proof of Theorem 0.3. Corollary 8.14. Suppose that u\,..., u ^k G H d(M) are linearly in-dependent. Given Ci > 1, then there exist a subset fai,..., fak and infinitely many m such that for i = 1 , . . . , k (8.15) fai m+1)<rtd fai(nm). Proof. This follows immediately by combining Corollary 7.9 and Proposition 8.6. q.e.d. Remark 8.16. We will continue to work with the functions f i. However since the way they are defined is a bit abstract, we will try to clarify their usefulness by explaining a particular consequence of Corol-lary 8.14. The reader should note however that this consequence will not be used later on, and rather we will need to use more of the information that the functions f i carry. Given a set of 2k linearly independent functions fug of H d, Corol-lary 8.14 allows us to find infinitely many m for which there exist k orthonormal (at b = Çlm+1) harmonic functions (in fact in the span of fu i g). Further, these k functions have growth of degree at most 2d on the annulus between b = Qm and b = Qm+1. Note however that for different m the set of harmonic functions with growth of degree at most 2d may be different. That is, from Corollary 8.14 together with (8.9) we have the following: Under the assumptions of Corollary 8.14 we have infinitely many m such that for i = 1, • • • , k, ( 8- 1 7) w inm+r (^ m + 1) < V4d I w ai t nm + 1 (^ m), and for 1 < i < j < k (8.18) Z waÇim+iwaÇim+ijrbj = 0. b=üm+1 52 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Note also that if we could show that these k harmonic functions were orthogonal at Çtm (and not at Çtm+l) then, after applying the gradient estimate together with the meanvalue inequality, Theorem 0.3 would follow immediately from the results of Section 6. 9. Towards the inductive step In this section we will use the results of Sections 7 and 8 to show a result (Proposition 9.1) that will be used in the inductive step of Theorem 0.3. Given a large annulus and a set of independent harmonic functions fug such that the corresponding functions f i (see Section 8) grow poly-nomially of order at most do on this annulus, we show how to get (a) a subset ai, (b) a subannulus, (c) a nonconstant harmonic function u in the span of fUj jj < a\g, (d) a constant d > do, and (e) a larger subannulus, such that on this subannulus (b) (1) fai grows polynomially of order at most d, (2) i < U u < 2d on double the subannulus, and (3) U u is almost constant, and on the larger subannulus (e) (i) u is orthogonal to fUj j j < a\g at the outer radius, and (ii) u grows polynomially of order at most d. It will be important that we will be able to take d very close to do and U u very close to being constant if we are willing to go to a relatively small subset of the functions and a relatively small subannulus. In the applications, Proposition 4.50 together with (2) and (3) will allow us to conclude that this u is very close to separating variables on this subannulus. This together with (i), (ii), and Proposition 5.1 will allow us to conclude that the harmonic functions that we define inductively in this way are almost orthogonal on a subannulus. We will now make this precise in the following proposition. Proposition 9.1. (Towards the inductive step of Theorem 0.3). Let m > 5, m > 9, Q > f2L(n, | ) > 2, and do > 1 be given. Here Î7L(n, | ) h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 53 is given by Corollary 3.29. Set (9.2) d = d m m m - Im A \m There exists R = R(p, m, do, Ci) > 0 such that if r > R and fi, • • -, f l are as in Definition 8.1 where u i(p) = 0 and (9.3) f i{Qmm r) < Qdomm f i{r) for all i, then we have the following: There exist l functions f a i , . . . , fal and integers h and j with 0 < h < m — 2 and mh < j < m(h + 1) — 1 such that for i = 2 , . . . , l, (9-4) fai(j+1r)<nd fai(j r), and setting u = wa tim(h+2)r, we have for Çtj r < s < Çtj+2r, (9.5) 2 <U u{s) < 2 d : (9.6) and (9.7) I u(nm {h +2r) < n2dm I u(nmh r), U u{j+1r) log' U u(j r) log(5d) Proof. First, we apply Corollary 7.21 to get lm functions fß1,..., fßlm such that for some h with 0 < h < m — 2 and i = 1 , . . . , lm, (9.8) and (9.9) where d f i(nm(h+1)r) < ndim f i(nmh r) f i (fi m(h+2) mh ) < Q2^ m f i(Qmh r) mjdo. Note that we will only use (9.8) for i > 1 and (9.9) only for i = 1. Set u = w^ tim(h+2)r and a\ = ß. From the barrier property, (8.9), it follows that (9.9) implies (9.10) I u(Çm(h+^r) < Çl2dlm I u(Çmh r) < Çl2dm I u(Çmh r) . 54 t o b i a s h. c o l d i n g & william p. minicozzi ii In particular by Proposition 3.36 (noting that Q > 2), there exists R0 = Rjp) > 0 such that for r > RQ, (9.11) D u(nm(-h+^-1r) < C2Q2dlm D u{tmh+lr) < C2Sl3d°m D u(nmh+1r) where C2 = C2(n) > 0. Set J l 6 1 9 logf e = min < — log —, — log —, =• | 3 5 ' 4 8 ' 2 m ^ (9.12) By Proposition 4.11, Proposition 3.3, and (9.11) we can choose Rx = Rx{pi C2n3d°m, e, Cl2m~2) > R0 so large so that for r > Rx and Qmh+lr s Qm(h+2)-2r (9.13) and (9.14) log DJs) E J s) < e çim(h + 2)-2r ümh+1r minjlogW u)'(t),0}dt > - e . Note that (9.13) is equivalent to that for Çlmh+lr <s< fìm(h+2)-2r U u s) (9.15) log < e. W u s) From (9.14) wehave for Qmh+1r <s<t< fìm(h+2)-2r, (9.16) W u s)<eW u(t), which together with (9.15) implies that for Çmh+lr < s <t < fìmCh+2)-2r (9.17) fu(s) < e W u{s) < e2'W u{t) < e3'U u{t) . Since u is nondecreasing, from (9.10) it follows that I unm(h + 2)- 2r) <I u Q ^ h + 2 r ) (9.18) <Çi2dlm I uÇmh r) <Cl2dlm I u{0m(h+lr) . h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 55 By (2.14) and (9.18), there exists some so with Çtm(h+l)r < s < Qm(h+2)-2r such that (9.19) m 5 U u(s0) < -di < -di . m — 2 3 Combining (9.17) and (9.19) we see that, for Çlmh+1r <s< Qmh+^r, (9.20) U u(s) < e3eU(s0) < e 3 e | d ! < 2dt. By Corollary 3.29 we can choose R2 = R2(p) > 0 so large that if r > R2 then there exists a si satisfying mh--l mh+1 Qmh+ir <sl< nL(n, -)ttmh+"r < Q o mh+2 with (9.21) 5 < U u(sl) We now set R = max{Ri, R2J- By (9.21) and (9.17) we have that for r > R and Çlmh+2r < s < Çlm(h+i)r (9.22) \ < e-3fl < e-^U(sl) < U u(s) . Combining (9.20) and (9.22) we get that for Qmh+2r < s < Çmh+^r, 1 (9.23) <U u{s) < 2 d i . Note that (9.23) implies that for Qmh+2r <s<t< Qm(h+1)r, U u(s) (9.24) log U u{t) < log(4di Consider the (m — 3) subintervals given by Qmh+j r to Qmh+j+1r for j = 2 , . . . , m — 2. From (9.15) and (9.24) it follows that there exist 56 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii at least (m — 3 — m) subintervals on which the variation is less than ^ ( M i ) . That i m-2 X j = 2 log W u(Q mh+j+l m - 2 j = 2 m-2 W u(Q mh+j+l W u(nmh+j r) W u{Qmh+j+l m — 4 (9.25) < log j = 2 W uiQmh+j r) -,o < log W u(fìmh+2r) nm(fe+l)-lr 2 Z minf(logW u)'(t),0gdt îî?ftfe+2r U u(nm(h+1)-1r) U u(Qmh+2r) + 4e < log(4di) + 4e 9 9 < l o g ( 4 d i ) + l o g - = l o g - d i By (9.15) and (9.25), there exist at least (m — 3 — p o ) many j between 2 and m — 2 such that log U u(îî hm+j + l U u(Qhm+j) < log W u(n hm+j + l (9.26) < + 2e < + 2e logddo logddo^logf m log(5di) m We will call such intervals good. We now consider the restriction of the ffßtg to the union of the good intervals. By (9.8), these restricted functions grow with exponent at most equal to d2 d. Again applying Corollary 7.21, this m — 3 — \m time to the union of these good intervals and the restrictions of the functions, fß{, we get a j with mh < j < m(h + 1) — 1 such that (9.27) log u(Q-j + i U u(j r) < log(5di h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 57 and l — 1 functions f a 2 , . . . , fal such that for i = 2 , . . . , l, (9.28) fai(j+1r) 0 and all i > 1 (9.29) Z w0iümih+2)r w0iys\Vb\ = 0. b=s This follows since wa tim(h+2)r lies in the linear span of fu k \k < a\g and waits is orthogonal to u k (k < a{) at b = s for i > 1. Note also that Ro, Ri, R2 (and hence R) are independent of l and also of the particular harmonic functions. The key for applications of Proposition 9.1 is, for given l and Ci > 2, to choose m and m so large that the degree of growth, d, of the functions fa2i '••)f«l from the inner radius b = r (= Çtj r) to the outer radius b = Çr is not much larger than do. In fact, in the applications, the more times that we need to iterate this step, the closer that d needs to be to do. Further, we will choose m and R so large that U u is almost constant (i.e., almost separate variables) on the annulus between b = r and b = Qr. Here if Ci is large then u has to be even closer to separating variables, so m needs to be even larger. The reason for this is that we want to apply Proposition 5.1 to get a definite separation at b = r. In Section 10 we will need to keep close track of these relationships. 10. Harmonic functions with polynomial growth As before, let M be an n-dimensional Riemannian manifold with nonnegative Ricci curvature and Euclidean volume growth, and let p G M be fixed. We are now prepared to prove the main theorem. After some pre-liminary remarks, the proof will consist of three steps. First, we will find annuli and a subspace of the harmonic functions with polynomial growth such that a basis for this subspace has controlled growth on these annuli. This step relies mainly on the properties of the functions f i constructed in Section 8 and the general properties of functions of one variable with polynomial growth. Next, we will construct a set of harmonic functions contained in this subspace which have controlled growth, almost separate variables, 58 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii and are pairwise almost orthogonal on a subannulus. We accomplish this through repeated applications of Propositions 9.1, 4.50, and 5.1. In essence, this step gives an effective version of the finiteness theorem, and it is here that we strongly use the results on the frequency function (and thus the Euclidean volume growth assumption). Finally, we will use the uniform bound on the growth and the mean-value inequality to get a Lipschitz bound for these harmonic functions on a subannulus. Proposition 6.5 gives a bound on the number of such functions, and the theorem will then follow since we can use this to control the number of functions that we started with. Proof. (Theorem 0.3). Fix Q > max{i7^(n, | ) , 4}. Here i7^(n, | ) is given by Corollary 3.29. Set ( \ — - j 16C'nd432d, where C = C(n) > 0 is the constant occuring in the meanvalue inequal-ity, Proposition 2.26. By (2.23) we can choose RQ = Ro(p) > 0 so large that for r > RQ, (10.2) If r > R then b l o g -r 2 < l o g p (10.3) X = {b=^r}cAn^ p (p). The relative volume comparison theorem, , , together with (10.3) imply that there exists an integer v = z/(n, k) such that there exist at most v disjoint balls of radius -^ with centers contained in X. We think of X as a metric space with distance function given by the restriction of the Riemannian distance on M. Let ß be the probability measure on X given by (10.4) fi(A) = Afr)1~n Z |Vb| . Note that the normalization in (10.4) comes from the fact that I\ = n V M) that is (2.15). Applying Proposition 6.5 to X with the probability measure /i, we get a constant N = N(k, n) such that for f] =\ any set of ^-almost orthonormal functions with Lipschitz bound j r o n X has at most 1 elements. h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 59 We will show that if dim H > C, where C = C(N), then for all R > 0 we can find an r > R and N ^-almost orthonormal functions with Lipschitz bound -^ on X. This contradiction yields the result. Choose integers m > 5 and m > 9 so large that (10.5) (-m f m = ) <2, and (1„.6) e x p f 2 p N ) < 2 . Further, let (?«i)i=i,...,N satisfy TOi > m, and set (10.7) (i = 1 d log< 4d ) Observe that (10.6) implies (10.8) — exp — < — . v ; 8d 8d - 4d To simplify notation, define inductively N\f, A/N-i> -A/N-2) • • • >Ni, N , by (10.9) Further, for i = 0 , . . . , N — 1, set (10.10) V = 0, N ; = (N i+i + 1) m + 1 m . M i = m N-iUN i+1m j , and (lo.ii) üi = nM. Finally, for i > 1 let {-\ n 1 o\ 2 o f~208 diht--4iht (10.12) i = 8ei_1ili Note that fìi m° m% depend not on m'i but only on m j for j > i. On the other hand e i i depends only on m - i . These two facts will allow 60 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii us to choose m'_i large so that Ti < r], where i] < ^- That is, choose m > ^m-i) • • • ) m1 inductively so large that Ti < TO Observe that this implies that if Ctm% is large then ei_i must be small. The numbers TOi are now fixed, as are the quantities N i, M i, fii, ei, Ti which are defined from the m i. We will show that we can take C(N) = 2N + 1- To see this, suppose that u = 1, u i , . . . , u2N G H d(M) are linearly independent. We may assume that u i(p) = 0 for all i > 0. Given this set of harmonic functions, we will now proceed, for all R > 0, to construct an r > R and a set, v i , of TO-almost orthogonal harmonic functions on the annulus fr < b f g lrg. In addition each i will ei-almost separate variables on an annulus, fr i < b < ^r i g, containing the annulus, fr < b < Çlrg. In fact, for any j with 1 < j < N the functions i with i < j will be pairwise ^-almost orthogonal on the larger annulus fr j < b < Qj r j g D fr < b < Çlrg. Note that in order to show that i and v j are ^-almost orthogonal (with Tj < r]) on fr j < b < Qj r j g if i < j we will need i to ei-almost separate variables on fr i < b < Cli rg D fr j < b < Çlj r j g. Since Qi-i is much larger than fij_i if i < j we will need i to be much closer to separating variables than j ; cf. Section 5. Note also that for different r the i may be different. It will be possible to do the following construction for infinitely many annuli; however, in the end we need only do our construction on a (single) sufficiently large annulus. As in Definition 8.1, we set (10.13) fi = I ul, (10.14) wl,r = ul, and let f ^-, • • •, f2N and w2,r, • • •, w2N,r be as in that definition (with respect to u\,..., u2N)- These f i and w i<r will be fixed from now on. The first step of the proof consists of finding a sequence of annuli where a subset of the functions, f 1 , . . . , f 2 N ) has controlled growth. To get the initial control, we apply Corollary 8.14 to get a subset f a i , . . . , faM and infinitely many integers jo such that for i = 1 , . . . , N , (10-15) fat(Qj+1)<Qtd f a j ) -Fix such a jo, and set ro = Çij0°. The next step is, for any such jo which is sufficiently large, to in-ductively construct an independent set of harmonic functions which are h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 61 pairwise ^-almost orthogonal and whose growth is controlled. At the i-th stage of the induction, we will be working with N i functions and i independent harmonic functions on an annulus r i < b < ^r i g C fr i-i < b < i_ir i_ig C • • • C fri < b < Çiirig f fr0 < b < ïï0r0g. These N i functions will grow at most like "id i on these annuli, where for i = l,...,N, ;io.i6) d = 2d m TTi n j = i m j — 4 m — 1 Note that from the choice of m and m (see (10.5)), we have (10.17) di < d2 < ••• < d N < Ad. Applying Proposition 9.1, we get a R\ = Ri(p, mi, 8d, Ç}) such that if jo is large enough to ensure that '10.18) r0 = W0j > Ri , then there exist N \ + 1 functions fß1,..., fßK +1 and integers h\ and j \ with 0 < h\ < m — 2 and m \ h \ < j \ < m{h\ + 1) — 1 such that for i = 2 , . . . , N i + l, (10.19) j l + i 2d fß,(jrLr0)<n^fßt(^r0), for fìj r0 <s< Qj1+2r0 (10.20) and (10.21) - < U v1 (s) < 4di < 16d, log U vi(nj1+1r0) U v î î j ro log(10di m i log(40d) ei m i 16d Here v\ = w m1(h1+2) • Set r\ = f j r - Note that ßi&i Q\ri < nm l{h l+2)r0 < n0r0. 62 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii The frequency bounds (10.20) combined with (2.9) and (2.14) give the following bound for the growth of D vl, D v 1(î2Ïr 1)=Cv 1(î2Ïr 1)v 1(î2Ïr 1) (10.22) <lQdI vl(Q2 iri) < l%dm\d I vl{ri) <32d^4d D vl(ri) . Now by Proposition 4.50 we have that (10.20), (10.21), and (10.22) together with (10.8) give the existence of an Rx = Rx{p, 32dn6 1 4d, 16d, ei, üj) > R0 such that if (10.23) r0 = Slj > Ri , then v\ ei-almost separates variables on the annulus fr\ < b < Çi\ri. We proceed inductively. Again by Proposition 9.1, we get a R g R2(p,m2,8d,ÇÎ2) (in fact, R2 = R\ will do) such that if jo is large enough to ensure that (10.24) r0 = nj0° > R 2 , then there exist N2 + 1 functions f 1 1 , . . . , flAf , where f i ^ f f 2 ) ' ' ' ) f t v i + l g' and integers h2 and j2 with 0 < h2 < m — 2 and m2h2 < j2 < m2(h2 + 1) - 1 such that for i = 2 , . . . ,N2 + 1, (10.25) f»(^2j + 1ri) < ^ f j ri) , for fìj r < s < Qj2 2+2ri (10.26) - < U v{s) < 4d2 < 16d, I vJQm(-h2+^r) <Q4d2m I vJQmh2r) h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 63 and '10.28) j + 1 log U v { ^ r i ) log(10d2 m 2 log(40d) e2 m 2 16d Here v2 = w m(h2+2) . Set r2 = fi2j ri. Note that (10.29) ri < 0m h ri < r2 < ^2r2 < ^r2 < Q m h m(h2+2) rl < ^ l r l Using Proposition 5.1 we will now show that v\ and v2 are 772-almost orthogonal on the annulus fQmh2ri < b < ttm{h2+2]rj D fr2 < b < Q2r2}. By definition, v\ is a linear combination of u\,--- ,u1 and at b = 0 m { h 2 + 2) (10.30) ri, v2 is orthogonal to all Ui with i < ji; therefore v1v2IVbI = 0 . b=nm{h2+2)ri Note also that by (2.14) and (10.20) we have (10.31) I v1 (ttm{h2+2]ri) < Qm64d I vl (Qmh2ri) . Since v\ ei-almost separates variables on the annulus, from Proposition 5.1, (10.27), (10.30), (10.31) it follows that m h m(h2+2)r | and by (10.20), we get for Qmh2r < s < ttm{h2+2) r i A-n v\v2\Vb\ b=s < 8eiO (64d+2)2m I / 0 m ( h + 2) )r1)v(n m ( h 2 + 2) v2 L 2 r) (10.32) < 8ei ^(64d+2)2m nl4dm I v l(nmh 2r i) Q1 2 6dm2I v(Qmh 2r 1) 64 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Here the last inequality follows from the monotonicity of I; that is, for ÇImh r\ < s we have I i(Qmh r) < I v i(s). This proves that v\ and v2 are ^-almost orthogonal on all level sets in the annulus fQmh2ri < b < ttm{h2+2]rg D fr2<b< Q2r2g. The frequency bounds (10.26) combined with (2.9) and (2.14) give the following bound for the growth of D v , (10.33) D v {Q2 2r2) = U v {tt2 2r2)I v {Q2 2r2) (10.34) < lQdI v(Q2 2r2) < ldÇifd I v{r 2) < 32dn6 2 4d D v (r2) . Now, as above, by Proposition 4.50 we have that (10.26), (10.28), and (10.33) together with (10.8) yield the existence of an R2 = R 2(p, 32d^ 4d, 16d, e2, Q2 2) > R0 (in fact R2 = R\ will do) such that if (10.35) r0 = Qj0° > R2 , then v2 62-almost separates variables on the annulus fr2 < b < Çl2r2g. For each j 0 satisfying (10.15), (10.18), and (10.23), after N stages we are left with N linearly independent harmonic functions v\,..., v N on the annulus fr < b < £lrg, where (10.36) r = r N = nj N •••Cij . Note that [r1: Qiri] D [r2, ^ 2r 2] D • • • D DN, ^IN^N] = [r, ^r]-On the annulus, fr < b < £lrg, these harmonic functions, f i g, must: (a) have U bounded by 16d, (b) ei-almost separate variables, and, most importantly, (c) be pairwise ^-almost orthogonal at all level sets. The last step is to get a gradient bound, and hence a Lipschitz bound, on X for these functions. This will allow us to apply the results of Section 6 to deduce a contradiction. To get the gradient bound, observe first that the uniform bound on U v i on the interval [r, Qr], (10.37) U v i(s)<16d, h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 65 implies by (2.14) that (10.38) I v i(Qr)<4d2d I Q -r 4 Furthermore, (10.37) yields that (10.39) D v i (Or) < 16dI v i (Or) . By (10.2) together with the meanvalue inequality, Proposition 2.26, we get a constant C = C{n) > 0 such that for r > R 2 / ^ Z 2 sup j r v i j < —— B j r v i j (10.40) p 2 " ^ Q r 4 ^ CD i(Qr) ß (Or)2V M Combining (10.38), (10.39), and (10.40) leads to the gradient estimate n ~ ' 4 \ ~ CD v i(Qr) sup jrv i j < B p V3 '10.41) < 3 (Or)2V M 4 \ n l6CdI v i(Qr) 3j (Or)2V M 4 ^ n 1 6 C d 4 3 2 I f r 3 (Or)2V M We now normalize the i to get N ^-almost orthogonal harmonic functions v i , . . . , v ^ on fr < b < Org with I i(^r) = n V M for i = l,...,N. From (10.41) it follows that (10.42) sup j r ^ j 2 < ( - 1 6 C n d 4 3 d ( O r ) " 2 . B p 3 Finally note that by the triangle inequality we have that if x,y G fb = j r g C B p r then the minimal geodesic (in M) between x and y 2 / 3 lies entirely inside B p . Since the v; are ^-almost orthonormal at b = -^r and satisfy the Lip-schitz estimate which follows from the gradient bound (10.42) together with the the above application of the triangle inequality, we can apply Proposition 6.5 to obtain the theorem. q.e.d. 66 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Remark 10.42. (Conical case). If M is C(N) (as it is in Section 1), then U is monotone nondecreasing by Lemma 1.18. Given a set of independent harmonic functions fug C H d, at r = 1 we can extract an orthonormal basis i for the space spanned by the Ui. By Lemma 1.28, the frequency of i (G H d) is uniformly bounded by d. Integrating this out to r = 2 leads to a uniform bound on I v i (2) and hence, given the bound on U v i, we get a uniform bound on D v i{2). By the Li-Schoen meanvalue inequality, we obtain a Lipschitz bound for r < 1 for the independent functions. Proposition 6.5 now yields a bound on the di-mension of H d(C(N)) just in terms of d and the lower bound on the Ricci curvature of N. In contrast to the results of Section 1, this bound is not sharp. 11. Examples In contrast to the Euclidean case, it is possible for M with non-negative Ricci curvature to admit harmonic functions with nonintegral rates of growth (cf. Example 1.70). Even if M is Ricci flat and Kahler, examples exist. Example 11.1. (Tian-Yau, ). There exist Ricci flat Kahler manifolds with Euclidean volume growth which have harmonic func-tions with growth strictly between one and two. We note that there are manifolds with positive sectional curvature which admit no nontrivial harmonic functions with polynomial growth. To our knowledge, no such example has been constructed with Euclidean volume growth even under the less restrictive assumption of nonnegative Ricci curvature; see for more on this. Example 11.2. There exist manifolds with nonnegative Ricci cur-vature (in fact, positive sectional curvature) which admit no nontrivial polynomial growth harmonic function. In fact, one may round off a metric of the form dr2 + r2ad92, where a < 1 and d92 is the standard metric on S n _ 1. In this case, since dB r(p) is connected, if u G d, then by the max-imum principle there exists x G dB r(p) with u(x H u(p). Integrating along curves beginning at x, we obtain u = u(p) by using the gradient estimate together with the facts that u has polynomial growth and that diam dB r(p)/r —> 0. This example was observed by Kasue in . h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 67 Example 11.3. (Klembeck, , and cf. ). In the holomorphic case, there exists a Kahler metric on C n of positive sectional curvature and quadratic curvature decay which does not admit any nonconstant holomorphic functions with polynomial growth. The next example reveals some of the difficulties in the general case compared with the model case of a cone. It shows in particular that unlike the model case of a cone the frequency of a harmonic function on a manifold with nonnegative Ricci curvature and Euclidean volume growth may not be monotone. Example 11.4. (). There exist manifolds with nonnegative Ricci curvature, Euclidean volume growth, and quadratic curvature decay which admit harmonic functions with polynomial growth whose fre-quency oscillate between two different numbers. Let us explain the idea behind Example 11.4. From Section 1 (The-orem 1.66) we know that for a cone the order of growth of u G H d (on a large annulus) is given in terms of an eigenvalue of the cross-section (see (1.6)). If we consider a manifold which on a large annulus looks roughly like an annulus in a cone centered at the vertex, then the growth of such a u will be given almost in terms of an eigenvalue of the cross-section of the cone. By changing the cross-section slowly into a different cross-section (see ) which is not isospectral to the original one we can change the growth of u. Oscillating back and forth between two non-isospectral cross-sections gives a harmonic function with polynomial growth on a manifold with nonnegative Ricci curvature, Euclidean volume growth, and for which the order at infinity is not well defined. We refer to for an extensive discussion of examples of manifolds with nonnegative Ricci curvature and nonuniqueness of tangent cones at infinity. Examples of manifolds with nonnegative Ricci curvature, Euclidean volume growth, quadratic curvature decay, and for which the tangent cone at infinity is not unique were first constructed by Perelman, . We end this section by showing that if one makes a small pertubation (in an appropriate norm) of the metric, then dim H remains unchanged. In fact we have the following proposition which says that any polynomial growth asymptotically harmonic function lies within a bounded distance of a harmonic function (this new harmonic function will of course be forced to have the same rate of growth). Proposition 11.5. Suppose that Ric M n > 0 and Vol(B r(p)) > Vr n 68 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii for some V > 0 (here we assume that n > 3). Suppose also that u is a smooth function on M, and u and | r u | have polynomial growth. If in addition (11.6) |Au| < f(r) for some bounded, integrable (on R n), and nonnegative function, f, then (11.7) lim e tAu(x) t—>-oo exists for all x, and further :n.8) \e tAu - uWoo < C(n) / f(s)s n-1ds + sup f o [0,oo) Proof. Since e t u(x) = R M H(x, y, t)u(y)dy, we have that d Z d — (e tAu)(x) = —H(x,y,t)u(y)dy AH(x,y,t)u(y)dy. M Integrating by parts gives (11.9) M AH(x,y,t)u(y)dy= r n H(x,y,t)u(y)dy B r(p) dB r(p) (11.10) dB r{p) H(x,y,t)r n udy + H(x1y1t)Audy. B r(p) By the Bishop volume comparison theorem and the fact that u and | r u | have polynomial growth together with the fact that H and R A , , \rH\2 decay exponentially we get that (11.11) lim Z AH(x,y,t)u(y)dy= lim Z H(x,y,t)Audy. Further, (11.12) B r(p) H(x, y, t)Audy H(x,y,t)f(r(y))dy. B r(p) h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 69 Using the Li-Yau estimate on the heat kernel [ see ; i.e., for any e > 0 there exists a constant C > 0 such that (11.13) we have (11.14) H(x,y,t)<Ct-texp-i-x-^] H(x,y,t)f(r(y))dy M < C t~~exp M \x - yy ( 4 + e ) t f(r(y))dy. We will now deal separately with the cases t > 1 and t < 1. For t < 1, we bound the right-hand side in terms of the sup of f and the integral of the Euclidean heat kernel (again using the Bishop volume comparison theorem). That is, H(x,y,t)f(r(y))dy (11.15) M _n \x — y\ <Csupf t 2 exp -M (4 + e)t < Csupf. dy For t at least one, we have that exp \x-y? At 1, and (11.16) H(x,y,t)f(r(y))dy<Ct-- f(r(y))dy. M M Now combining (11.15) and (11.16) yields the bound | ( e t u x (11.17) < (Csupf) + f{r{y))dy Ct~-M l Z o o / f(s)s n-1ds+ sup f Jo [0,oo) which proves (11.7). Since e t u(x) — u(x) = H(x,y,t)(u{y) — u(x))dy, M (11.8) follows by the same argument. q.e.d. 70 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Appendix A. The first variation of energy In this appendix, we will, for the sake of completeness, collect some well-known consequences of the first variation of energy that we need for this paper. In the following, we will take M to be a complete Riemannian man-ifold and u to be a smooth function on M. Given a one-parameter family (f>t of diffeomorphisms of M we define a one-parameter family of functions u t = u o t. We let v denote -dt. Let B be a bounded domain in M. Henceforth, we suppose that the diffeomorphisms are the identity outside of B; equivalently, we take the vector field v to have support in B. Now we define E t to be the Dirichlet energy of u t in B, that is, (A.l) E t= Z \Vu t\2. B Lemma A.2. (First Variation). Let M and v be as above. Then the first variation of energy is given by (A.3) E ( 0 ) = 2 Z V(v)(Vu,Vu) - Z \Vu\2div(v) . B B In the following, we will work in normal coordinates. We can then rewrite (A.3) as (A.4) E'(0) = 2 Z v ij u i u j - Z u2v jj , B B where additional indices refer to covariant derivatives, and the usual summation conventions are to be understood. Proof. By definition, we have (A.5) E t = dt Z t -Differentiating under the integral sign gives (A.6) E't = Z d-\Vu t|2 = 2 Z {Vd {Vu t) , Vu t) . B dt B dt h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 71 We will now calculate the integrand above in normal coordinates. By the chain rule, (A.7) d ~dt t=o Vu t\2 = 2 u j(f>otji + u jotji (u j(f>otji) . By construction, we have (A.8) (j>oji = Sij and t). t=o Integrating equation (A.10), we get that the integral of the first term in equation (A.9) is given by (A.ll) d ~dt IvulVt) t=QB which becomes, by the change of variables formula, (A.12) d ~dt |Vu| 2Jac(t _ 1) ^ i9 d B dt t=0 t=0B To first order, we have that <f)t)ij = Sij + tv ij, therefore Jac(^ t) =det(Sij + tv ij) = 1 + t(v jj)+O(t2). Thus d Jac(ç!t x) . (A.13) (A.14) dt iVupJac^"1) Vu\2(-div(v)) . t=QB B Putting the above all together, the lemma now follows. q.e.d. We will now derive some general identities from the first variation formula, by making careful choices of the domain B and the variation vector field v. Let b be a Lipschitz function such that (A.15) B = {x e M\b(x) < r}. 72 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii Take 7 : R — > R to be a cut-off function with support in {|x| < r} such that (A. 16) j(x) = 1 for x < (r — e) and 7—7-0 linearly otherwise. Let Ae denote the region in B on which b > (r — e). Finally, we choose the variation vector field v to be (A.17) v(x) = -j(b(x)) rb2. We will often write 7 for job. Note that by (A.17), (A.18) div(v) = ^7A(b2) + \l'{rb, rb2i . It follows from (A.18) that given any function u, (A.19) ru|2div(v) = - 7 | r u | 2 A b 2 B B 1 irurbirbr. e A £ Similarly, (A.20) v = ^ [ T ' b - ( ^ + 7 ( ^ (A.21) 2 Z r v ( r u , r u ) = Z Z Hess(b2)(ru, r B B - b(ru,rbi2 Combining (A.19) and (A.21), the first variation formula (A.3) implies the following: E'(0) = - - Z 7 | r u | 2 A b 2 B + - |ru| 2b|rb| 2 (A.22) 6 A + Z Hess(b2)(ru, r u ) B - - b(ru,rbi2. e AC h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 73 If we now let e approach zero, we get the following proposition. Proposition A . 2 3 . If u is harmonic, and b and B are as above, then 1 (A.24) j u j 2 A b 2 Hess(b2)(ru,ru) r jruj jrbj dB 2r dB du dn jrbj, where -u is the normal derivative of u on dB. Recall that dB = on {xjb{x) = r } . Proof. Since u is harmonic, it is a critical point for the energy functional and E ( 0 ) = 0. By (A.22), we have 1 (A.25) 7 j r u j 2 A b 2 - 7Hess(b 2)(ru, ru) 2 B B = - I jruj2bjrbj2 - - f b(ru,rbi2. As e — T - 0, the left-hand side in (A.25) clearly approaches the left-hand side in (A.24). Furthermore, the tube Ae is to first order a subdomain of the normal bundle of dB of width T ^ b- It follows that the right-side of (A.25) approaches the right-side of (A.24). q.e.d. Corollary A.26. If u is harmonic, and p is the distance function from a fixed point p G M, then 1 jruj2Ap2 - [ Hess(p2)(ru,ru) (A.27) r jruj fJB r 2r dB r du dr where B r is the ball of radius r, and -u is the radial derivative of u on dB r. R e m a r k A.28. If we take M to be C(N), the cone on a compact manifold N, then Hess(/>2) = 25ij where p is the distance from the vertex. For example, C(S'n _ 1') is R n. Therefore, Corollary A.26 implies that (A.29) In 2) jruj B r r jruj dB r 2r dB r du dr 74 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii where B r is the ball of radius r centered at the vertex of the cone. If we let D(r) denote the scaled energy on the ball of radius r centered at the vertex, then j u j 2 (A.30) D'(r) = (2 - n ) r _ 1 D ( r ) + r2~n Z JfJB r By substituting equation (A.29), equation (A.30) becomes 2 - n (A.31) D'(r) = 2r2 fJB r du > 0. dr Equation (A.31) is the usual monotonicity of scaled energy. References F. Almgren, Jr., Q-valued functions minimizing Dirichlet's integral and the regular-ity of area minimizing rectifiable currents up to codimension two, Preprint. M. Avellenada & F.-H. Lin, Un theoreme de Liouville pour des equations elliptiques a coefficients periodiques, C.R. Acad. Sci. Paris, Ser. 1, 309 (1989) 245-250. R. Bishop & R. Crittenden, Geometry of manifolds, Academic Press, New York, 1964. Yu. Burago, M. Gromov & G. Perelman, A.D. Alexandrov spaces with curvature bounded below, Uspekhi Mat. Nauk 47:2 (1992) 3-51, Russian Math. Surveys 47:2 (1992) 1-58. J. Cheeger, On the spectral geometry of spaces with cone-like singularities, Proc. Nat. Acad. Sci. U.S.A. 76 (1979) 2103-2106. J. Cheeger & T.H. Colding, Lower bounds on the Ricci curvature and the almost rigidity of warped products, Ann. of Math. 144 (1996) 189-237. , On the structure of spaces with Ricci curvature bounded below; I , J. Dif-ferential Geom, to appear. , On the structure of spaces with Ricci curvature bounded below; II, to appear. , Almost rigidity of warped products and the structure of spaces with Ricci curvature bounded below, C.R. Acad. Sci. Paris, Ser. 1, 320 (1995) 353-357. J. Cheeger, T.H. Colding & W.P. Minicozzi II, Linear growth harmonic functions on complete manifolds with nonnegative Ricci curvature, Geom. Funct. Anal. 5 (1995) 948-954. h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 75 J. Cheeger, T.H. Colding & G. Tian, On the singularities of spaces with Ricci curvature bounded , to appear. J. Cheeger & G. Tian, On the cone structure at infinity of Ricci flat manifolds with Euclidean volume growth and quadratic curvature decay, Invent. Math. 118 (1994) 493-571. S.Y. Cheng, Liouville theorem for harmonic maps, Proc. Sympos. Pure Math. Amer. Math. Soc. 36 (1980) 147-151. , Eigenfunctions and nodal sets, Comment. Math. Helv. 51 (1976) 43-55. S.Y. Cheng & S.T. Yau, Differential equations on Riemannian manifolds and their geometric applications, Comm. Pure Appl. Math. 28 (1975) 333-354. T. Christiansen & M. Zworski, Harmonic functions of polynomial growth on certain complete manifolds, Geom. Funct. Anal. 6 (1996) 619-627. T. H. Colding, Shape of manifolds with positive Ricci curvature, Invent. Math. 124 (1996) 175-191. , Large manifolds with positive Ricci curvature, Invent. Math. 124 (1996) 193-214. , Ricci curvature and volume convergence, Ann. of Math., to appear. , Stability and Ricci curvature, C.R. Acad. Sci. Paris, Ser. 1, 320 (1995) 1343-1347. T. H. Colding & W.P. Minicozzi II, On function theory on spaces with a lower Ricci curvature bound, Math. Res. Lett. 3 (1996) 241-246. , Large scale behavior of kernels of Schrodinger operators, Amer. J. Math. 117 (1997). , Harmonic functions on manifolds, Ann. of Math., to appear. , Weyl type bounds for harmonic functions, Invent. Math., to appear. , Generalized Liouville properties of manifolds, Math. Res. Lett. 3 (1996) 723-729. , (in preparation) H. Donnelly & C. Fefferman, Nodal domains and growth of harmonic functions on noncompact manifolds, J. Geom. Anal. 2 (1992) 79-93. M. Gromov, J. Lafontaine & P. Pansu, Structures métriques pour les varieties riemanniennes, Cedic/Fernand Nathan, Paris, 1981. M. Gromov & R. M. Schoen, Harmonic maps into singular spaces and p-adic superrigidity for lattices in groups of rank one, Inst. Hautes Etudes Sci. Publ. Math. 76 (1992) 165-246. 76 t o b i a s h. c o l d i n g & w i l l i a m p. m i n i c o z z i ii R. Hardt & L. Simon, Nodal sets for solutions of elliptic equations, J. Differential Geom. 30 (1989) 505-522. A. Kasue, Harmonic functions of polynomial growth on complete manifolds, Proc. Sympos. Pure Math., Amer. Math. Soc. Vol. 54, Part 1, (Ed. R. Greene and S.T. Yau), 1993. , Harmonic functions of polynomial growth on complete manifolds II, J. Math. Soc. Japan 47 (1995) 37-65. A. Kasue & T. Washio, Growth of equivariant harmonic maps and harmonic mor-phism, Osaka J. Math. 27 (1990) 899-928; Errata, Osaka J. Math. 29 (1992) 419-420. J. Kazdan, Parabolicity and the Liouville property on complete Riemannian man-ifolds, Aspects of Math. Vol. E10 (1987) 153-166. P. Klembeck, A complete Kahler metric of positive sectional curvature on C n, Proc. Amer. Math. Soc. 64 (1977) 313-316. P. Li, Large time behavior of the heat equation on complete manifolds with non-negative Ricci curvature, Ann. of Math. (2) 124 (1986) 1-21. , The theory of harmonic functions and its relation to geometry, Proc. Sym-pos. Pure Math., Amer. Math. Soc. Vol. 54, Part 1, (Ed. R. Greene and S.T. Yau), 1993. , Linear growth harmonic functions on Kahler manifolds with nonnegative Ricci curvature, Math. Res. Lett. 2 (1995) 79-94. , A lower bound for the first eigenvalue of the Laplacian on a compact manifold, Indiana Univ. Math. J. 28 (1979) 1013-1019. P. Li & R. M. Schoen, L p and mean value properties of subharmonic functions on Riemannian manifolds, Acta Math. 153 (1984) 279-301. P. Li & L-F. Tam, Linear growth harmonic functions on a complete manifold, J. Differential Geom. 29 (1989) 421-425. , Complete surfaces with finite total curvature, J. Differential Geom. 33 (1991) 139-168. P. Li & S.T. Yau, Estimates of eigenvalues of a compact riemannian manifold, Proc. Sympos. Pure Math., Amer. Math. Soc. 36 (1980) 205-239. , On the parabolic kernel of the Schrodinger operator, Acta Math. 156 (1986) 153-201. A. Lichnerowicz, Geometric des groupes des transformationes, Dunod, Paris, 1958. F.-H. Lin, Nonlinear theory of defects in nematic liquid crystals; phase transition and flow phenomena, Comm. Pure Appl. Math. 42 (1989) 789-814. h a r m o n i c f u n c t i o n s w i t h p o l y n o m i a l g r o w t h 77 N. Mok, A survey on complete noncompact Kahler manifolds of positive curvature, Proc. Sympos. Pure Math., Amer. Math. Soc. 41 (1984) 151-162. N. Mok, Y.-T. Siu & S.T. Yau, The Poincaré-Lelong equation on complete Kahler manifolds, Composito Math. 44 (1981) 188-218. G. Perelman (personal communication). , Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers, Preprint. R.M. Schoen, A report on some recent progress on nonlinear problems in geometry, Surveys Differential Geom. I (1991) 201-241. R.M. Schoen & S.T. Yau, Lectures on differential geometry, Internat. Press, Hong Kong, 1995. G. Tian & S.T. Yau, Complete Kahler manifolds with zero Ricci curature. II, Invent. Math. 106 (1991) 27-60. J. Wang, Linear growth harmonic functions on complete manifolds, Comm. Anal. Geom. 3 (1995) 683-698. H. Wu, Polynomial functions on complete Kahler manifolds, Proc. Sympos. Pure Math., Amer. Math. Soc. Vol 52, Part 2, 1991, 601-610. S. T. Yau, Harmonic functions on complete Riemannian manifolds, Comm. Pure Appl. Math. 28 (1975) 201-228. , Some function-theoretic properties of complete Riemannian manifold and their applications to geometry, Indiana Univ. Math. J. 25 (1976) 659-670. , Nonlinear analysis in geometry, Enseign. Math. (2) 33 (1987) 109-158. , Open problems in geometry, Proc. Sympos. Pure Math., Amer. Math. Soc. Vol 54, Part 1, 1993, (Ed. R. Greene and S.T. Yau). , Open problems in geometry, Chern - A Great Geometer of the Twentieth Century, (Ed. S.T. Yau), Internat. Press, Hong Kong, 1992. Courant Institute of Mathematical Sciences, New York Johns Hopkins University, Baltimore |
190225 | https://www.thyroid.org/professionals/ata-publications/clinical-thyroidology/february-2013-volume-25-issue-2/clin-thyroidol-20132543-44/ | Subacute Thyroiditis Is Treated Effectively by a Low Dose of Prednisolone
Clin Thyroidol 2013;25:43–44 | American Thyroid Association
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Home » Professionals Portal » ATA Publications » Clinical Thyroidology Archive » February 2013 Volume 25 Issue 2 » Clin Thyroidol 2013;25:43–44
Subacute Thyroiditis Is Treated Effectively by a Low Dose of Prednisolone
Jerome M. Hershman
Kubota S, Nishihara E, Kudo T, Ito M, Amino N, Miyauchi A. Initial treatment with 15 mg of prednisolone daily is sufficient for most patients with subacute thyroiditis in Japan. Thyroid. December 10, 2012 [Epub ahead of print].
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SUMMARY • • • • • • • • • • • • • • • • • • • • • • • •
Background
Subacute (granulomatous, DeQuervain’s) thyroiditis is an uncommon condition that has been treated with either nonsteroidal antiinflammatory drugs (NSAIDs) or corticosteroids for many years. The response to steroids is often more dramatic and quicker than the response to NSAIDs, but physicians are reluctant to use corticosteroids for this usually self-limited disorder because of their well-known side effects. The usual initial dose is 40 mg of prednisone. The basis for this dose has not been established by prospective studies. The current report is an evaluation of the efficacy of a prednisolone dose of 15 mg per day for 2 weeks, with reduction of the dosage by 5 mg every 2 weeks, as patients are carefully followed.
Methods
Subacute thyroiditis was diagnosed based on the criteria of swelling, pain, and tenderness within the thyroid gland associated with increased FT4, decreased TSH, increased C-reactive protein (CRP), and a hypoechoic area in the thyroid ultrasonogram corresponding to the tender area. Patients were treated with 15 mg of prednisolone per day for 2 weeks with reduction of the dose by 5 mg every 2 weeks for 6 weeks. If pain continued or the CRP remained high, prednisolone treatment was extended and then tapered over 12 weeks. All patients received "anti-ulcer" drugs.
Results
From February 2005 through December 2008, the diagnosis of subacute thyroiditis was made in 384 patients; 54 were not treated with medication, 33 were treated with NSAIDs, 9 dropped out of the treatment protocol, 69 violated the protocol, and 219 followed the protocol and are the subjects of this report. Patients were followed every 2 weeks.
The mean age of the patients was 49 years, 88% were women, and the mean weight was 55 kg. The mean FT4 was 2.5 ng/dl (normal range, 0.7 to 1.6) and the mean FT3 was 7.15 pg/ml (normal range, 1.70 to 3.70); FT4 was elevated in 80% of the patients.
Thyroiditis improved in 6 weeks and did not recur in 113 patients (51.6%); 106 patients took prednisolone for 7 weeks or longer and 27 of them took prednisolone for more than 12 weeks. Seven patients required >15 mg per day; 2 of these patients were treated with 30 mg per day and 5 with 20 mg per day. About 20% took more than 8 weeks to recover. There was a significant negative correlation between the FT4 and the duration of therapy and between the FT3 and the duration of therapy.
Transient hypothyroidism occurred in 31% of patients, and permanent hypothyroidism was found in only 3.6% of patients.
Conclusions
Subacute thyroiditis can be treated effectively with a daily dose of 15 mg of prednisolone for 2 weeks and subsequently tapering by 5 mg per day every 2 weeks.
ANALYSIS AND COMMENTARY • • • • • •
This study is a valuable clinical contribution to thyroidology because it is the first study that analyzed the response to corticosteroid therapy in a large population of patients with subacute thyroiditis. Treatment with about half of the usually recommended steroid dose was effective in ameliorating the disorder in 80% of patients within 8 weeks. Because the mean weight of these Japanese patients, mainly women, was only 55 kg, the 15-mg dose (0.27 mg per kilogram) would probably be equivalent to at least 20 mg of prednisone in a Western population.
The late Robert Volpé was an expert in this disorder and wrote an excellent review of its management (1). Volpé advocated a dose of 40 mg of prednisone, tapering it over 6 weeks. He noted that about 20% of patients will have a recurrence, necessitating the restoration of a higher dose, similar to the findings of the current report. Volpé expressed a preference for early initiation of steroid therapy, which is also the preferred therapy of the authors of this paper rather than initiating therapy with NSAIDs, as recommended by the guideline 96 of the ATA, before using prednisone therapy (2). The 6-week duration of corticosteroid therapy in this study is somewhat longer than that reported with empirical therapy of 49 patients in Minnesota with tapering of 40 mg of prednisone in 7 days and continuation of the reduced dose for only 30 days (3). However, two thirds of the group also received other therapy, probably NSAIDs.
It is interesting that the patients with higher thyroid hormone levels had faster restoration of normal levels with the glucocorticoid therapy and were more likely to be in the short-term medication group (6 weeks). The explanation suggested by the authors is that these patients had more destruction of their thyroid glands as compared with those who required a longer duration of therapy for resolution of the disorder. Presumably, the destruction reversed more quickly because the maximum destruction occurred at an earlier time—a unique hypothesis.
It is interesting that the authors followed the guidelines of the Japan Thyroid Association (www.japanthyroid.jp/doctor/guideline/english.html#akyuu) and did not perform a radioactive iodine–uptake test to confirm the diagnosis. This is another instance in which ultrasonography is replacing the use of radioisotopes in clinical diagnosis.
References
Volpé R. The management of subacute (DeQuervain’s) thyroiditis. Thyroid 1993;3:253-5.
Bahn Chair RS, Burch HB, Cooper DS, Garber JR, Greenlee MC, Klein I, Laurberg P, McDougall IR, Montori VM, Rivkees SA, Ross DS, et al. Hyperthyroidism and other causes of thyrotoxicosis: management guidelines of the American Thyroid Association and American Association of Clinical Endocrinologists. Thyroid 2011;21:593-646. Epub April 21, 2011; doi: 10.1089/thy.2010.0417.
Fatourechi V, Aniszewski JP, Fatourechi GZ, Atkinson EJ, Jacobsen SJ. Clinical features and outcome of subacute thyroiditis in an incidence cohort: Olmsted County, Minnesota, study. J Clin Endocrinol Metab 2003;88:2100-5.
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Int J Food Microbiol
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. 2009 Oct 31;135(2):144-51.
doi: 10.1016/j.ijfoodmicro.2009.07.031. Epub 2009 Aug 4.
Cell membrane damage induced by phenolic acids on wine lactic acid bacteria
F M Campos1,J A Couto,A R Figueiredo,I V Tóth,A O S S Rangel,T A Hogg
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1 CBQF/Escola Superior de Biotecnologia, Universidade Católica Portuguesa, Rua Dr. António Bernardino de Almeida, 4200-072 Porto, Portugal. fmcampos@mail.esb.ucp.pt
PMID: 19733929
DOI: 10.1016/j.ijfoodmicro.2009.07.031
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Cell membrane damage induced by phenolic acids on wine lactic acid bacteria
F M Campos et al. Int J Food Microbiol.2009.
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Int J Food Microbiol
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. 2009 Oct 31;135(2):144-51.
doi: 10.1016/j.ijfoodmicro.2009.07.031. Epub 2009 Aug 4.
Authors
F M Campos1,J A Couto,A R Figueiredo,I V Tóth,A O S S Rangel,T A Hogg
Affiliation
1 CBQF/Escola Superior de Biotecnologia, Universidade Católica Portuguesa, Rua Dr. António Bernardino de Almeida, 4200-072 Porto, Portugal. fmcampos@mail.esb.ucp.pt
PMID: 19733929
DOI: 10.1016/j.ijfoodmicro.2009.07.031
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The aim of this work was to investigate the effect of phenolic acids on cell membrane permeability of lactic acid bacteria from wine. Several phenolic acids were tested for their effects on the cell membrane of Oenococcus oeni and Lactobacillus hilgardii by measuring potassium and phosphate efflux, proton influx and by assessing culture viability employing a fluorescence technique based on membrane integrity. The experimental results indicate that hydroxycinnamic acids (p-coumaric, caffeic and ferulic acids) induce greater ion leakages and higher proton influx than hydroxybenzoic acids (p-hydroxibenzoic, protocatechuic, gallic, vanillic, and syringic acids). Among the hydroxycinnamic acids, p-coumaric acid showed the strongest effect. Moreover, the exposure of cells to phenolic acids caused a significant decrease in cell culture viability, as measured by the fluorescence assay, in both tested strains. The results agree with previous results obtained in growth experiments with the same strains. Generally, phenolic acids increased the cell membrane permeability in lactic acid bacteria from wine. The different effects of phenolic acids on membrane permeability could be related to differences in their structure and lipophilic character.
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Bounding sum of reciprocals of the square roots of the first N positive integers
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I am trying to derive the following inequality: $$2\sqrt{N}-1<1+\sum_{k=1}^{N}\frac{1}{\sqrt{k}}<2\sqrt{N}+1,\; N>1.$$
I understand for $N\rightarrow \infty$ the summation term diverges (being a p-series with p=1/2), which is consistent with the lower bound in this inequality being an unbounded function in $N$. With respect to deriving this inequality, it is perhaps easier to rewrite it as $$2\sqrt{N}-2<\sum_{k=1}^{N}\frac{1}{\sqrt{k}}<2\sqrt{N},\; N>1.$$
In this expression, the fact that the upper and lower bounds are so concisely expressed made me wonder whether there is a more basic inequality I can begin with of the form $$a_{k}<\frac{1}{\sqrt{k}}1,$$ from which summing through would yield the desired inequality. That is $a_k \text{ and } b_k$ should satisfy $\sum_{k=1}^{N}a_k=2\sqrt{N}-2$ and $\sum_{k=1}^{N}b_k=2\sqrt{N}$. However, I am stuck on how to find the appropriate $a_{k}$ and $b_{k}$ to obtain the desired bounds. Alternatively, what other methods are at my disposal?
calculus
sequences-and-series
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asked Sep 28, 2016 at 22:03
OliviaOlivia
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I will give you a proof of a stronger inequality through creative telescoping.Let we start by noticing that
$$ \sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}} = \frac{1}{\sqrt{n+1/2}+\sqrt{n-1/2}}\geq\frac{1}{2\sqrt{n}}$$ by the concavity of the function $f(x)=\sqrt{x}$ over $\mathbb{R}^+$. It follows that:
$$ \sum_{n=1}^{N}\frac{1}{\sqrt{n}}\color{red}{\leq} 2\sum_{n=1}^{N}\left(\sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}}\right) = \color{red}{\sqrt{4N+2}-\sqrt{2}}.$$ On the other hand, $ 2\sqrt{n+\frac{1}{2}}-2\sqrt{n-\frac{1}{2}}-\frac{1}{\sqrt{n}}=\Theta\left(\frac{1}{n^{5/2}}\right)$, hence $$ \sum_{n=1}^{N}\frac{1}{\sqrt{n}} \color{red}{\geq \sqrt{4N+2}-\sqrt{2}-C}$$ where: $$ 0\leq C = \sum_{n\geq 1}\left(2\sqrt{n+\frac{1}{2}}-2\sqrt{n-\frac{1}{2}}-\frac{1}{\sqrt{n}}\right)=\color{green}{-\left[\sqrt{2}+\zeta\left(\frac{1}{2}\right)\right]}\leq \frac{1}{21}. $$
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answered Sep 28, 2016 at 23:14
Jack D'AurizioJack D'Aurizio
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$\begingroup$ Accidentally, this also proves $$\zeta\left(\frac{1}{2}\right)\approx -\sqrt{2}$$ :D $\endgroup$
Jack D'Aurizio
– Jack D'Aurizio
2016-09-28 23:23:11 +00:00
Commented Sep 28, 2016 at 23:23
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To get the lower bound, you can convert the sum to an integral. $\frac 1{\sqrt k} \gt \int_k^{k+1}\frac 1{\sqrt x}dx=2\sqrt x|_k^{k+1}=2(\sqrt {k+1}-\sqrt k)$, so $\sum_{k=1}^N \frac 1{\sqrt k} \gt \int_1^N \frac 1{\sqrt x}dx=2\sqrt N -2$ Then to get the upper limit you have the same approach. $\frac 1{\sqrt k} \lt \int_{k-1}^{k}\frac 1{\sqrt x}dx=2\sqrt x|_{k-1}^{k}=2(\sqrt {k}-\sqrt {k-1})$ so $\sum_{k=1}^N \frac 1{\sqrt k} \lt \int_0^{N-1} \frac 1{\sqrt x}dx=\int_0^{1} \frac 1{\sqrt x}dx+\int_1^{N-1} \frac 1{\sqrt x}dx=2+2\sqrt N -2=2\sqrt N$
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answered Sep 28, 2016 at 22:15
Ross MillikanRoss Millikan
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Use induction on $N$, and compose inequalities. For instance, on the lower bound, you might get that
$$y<\sum$$
where $\sum$ is the sum by the inductive hypothesis, and you want to prove that $$x < \sum$$ so you prove that $x
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answered Sep 28, 2016 at 22:06
Carl SchildkrautCarl Schildkraut
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Drawing a perpendicular line to another line which is drawn parallelly from a point to another line
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I know it's a bit messy to describe, but I need to find an algorithm to provide me with the distance between two points, Only point we don't know is x5, and y5, Can you implement an algorithm that only includes other coordinates and some math stuff?? I am sending an example of my problem.
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asked Mar 17, 2023 at 18:35
ÖzdenÖzden
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stackoverflow.com/a/65791098/844416 - you need nlenMBo –MBo 2023-03-17 19:09:27 +00:00 Commented Mar 17, 2023 at 19:09
2 This is really a geometry question.Raymond Chen –Raymond Chen 2023-03-17 19:20:58 +00:00 Commented Mar 17, 2023 at 19:20
Yes, I am not an engineer and am not able to easily comprehend to create a formula, any help would be appreciated Özden –Özden 2023-03-17 19:51:58 +00:00 Commented Mar 17, 2023 at 19:51
Is linked answer clear?MBo –MBo 2023-03-18 06:17:08 +00:00 Commented Mar 18, 2023 at 6:17
1 use vector math... define your line as start point p0=(x3,y3); and unit direction vector dp=(x2-x1,y2-y1); dp/=|dp|; now do perpendicular projection of p=(x4,y4) using dot product (x5,y5)=p0+dpdot(p-p0,dp); no trigonometry is needed (the dot behaves like cos(a))Spektre –Spektre 2023-03-19 07:42:59 +00:00 Commented Mar 19, 2023 at 7:42
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use vector math...
define your line as start point:
p0=vec2(x3,y3);
and unit direction vector:
dp=vec2(x2-x1,y2-y1);
dp/=|dp|;
now do perpendicular projection of p=(x4,y4) using dot product:
vec2(x5,y5)=p0+dpdot(p-p0,dp);
As you can see no trigonometry is needed (the dot behaves like cos(a)) so now the distance is:
dist=|p0-p+dpdot(p-p0,dp)|
In case you do not need the vec2(x5,y5) and you want just the distance then in 2D you can simply dot product with perpendicular vector to dp...
cpp
p0=vec2(x3,y3);
dp=vec2(x2-x1,y2-y1); // direction vector
dp/=|dp|; // unit
dq=vec2(dp.y,-dp.x); // rotate it by 90 deg
dist=dot(p-p0,dq); // distance between p0 and p in dq direction
by simply swap x,y coordinates of dp and negate one of them (which just determine if the dp is rotated CW or CCW by 90 deg). However note that such dist result can be negative (will tell you from which side of the line (x3,y3),(x5,y5) you coming from so if you need unsigned distance just use abs:
dist=|dist|;
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answered Mar 19, 2023 at 12:08
SpektreSpektre
52.2k 12 12 gold badges 125 125 silver badges 408 408 bronze badges
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Rydberg constant
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Symbol R. A constant that occurs in the formulae for atomic spectra and is related to the binding energy between an electron and a nucleon. It is connected to other constants by the relationship R=μ 0 2 me 4 c 3/8 h 3, where μ 0 is the magnetic constant, m and e are the mass and charge of an electron, c is the speed of light, and h is the Planck constant. It has the value 1.097 × 10 7 m−1. It is named after the Swedish physicist Johannes Robert Rydberg (1854–1919), who developed a formula for the spectrum of hydrogen.
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They determine that each 150 mL serving will contain 0.66 grams of NaCl, which can replenish sodium ions essential for regulating nerve and muscle function. The molar mass of NaCl is 58.44 g/mol.
Calculate the concentration of NaCl in the solution in mol/L.
Choose 1 answer:
Choose 1 answer:
(Choice A) 4.4 mol/L A 4.4 mol/L
(Choice B) 0.044 mol/L B 0.044 mol/L
(Choice C) 0.075 mol/L C 0.075 mol/L
(Choice D) 0.000075 mol/L D 0.000075 mol/L
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RAD DEG
sin⁻¹ sin del ac cos⁻¹ cos()tan⁻¹ tan π ans ln log eˣ EXP
+-=
1 2 3 4 5 6 7 8 9 0.
×÷xʸ√
Periodic table
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
Atomic number 1
Symbol H
Average atomic mass 1.008
Atomic number 2
Symbol He
Average atomic mass 4.00
Atomic number 3
Symbol Li
Average atomic mass 6.94
Atomic number 4
Symbol Be
Average atomic mass 9.01
Atomic number 5
Symbol B
Average atomic mass 10.81
Atomic number 6
Symbol C
Average atomic mass 12.01
Atomic number 7
Symbol N
Average atomic mass 14.01
Atomic number 8
Symbol O
Average atomic mass 16.00
Atomic number 9
Symbol F
Average atomic mass 19.00
Atomic number 10
Symbol Ne
Average atomic mass 20.18
Atomic number 11
Symbol Na
Average atomic mass 22.99
Atomic number 12
Symbol Mg
Average atomic mass 24.31
Atomic number 13
Symbol Al
Average atomic mass 26.98
Atomic number 14
Symbol Si
Average atomic mass 28.09
Atomic number 15
Symbol P
Average atomic mass 30.97
Atomic number 16
Symbol S
Average atomic mass 32.06
Atomic number 17
Symbol Cl
Average atomic mass 35.45
Atomic number 18
Symbol Ar
Average atomic mass 39.95
Atomic number 19
Symbol K
Average atomic mass 39.10
Atomic number 20
Symbol Ca
Average atomic mass 40.08
Atomic number 21
Symbol Sc
Average atomic mass 44.96
Atomic number 22
Symbol Ti
Average atomic mass 47.87
Atomic number 23
Symbol V
Average atomic mass 50.94
Atomic number 24
Symbol Cr
Average atomic mass 52.00
Atomic number 25
Symbol Mn
Average atomic mass 54.94
Atomic number 26
Symbol Fe
Average atomic mass 55.85
Atomic number 27
Symbol Co
Average atomic mass 58.93
Atomic number 28
Symbol Ni
Average atomic mass 58.69
Atomic number 29
Symbol Cu
Average atomic mass 63.55
Atomic number 30
Symbol Zn
Average atomic mass 65.38
Atomic number 31
Symbol Ga
Average atomic mass 69.72
Atomic number 32
Symbol Ge
Average atomic mass 72.63
Atomic number 33
Symbol As
Average atomic mass 74.92
Atomic number 34
Symbol Se
Average atomic mass 78.97
Atomic number 35
Symbol Br
Average atomic mass 79.90
Atomic number 36
Symbol Kr
Average atomic mass 83.80
Atomic number 37
Symbol Rb
Average atomic mass 85.47
Atomic number 38
Symbol Sr
Average atomic mass 87.62
Atomic number 39
Symbol Y
Average atomic mass 88.91
Atomic number 40
Symbol Zr
Average atomic mass 91.22
Atomic number 41
Symbol Nb
Average atomic mass 92.91
Atomic number 42
Symbol Mo
Average atomic mass 95.95
Atomic number 43
Symbol Tc
Average atomic mass(97)
Atomic number 44
Symbol Ru
Average atomic mass 101.07
Atomic number 45
Symbol Rh
Average atomic mass 102.91
Atomic number 46
Symbol Pd
Average atomic mass 106.42
Atomic number 47
Symbol Ag
Average atomic mass 107.87
Atomic number 48
Symbol Cd
Average atomic mass 112.41
Atomic number 49
Symbol In
Average atomic mass 114.82
Atomic number 50
Symbol Sn
Average atomic mass 118.71
Atomic number 51
Symbol Sb
Average atomic mass 121.76
Atomic number 52
Symbol Te
Average atomic mass 127.60
Atomic number 53
Symbol I
Average atomic mass 126.90
Atomic number 54
Symbol Xe
Average atomic mass 131.29
Atomic number 55
Symbol Cs
Average atomic mass 132.91
Atomic number 56
Symbol Ba
Average atomic mass 137.33
Atomic number 72
Symbol Hf
Average atomic mass 178.49
Atomic number 73
Symbol Ta
Average atomic mass 180.95
Atomic number 74
Symbol W
Average atomic mass 183.84
Atomic number 75
Symbol Re
Average atomic mass 186.21
Atomic number 76
Symbol Os
Average atomic mass 190.23
Atomic number 77
Symbol Ir
Average atomic mass 192.22
Atomic number 78
Symbol Pt
Average atomic mass 195.08
Atomic number 79
Symbol Au
Average atomic mass 196.97
Atomic number 80
Symbol Hg
Average atomic mass 200.59
Atomic number 81
Symbol Tl
Average atomic mass 204.38
Atomic number 82
Symbol Pb
Average atomic mass 207.2
Atomic number 83
Symbol Bi
Average atomic mass 208.98
Atomic number 84
Symbol Po
Average atomic mass(209)
Atomic number 85
Symbol At
Average atomic mass(210)
Atomic number 86
Symbol Rn
Average atomic mass(222)
Atomic number 87
Symbol Fr
Average atomic mass(223)
Atomic number 88
Symbol Ra
Average atomic mass(226)
Atomic number 104
Symbol Rf
Average atomic mass(267)
Atomic number 105
Symbol Db
Average atomic mass(268)
Atomic number 106
Symbol Sg
Average atomic mass(269)
Atomic number 107
Symbol Bh
Average atomic mass(270)
Atomic number 108
Symbol Hs
Average atomic mass(269)
Atomic number 109
Symbol Mt
Average atomic mass(278)
Atomic number 110
Symbol Ds
Average atomic mass(281)
Atomic number 111
Symbol Rg
Average atomic mass(282)
Atomic number 112
Symbol Cn
Average atomic mass(285)
Atomic number 113
Symbol Nh
Average atomic mass(286)
Atomic number 114
Symbol Fl
Average atomic mass(289)
Atomic number 115
Symbol Mc
Average atomic mass(290)
Atomic number 116
Symbol Lv
Average atomic mass(293)
Atomic number 117
Symbol Ts
Average atomic mass(294)
Atomic number 118
Symbol Og
Average atomic mass(294)
Metals
Transition Metals
Nonmetals
Metalloids
36
Kr
83.80
Atomic number Symbol Average atomic mass Hover over an element to see its full name←←←
Atomic number 57
Symbol La
Average atomic mass 138.91
Atomic number 58
Symbol Ce
Average atomic mass 140.12
Atomic number 59
Symbol Pr
Average atomic mass 140.91
Atomic number 60
Symbol Nd
Average atomic mass 144.24
Atomic number 61
Symbol Pm
Average atomic mass(145)
Atomic number 62
Symbol Sm
Average atomic mass 150.36
Atomic number 63
Symbol Eu
Average atomic mass 151.96
Atomic number 64
Symbol Gd
Average atomic mass 157.25
Atomic number 65
Symbol Tb
Average atomic mass 158.93
Atomic number 66
Symbol Dy
Average atomic mass 162.50
Atomic number 67
Symbol Ho
Average atomic mass 164.93
Atomic number 68
Symbol Er
Average atomic mass 167.26
Atomic number 69
Symbol Tm
Average atomic mass 168.93
Atomic number 70
Symbol Yb
Average atomic mass 173.05
Atomic number 71
Symbol Lu
Average atomic mass 174.97
Atomic number 89
Symbol Ac
Average atomic mass(227)
Atomic number 90
Symbol Th
Average atomic mass 232.04
Atomic number 91
Symbol Pa
Average atomic mass 231.04
Atomic number 92
Symbol U
Average atomic mass 238.03
Atomic number 93
Symbol Np
Average atomic mass(237)
Atomic number 94
Symbol Pu
Average atomic mass(244)
Atomic number 95
Symbol Am
Average atomic mass(243)
Atomic number 96
Symbol Cm
Average atomic mass(247)
Atomic number 97
Symbol Bk
Average atomic mass(247)
Atomic number 98
Symbol Cf
Average atomic mass(251)
Atomic number 99
Symbol Es
Average atomic mass(252)
Atomic number 100
Symbol Fm
Average atomic mass(257)
Atomic number 101
Symbol Md
Average atomic mass(258)
Atomic number 102
Symbol No
Average atomic mass(259)
Atomic number 103
Symbol Lr
Average atomic mass(266) |
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Math.Atan2(Double, Double) Method
Definition
Namespace:
: System
Assemblies:
: mscorlib.dll, System.Runtime.Extensions.dll
Assemblies:
: netstandard.dll, System.Runtime.dll
Assembly:
: System.Runtime.Extensions.dll
Assembly:
: mscorlib.dll
Assembly:
: netstandard.dll
Source:
: Math.cs
Source:
: Math.cs
Source:
: Math.cs
Source:
: Math.cs
Important
Some information relates to prerelease product that may be substantially modified before it’s released. Microsoft makes no warranties, express or implied, with respect to the information provided here.
Returns the angle whose tangent is the quotient of two specified numbers.
public: static double Atan2(double y, double x);
public static double Atan2(double y, double x);
static member Atan2 : double double -> double
Public Shared Function Atan2 (y As Double, x As Double) As Double
Parameters
y
: Double
The y coordinate of a point.
x
: Double
The x coordinate of a point.
Returns
Double
An angle, θ, measured in radians, such that tan(θ) = y / x, where (x, y) is a point in the Cartesian plane. Observe the following:
For (x, y) in quadrant 1, 0 < θ < π/2.
For (x, y) in quadrant 2, π/2 < θ ≤ π.
For (x, y) in quadrant 3, -π ≤ θ < -π/2.
For (x, y) in quadrant 4, -π/2 < θ < 0.
For points on the boundaries of the quadrants, the return value is the following:
If y is 0 and x is not negative, θ = 0.
If y is 0 and x is negative, θ = π.
If y is positive and x is 0, θ = π/2.
If y is negative and x is 0, θ = -π/2.
If y is 0 and x is 0, θ = 0.
If x or y is NaN, or if x and y are either PositiveInfinity or NegativeInfinity, the method returns NaN.
Examples
The following example demonstrates how to calculate the arctangent of an angle and a vector. The resulting value is displayed in the console.
// This example demonstrates Math.Atan() // Math.Atan2() // Math.Tan() using System; class Sample { public static void Main() { double x = 1.0; double y = 2.0; double angle; double radians; double result; // Calculate the tangent of 30 degrees. angle = 30; radians = angle (Math.PI/180); result = Math.Tan(radians); Console.WriteLine("The tangent of 30 degrees is {0}.", result); // Calculate the arctangent of the previous tangent. radians = Math.Atan(result); angle = radians (180/Math.PI); Console.WriteLine("The previous tangent is equivalent to {0} degrees.", angle); // Calculate the arctangent of an angle. String line1 = "{0}The arctangent of the angle formed by the x-axis and "; String line2 = "a vector to point ({0},{1}) is {2}, "; String line3 = "which is equivalent to {0} degrees."; radians = Math.Atan2(y, x); angle = radians (180/Math.PI); Console.WriteLine(line1, Environment.NewLine); Console.WriteLine(line2, x, y, radians); Console.WriteLine(line3, angle); } } / This example produces the following results: The tangent of 30 degrees is 0.577350269189626. The previous tangent is equivalent to 30 degrees. The arctangent of the angle formed by the x-axis and a vector to point (1,2) is 1.10714871779409, which is equivalent to 63.434948822922 degrees. /
// This example demonstrates Math.Atan() // Math.Atan2() // Math.Tan() // Functions 'atan', 'atan2', and 'tan' may be used instead. open System [] let main _ = let x = 1. let y = 2. // Calculate the tangent of 30 degrees. let angle = 30. let radians = angle (Math.PI / 180.) let result = Math.Tan radians printfn $"The tangent of 30 degrees is {result}." // Calculate the arctangent of the previous tangent. let radians = Math.Atan result let angle = radians (180. / Math.PI) printfn $"The previous tangent is equivalent to {angle} degrees." // Calculate the arctangent of an angle. let radians = Math.Atan2(y, x) let angle = radians (180. / Math.PI) printfn $"""The arctangent of the angle formed by the x-axis and a vector to point ({x},{y}) is {radians}, which is equivalent to {angle} degrees.""" 0 //This example produces the following results: // The tangent of 30 degrees is 0.577350269189626. // The previous tangent is equivalent to 30 degrees. // // The arctangent of the angle formed by the x-axis and // a vector to point (1,2) is 1.10714871779409, // which is equivalent to 63.434948822922 degrees.
' This example demonstrates Math.Atan() ' Math.Atan2() ' Math.Tan() Class Sample Public Shared Sub Main() Dim x As Double = 1.0 Dim y As Double = 2.0 Dim angle As Double Dim radians As Double Dim result As Double ' Calculate the tangent of 30 degrees. angle = 30 radians = angle (Math.PI / 180) result = Math.Tan(radians) Console.WriteLine("The tangent of 30 degrees is {0}.", result) ' Calculate the arctangent of the previous tangent. radians = Math.Atan(result) angle = radians (180 / Math.PI) Console.WriteLine("The previous tangent is equivalent to {0} degrees.", angle) ' Calculate the arctangent of an angle. Dim line1 As [String] = "{0}The arctangent of the angle formed by the x-axis and " Dim line2 As [String] = "a vector to point ({0},{1}) is {2}, " Dim line3 As [String] = "which is equivalent to {0} degrees." radians = Math.Atan2(y, x) angle = radians (180 / Math.PI) Console.WriteLine(line1, Environment.NewLine) Console.WriteLine(line2, x, y, radians) Console.WriteLine(line3, angle) End Sub End Class ' 'This example produces the following results: ' 'The tangent of 30 degrees is 0.577350269189626. 'The previous tangent is equivalent to 30 degrees. ' 'The arctangent of the angle formed by the x-axis and 'a vector to point (1,2) is 1.10714871779409, 'which is equivalent to 63.434948822922 degrees. '
Remarks
The return value is the angle in the Cartesian plane formed by the x-axis, and a vector starting from the origin, (0,0), and terminating at the point, (x,y).
This method calls into the underlying C runtime, and the exact result or valid input range may differ between different operating systems or architectures.
Applies to
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190234 | https://unacademy.com/content/ssc/study-material/mathematics/how-speed-time-and-distance-can-be-applied-in-real-life-situations/ | How Speed, Time and Distance can be Applied in Real Life Situations
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Speed, time and distance are fundamental concepts used in real-life applications. These concepts use physics as well as other quantitative concepts to solve and provide the desired solutions. This article provides concepts and shortcuts for time speed and distance for a better and easy understanding of the topic.
Speed, Time & Distance – Introduction
Speed
The rate at which an object moves from one location to another in a certain amount of time is known as speed. It is a scalar quantity since it defines the magnitude of a moving item, not its directions. The SI unit of speed is m/s.
Time
In physics, time is defined by its unit of measure – time is the reading on a clock. It is a scalar quantity in classical, non-relativistic physics.
Distance
The overall movement of an object, regardless of direction, is referred to as distance. We can define distance as the amount of ground covered by an item, regardless of its starting or ending position.
Different units can be used to indicate speed, distance, and time:
Time is measured in seconds (s), minutes (min), and hours (h) (hr)
Distance is measured in metres (m), kilometres (km), miles (miles), and feet
m/s and km/hr are the units of measurement for speed
So, if Distance = km and Time = hr, Speed = Distance/ Time, and Speed units are km/ hr
Relationship between Speed, Time & Distance
Distance/Time = Speed — This shows us how fast or slow an object moves. It is defined as the distance travelled divided by the time it took to travel that distance
Distance is directly proportional to the speed, but time is inversely proportional
Distance = Speed x Time.
Time = Distance / Speed, the time taken will reduce as the speed increases and vice versa
Conversion rates of Speed, Time & Distance
To convert from m / sec to km / hour, we multiply by 18 / 5. So, 1 m / sec = 18 / 5 km / hour = 3.6 km / hour
1 yard = 3 feet
Similarly, 1 km/hr = 5/8 miles/hour
1 mile = 1760 yards
1 mile = 5280 feet
To convert from km / hour to m / sec, we multiply by 5 / 18. So, 1 km / hour = 5 / 18 m / sec
1 kilometer= 1000 metres = 0.6214 mile
1 hour= 60 minutes= 6060 seconds= 3600 seconds
1 mile= 1.609 kilometer
1 yard = 3 feet
Examples
How many kilometres can a man travel in 3 hours 45 minutes, if he covers 12 metres in a second?
Solution:
12 m/s = 12 18/5 kmph
Time can be written as,
3 hours 45 minutes = 3 3/4 hours = 15/4 hours
Therefore,
Distance = speed time = 12 18/5 15/4 km = 162 kms.
A person can cross a 600-meter-long street in 5 minutes. What is his current speed in kilometres per hour?
Solution:
Speed = (600560) m/sec
= 2 m/sec
= Converting m/sec into km/sec
(2185) km/hr.
= 7.2 km/hr
A man walks 6 km at 1 1/2 kmph, runs 8 km at 2 kmph, and then takes the bus for another 32 km. The bus travels at a speed of 8 kilometres per hour. Calculate the man’s average speed.
Solution:
The man walked 6 kilometres at 1.5 kilometres per hour, then 8 kilometres at 2 kilometres per hour, and 32 kilometres at 8 kilometres per hour.
Time taken individually:
=> 6 m / 1.5 m = 4 m
=> 4 m = 8/2
=> 4 m = 32/8
Man’s average speed equals total distance divided by total time.
=> 3 (5/6) = 46/12
T1 train from Gujarat to Mumbai departs at 7 a.m. and arrives at 12 p.m. T2 is a second train that departs Mumbai at 7 a.m. and arrives in Ahmedabad at 1 p.m. When did the two trains come into contact?
Solution:
Allow x to be the distance between two stations.
Mumbai is reached in 5 hours by train T1.
Gujarat is reached in 6 hours by train T2.
Both trains’ relative speeds will be summed = x/5 + x/ 6 = 11x/30
Time taken = x/11x/30 = 30/11 or (30 X 60)/11 = 2.43 hours, resulting in 7 + 2.43 = 9.43 am.
Applications of Speed, Time & Distance
The concepts of time, speed, and distance are used for calculating the speed of trains and the distance covered. It is also used for finding out the length of a train.
It is also used in solving problems related to boats and streams, which are useful in determining the flow of a stream.
Also, various problems involving circular motion can be solved using formulas based on time, speed, and distance.
Conclusion
Speed, time, and distance are crucial concepts in physics and mathematics. These concepts are used in various real life applications, and calculated for different research and other purposes. Besides that, its questions are common in aptitude exams, which can be useful for students who prepare well.
Frequently Asked Questions
Get answers to the most common queries related to the SSC Examination Preparation.
Explain the formula for time.
Since Speed = DistanceTime...Read full
What is the formula for time & work?
The formula for time taken = 1Rate of wo...Read full
How to solve Time, speed & distance easily?
The comparable formula is d = rt, which equals distance times rate times time. To calculate speed or rate, use the s...Read full
Since Speed = DistanceTime , Formula for time can be mentioned as Time = DistanceSpeed
The formula for time taken = 1Rate of work .
Work Done = Time Taken × Rate of Work
The comparable formula is d = rt, which equals distance times rate times time. To calculate speed or rate, use the speed formula, s = d/t, which equals distance divided by the time. To find the time, use the formula t = d/s, which equals time divided by distance divided by speed.
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190235 | https://math.stackexchange.com/questions/2033509/solve-equation-for-z-z-12i | complex numbers - Solve equation for $|z| - z = 1+2i$ - Mathematics Stack Exchange
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Solve equation for |z| - z = 1+2i
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Asked 8 years, 10 months ago
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\begingroup
Solve equation:
|z| - z = 1+2i
I know that we can express |z| as \sqrt{x^2+y^2} and z as x+yi but I get stuck after some steps.
\sqrt{x^2+y^2}-(x+yi) = 1+2i
If going by that can we say that?: \sqrt{x^2+y^2}-x = 1 and -yi=2i
I'm getting stuck after this part (which I hope was not wrong).
complex-numbers
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asked Nov 27, 2016 at 22:08
kushtrimhkushtrimh
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\endgroup
1
1 \begingroup Yes, so y = -2, and you want to find all x such that \sqrt{x^2+4} = x+1.\endgroup Daniel Fischer –Daniel Fischer 2016-11-27 22:11:45 +00:00 Commented Nov 27, 2016 at 22:11
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\begingroup
The equation can be rewritten as: \;z = |z| -1-2i\;\tag{1}
Then:
\begin{align} |z|^2 = z \bar z & = (|z| - 1 - 2i)\,(|z| - 1 + 2i) \ &= |z|^2 +|z|(-1+2i-1-2i) +(-1-2i)(-1+2i)\ &= |z|^2 - 2 |z| + 5 \end{align}
Canceling the |z|^2 term between the first and last expressions gives |z|= \frac{5}{2} then substituting in (1):
z = \frac{5}{2} - 1 - 2i = \frac{3}{2} - 2i
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answered Nov 27, 2016 at 23:16
dxivdxiv
78k 6 6 gold badges 69 69 silver badges 127 127 bronze badges
\endgroup
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\begingroup
Since two complex numbers are equal iff their real and imaginary part are equal, compare them in both sides:
\sqrt{x^2+y^2}-x-iy=1+2i\iff \begin{cases}\sqrt{x^2+y^2}-x=1\{}\y=-2\end{cases}
and etc.
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answered Nov 27, 2016 at 22:25
DonAntonioDonAntonio
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\endgroup
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
StatPearls [Internet].
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
Anatomy, Abdomen and Pelvis, Scrotum
Rosa A. Garcia; Hussain Sajjad.
Author Information and Affiliations
Authors
Rosa A. Garcia; Hussain Sajjad1.
Affiliations
1 RMU and Allied Hospitals
Last Update: July 24, 2023.
Introduction
The scrotum is a male reproductive structure located under the penis. It has the shape of a sac and divides into two compartments. The structures contained in this sac include external spermatic fascia, testes, epididymis, and spermatic cord. The scrotum derives from the labioscrotal swelling, which is an embryonic structure that appears in the fourth week of embryonic development. Congenital malformations may occur during the development of the scrotum. These malformations can present in conjunction with other defects due to a common embryologic origin.
Structure and Function
Structure:
The scrotum is a thin external sac that is located under the penis and is composed of skin and smooth muscle. This sac is divided into two compartments by the scrotal septum. The average wall thickness of the scrotum is about 8 mm. It has a parietal and a visceral layer. The parietal layer has the function of covering the inner aspect of the scrotal wall and the visceral layer coats the testis and epididymis. The structures contained in the scrotal sac are the external spermatic fascia, testes, epididymis, and spermatic cord.
Function:
The scrotum is responsible for protecting the testes. It helps with the thermoregulation of the testicles. It keeps the temperature of the testis several degrees below the average body temperature, which is an essential factor for sperm production.
Embryology
The labioscrotal swellings are the structures that appear during the fourth week of gestations and give origin to the scrotal tissue. These two structures are present lateral to the genital tubercle. The migration of the labioscrotal swellings takes place in the 9th to 11th weeks of gestation and follows in a caudal and medial direction until they fuse at the 12th-week of gestation and form the scrotum.
Blood Supply and Lymphatics
Blood Supply:
Arteriovenous anastomoses and subcutaneous plexuses are responsible for providing blood supply to the scrotum. The external pudendal branches arising from the femoral arteries and the scrotal branches of the in the internal pudendal arteries play a significant role in supplying blood to the scrotum. The nomenclature of the venous drainage of the scrotum mirrors the previously mentioned corresponding arteries.
Lymphatic drainage:
The arrangement of the lymphatic drainage of the scrotum is related to the development of this sac. The wall of the scrotum drains into the superficial inguinal lymph nodes. On the other hand, the contents of the scrotum drain to the lumbar lymph nodes. It is important to recall that the testes migrate from the abdominal wall through the inguinal canal and into the scrotum. During this path, the testis will drag its blood supply and lymph vessels into the scrotal sac.
Nerves
The innervation of the scrotum derives from the branches of the following four nerves: genitofemoral, pudendal, posterior femoral cutaneous, and ilioinguinal nerves.
Anterior Innervation:
The cremaster muscle and anterior scrotum receive their innervation by the genital branch of the genitofemoral nerve. This nerve arises from the L1-L2 segments of the lumbosacral plexus and then travels through the inguinal canal to supply the anterior skin of the scrotum.
Posterior Innervation:
The posterior scrotum is innervated by branches of the pudendal nerve and by the posterior femoral cutaneous nerve.
The ilioinguinal nerve also aids the genitofemoral nerve in the innervation of the cremasteric muscle.
Cremasteric Reflex:
Both the ilioinguinal and the genitofemoral nerves provide a sensory synapse that activates the motor neurons responsible for the cremasteric reflex. This physiologic reflex has both protective and thermoregulatory testicular functions.
Muscles
The scrotum is a fibromuscular structure. The muscle fibers contained in the scrotum are the dartos muscle and the cremasteric muscle. The dartos muscle is a smooth muscle sheet located underneath the skin of the scrotum. The cremasteric muscle is a paired muscle that has many protective functions. This paired muscle is composed of two parts, a medial cremaster portion originating from the pubic tubercle and the lateral cremaster portion originating from the internal oblique muscle.
Physiologic Variants
The scrotum may present with congenital malformations that are due to abnormal development of the labioscrotal swellings, which occur during intrauterine development. These malformations may include accessory scrotum, bifid scrotum, ectopic scrotum, and penoscrotal transposition.
Accessory Scrotum:
It is the rarest of congenital abnormalities of the scrotum. Although exact etiology is not clear, it is thought to occur due to the abnormal migration of labioscrotal swellings during the 4th to 12th weeks of development. The most common location of an accessory scrotum is mid perineum. The accessory scrotum does not contain testes. This additional tissue will have the same histologic features present in a typically developed scrotum and does not interfere with the development of the customarily positioned scrotum.
Bifid Scrotum:
A bifid scrotum is an abnormal cleft that presents in the midline of the scrotum. This cleft is caused by an incomplete fusion of the labioscrotal folds. The incomplete fusion can be attributed to an under-secretion of androgens during the first trimester. This androgenic under-secretion can also cause hypospadias and maybe the reason why most patients that are born with a bifid scrotum also present with hypospadias.
Ectopic Scrotum:
An ectopic scrotum occurs when the normal scrotum that is usually located under the penis is found in a different location. The main factor of this abnormality is a defect in the formation of the gubernaculum. The most common sites in which you can find an ectopic scrotum include a suprainguinal, inguinal, infra-inguinal, or perineal region.
Penoscrotal Transposition:
This condition is a genital malpositioning in which the penis is lying in an abnormal location in relation to the scrotum. This rare malformation is due to a defect of the scrotum during its caudal migration in the 9th to 11th week of gestation. This inversion can be classified as either partial or complete. A partial transposition occurs when the penis lies in the middle of the scrotum. A complete penoscrotal transposition, on the other hand, is seen when the penis emerges beneath the scrotum. The literature has shown that this anomaly can present in conjunction with central nervous system, cardiovascular and gastrointestinal tract congenital malformations
Surgical Considerations
Cryptorchidism:
One of the most common neonatal malformations is an undescended testicle. It occurs in about 2-4% of male neonates. Given the numerous associated complications of cryptorchidism, those patients with failure of the testes to spontaneously descend require a surgical repair called orchiopexy. The traditional approach of the orchiopexy is by making an inguinal incision. Given the numerous structures that can suffer injury through this incision, a trans scrotal approach was introduced. The advantages of this effective technique include uncomplicated dissection, better wound healing, and shorter operation time. The disadvantages of this approach include wound infections, testicular hypotrophy, hydrocele, and hernias.
Testicular Torsion:
Testicular torsion is a vascular emergency defined as a rotation of the spermatic cord that causes strangulation of the testicular artery, which leads to loss of blood supply to the testicle. This pathology requires a prompt surgical approach. A defect in embryologic development of the tunica vaginalis is mainly responsible for testicular torsion. Patients with testicular torsion may present with acute pain, nausea, and vomiting. Testicular torsion should be approached urgently, given that after the first 24 hours of the loss of blood supply, the tissue can suffer necrosis. Once the tissue is necrotic, there is very little likelihood of saving the testis.
Hematocele:
Hematocele is a blood collection in the scrotum that can be associated with a traumatic injury or a surgical procedure. It can be diagnosed using a scrotal ultrasound. This blood collection usually self resolves, but in rare cases, it may develop calcifications, which are extra-testicular lesions palpated on physical exams.
Clinical Significance
The genitourinary male physical examination has great significance in diagnosing many pathologies. Some of the clinical diseases that may show local signs detected on a physical exam are noninflammatory edema, cellulitis, and Fournier gangrene.
Noninflammatory Edema:
Noninflammatory edema may be visible on a scrotal ultrasound as a fluid collection that is not associated with scrotal wall hyperemia. The pathophysiology attributed to this edema could be heart failure, liver failure, lymphatic obstruction, or a fluid imbalance.
Cellulitis:
Cellulitis should be included in the differential diagnoses in any patient presenting with scrotal pain and a thickened erythematous scrotum on a physical exam. This diagnosis should be considered in any patient with a past medical history of immunosuppression, diabetes, or obesity, presenting with local signs of inflammation in the scrotum.
Infectious Diseases:
Many microorganisms can colonize in the scrotum and cause an infection. From the components of the scrotum, the tail of the epididymis is one of the first sites of infections, given that it is the most vascularized portion of the scrotum. Any infection originating from a surrounding anatomical structure can migrate and eventually infect the scrotum. The most common microorganisms infecting the scrotum are bacterial pathogens, among which Escherichia coli, Proteus, Neisseria gonorrhea, and Chlamydia trachomatis predominate.
Review Questions
Access free multiple choice questions on this topic.
Comment on this article.
Figure
Median Sagittal Section of Male Pelvis Anatomy. Anatomy includes pelvis, sacrum, peritoneum, vesical layer, fascia of urogenital diaphragm, superior and inferior layer, bladder, prostate, symphysis pubis, rectum, anal canal, corpus cavernosum penis and (more...)
Figure
The Male Genital Organs, The scrotum. On the left side the cavity of the tunica vaginalis has been opened; on the right side only the layers superficial to the Cremaster have been removed Henry Vandyke Carter, Public Domain, via Wikimedia Commons
Figure
Scrotum anatomy Image courtesy S Bhimji MD
References
1.
: Lee JI, Jung HG. Perineal accessory scrotum with a lipomatous hamartoma in an adult male. J Korean Surg Soc. 2013 Dec;85(6):305-8. [PMC free article: PMC3868684] [PubMed: 24368990]
2.
: Mikuz G. Ectopias of the kidney, urinary tract organs, and male genitalia. Pathologe. 2019 Jun;40(Suppl 1):1-8. [PubMed: 30446779]
3.
: Rebik K, Wagner JM, Middleton W. Scrotal Ultrasound. Radiol Clin North Am. 2019 May;57(3):635-648. [PubMed: 30928082]
4.
: Gupton M, Varacallo M. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Oct 24, 2022. Anatomy, Abdomen and Pelvis: Genitofemoral Nerve. [PubMed: 28613484]
5.
: Mellick LB, Mowery ML, Al-Dhahir MA. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Apr 19, 2023. Cremasteric Reflex. [PubMed: 30020720]
6.
: Iida K, Mizuno K, Nishio H, Moritoki Y, Kamisawa H, Kurokawa S, Kohri K, Hayashi Y. Accessory Scrotum With Perineal Lipoma: Pathologic Evaluation Including Androgen Receptor Expression. Urol Case Rep. 2014 Nov;2(6):191-3. [PMC free article: PMC4782125] [PubMed: 26958486]
7.
: Swartz JM, Ciarlo R, Denhoff E, Abrha A, Diamond DA, Hirschhorn JN, Chan YM. Variation in the clinical and genetic evaluation of undervirilized boys with bifid scrotum and hypospadias. J Pediatr Urol. 2017 Jun;13(3):293.e1-293.e6. [PMC free article: PMC5483185] [PubMed: 28215832]
8.
: Nazem M, Hosseinpour M, Alghazali A. Trans-scrotal Incision Approach versus Traditional Trans-scrotal Incision Orchiopexy in Children with Cryptorchidism: A Randomized Trial Study. Adv Biomed Res. 2019;8:34. [PMC free article: PMC6543866] [PubMed: 31259163]
9.
: Alonso V, Rodriguez LE, Rodriguez MM. Conservative Management of Scrotal Hematoma Secondary to Adrenal Hemorrhage in Newborns. Urology. 2019 Nov;133:e1-e2. [PubMed: 31465793]
10.
: Adlan T, Freeman SJ. Can ultrasound help to manage patients with scrotal trauma? Ultrasound. 2014 Nov;22(4):205-12. [PMC free article: PMC4760557] [PubMed: 27433221]
: Disclosure: Rosa Garcia declares no relevant financial relationships with ineligible companies.
: Disclosure: Hussain Sajjad declares no relevant financial relationships with ineligible companies.
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
Bookshelf ID: NBK549893PMID: 31751083
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Garcia RA, Sajjad H. Anatomy, Abdomen and Pelvis, Scrotum. [Updated 2023 Jul 24]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
In this Page
Introduction
Structure and Function
Embryology
Blood Supply and Lymphatics
Nerves
Muscles
Physiologic Variants
Surgical Considerations
Clinical Significance
Review Questions
References
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190237 | https://www.hsto.info/iis-ru/bm600/dl/thm_ch06_elast_b4.pdf | Business Economics Elasticity and its applications Thomas & Maurice, Chapter 6 Herbert Stocker herbert.stocker@uibk.ac.at Institute of International Studies University of Ramkhamhaeng & Department of Economics University of Innsbruck Elasticities Question: How ‘strongly’ reacts one variable in response of a change in another variable?
→slope of a curve.
e.g. Demand for wheat in the USA: Qd = 3550 −266P where Qd is measured in million ‘bushels’ per year and P is measured in US$.
Slope: dQd dP = −266 i.e. if the price increases by one dollar the quantity demanded decreases by 266 mio ‘bushels’!???
Elasticities Problem: For the interpretation of the slope we have to know the dimensions in which the units are measured!
Alfred Marshall (1842-1924): proposed the use of relative changes!
Price-elasticity = Percentage change in quantity Percentage change in price = %∆Qd %∆P e.g. what is the percentage decrease in the quantity demanded of wheat if the price of wheat raises by one percent?
Elasticities Elasticity: Percentage change in the dependent variable resulting from a one percent increase in the independent variable.
Elasticities are a very general concept to express the ‘strength of reaction’ of one variable in response to the change of another variable.
Main advantage of elsticities: free of dimensions!
Two concepts: Arc-Elasticity: discrete change between two observed points.
Point-Elasticity: infinitesimal change (function must be known).
Arc-Elasticity When only two price-quantity pairs are known: (for discrete changes) E B = %-Change of Y %-Change of X = Y2−Y1 Y1 X2−X1 X1 ≡ ∆Y Y1 ∆X X1 ≡ ∆Y ∆X Y1 X1 = ∆Y ∆X X1 Y1 Y X b before X1 Y1 bc after X2 Y2 ∆X ∆Y Arc-elasticity For discrete changes: e.g.: price raises from 2$ to 4$ ⇒quantity decreases from 20 kg to 15 kg.
P Qd bc b ∆P P2 = 4 P1 = 2 bc bc ∆Qd 15 = Qd2 Qd1 = 20 ∆Qd = Qd1 −Qd2 = 20 −15 = 5 ∆P = P1 −P2 = 2 −4 = −2 E B Qd,P = ∆Qd Qd1 ∆P P1 = 5 20 −2 2 = −1 4 Arc-Elasticity Problem: the values of the arc-elasticity depend on whether the price increases or decreases!
E 1 Qd,P = Qd1−Qd2 Qd1 × 100 P1−P2 P1 × 100 ↔E 2 Qd,P = Qd2−Qd1 Qd2 × 100 P2−P1 P2 × 100 Solution: Mid-Point Method EQd,P = Qd1−Qd2 (Qd1+Qd2)/2 × 100 P1−P2 (P1+P2)/2 × 100 Point-Elasticity For curvilinear demand curves: Slope is calculated by using the derivative (∆→d).
Point Elasticity: EQd,P = dQd dP Qd P = dQd dP P Qd P Qd bc Qd1 P1 b Point-Elasticity Example: What is the price elasticity of demand for the following demand function: Qd = 25 −2.5P at the price P1 = 2, (⇒Qd1 = 25 −2.5 × 2 = 20): EQd,P = dQd dP P1 Qd1 = −2.5 2 20 = −0.25 Point-Elasticity Example Qd = 25 −2.5P at a different price P2 = 4 demand is Qd2 = 15, and the elasticity is therefore EQd,P = dQd dP P2 Qd2 = −2.5 4 15 = −2 3 ≈−0.66˙ 6 For linear functions, the value of the elasticity is different at each point!
Point-Elasticity Problem: For linear functions the elasticity is not constant along the line!
EQd,P = dQd dP Pi Qdi Solution: the elasticity is often calculated at the mean value of the variable.
EQd,P = dQd dP P Qd mit: P ≡1 N PN i=1 Pi und Qd ≡1 N PN i=1 Qdi Elasticities An elasticity shows the percentage change in the dependent variable (Y ) when the independent variable (X) increases by one percent.
Elasticities are positive if the derivative is positive, i.e. if the variables move in the same direction.
Elasticities are negative if the variables move in opposite directions.
An elasticity measures how ‘strongly’ one variable reacts in response of a change in another variable.
Rule of Thumb Managers can get a rough estimate of price elasticity by asking two questions: What price P customers you currently pay for the product?
At what price A would customers stop buying my product altogether?
The answers to this questions can be used to calculate a rough estimate of the demand elasticity, since EQd,P = P (P −A) where P is the price and A is the vertical intercept of the plotted demand curve (the P-axis).
Rule of Thumb Why?
P Q A b P P = A −s Q Q = 1 s [A −P] dQ dP = −1 s E = dQ dP P Q = −1 s P 1 s [A −P] = P P −A Graphical Derivation Elasticity and Angle Slope and Angle Remember: Hypotenuse Adjacent Opposite leg α tan α = Opposite leg Adjacent = Slope Elasticity and Angle Y X A bc α XA XA YA YA β Y = f (X) dY dX = tan α YA XA = tan β EA = dY dX YA XA = −tan α tan β Elasticity and Angle Y X α β A bc EA = dY dX YA XA = −tan α tan β Y = f ( X ) XA YA Elasticity and Angle P Qd α β B bc EA = dQd dP QdB PB = −tan α tan β Qd = f ( P ) QdB PB Elasticity and Angle 0 1 2 3 4 0 1 2 3 4 P Qd α bc β E1 = −tan α tan β = − 4 4 3 1 = −1 3 bc γ E2 = −tan α tan γ = − 4 4 2 2 = −1 bc δ E3 = −tan α tan δ = − 4 4 1 3 = −3 bc E4 = −tan α tan(0) = − 4 4 0 4 = −1 0 = −∞ bc E0 = −tan α tan(90) = − 4 4 4 0 = −1 ∞= 0 Elasticity and Angle P Qd α β A bc Positive Intercept: tan α > tan β EA = dQd dP QdA PA = tan α tan β > 1 Question: which value has the elasticity of a linear function that goes through the origin?
Functions with constant elasticity Y = f (X) = AX b dY dX = bAX b−1 Y X = AX b X = AX b−1 EY ,X = dY dX Y X = bAX b−1 AX b−1 = b ⇒Elasticities of power-functions are always constant!
Functions with constant elasticity Example: Y = f (X) = 3X 0.5 dY dX = 0.5 × 3X 0.5−1 = 0.5 × 3X −0.5 Y X = 3X 0.5 X = 3X 0.5−1 = 3X −0.5 EY ,X = dY dX Y X = 0.5 × 3X −0.5 3X −0.5 = 0.5 Elasticities of log-linear Functions ln Y = f (ln X) EY ,X = d ln Y d ln X = dY Y dX X = dY dX X Y Example: Y = f (X) = 3X 0.5 ln Y = ln 3 + 0.5 ln X EY ,X = d ln Y d ln X = 0.5 Elasticities of log-linear Functions EY ,X = d ln Y d ln X = dY Y dX X Intuition: d ln Y dY = 1 Y ⇒ d ln Y = dY Y d ln X dX = 1 X ⇒ d ln X = dX X EY ,X = dY Y dX X = d ln Y d ln X Elasticities of log-linear Functions Example: Y = AX b ln Y = ln A + b ln X EY ,X = d ln Y d ln X = b Example: Y = f (X) = 0.25X −2 ln Y = ln 0.25 −2 ln X EY ,X = d ln Y d ln X = −2 Demand with constant elasticity P Qd Qd = 2P−0.5 Qd = 0.25P−2 Elastic: Qd = 0.25P−2 ln Qd = ln 0.25 −2 ln P EQd,P = −2 Inelastic: Qd = 2P−0.5 ln Qd = ln 2 −0.5 ln P EQd,P = −0.5 Applications of Elasticities Theory of Demand Price Elasticity of Demand Demand Function: Qd = Qd(P, M, PS, PC, . . .) Price Elasticity of Demand: (sometimes called price or demand elasticity) EQd,P = dQd dP Qd P = dQd dP P Qd ≈%∆Qd %∆P The Price Elasticity of Demand shows the percentage decrease of the quantity demanded if price ceteris paribus increases by one percent.
Price Elasticity of Demand Demand is elastic if |EQd,P| > 1, or %∆Q > %∆P (The quantity demanded responds more than proportionally to a a change in price.
Demand is unit elastic if |EQd,P| = 1, or %∆Q = %∆P Demand is inelastic if 0 < |EQd,P| < 1, or %∆Q < %∆P (The quantity demanded responds less than proportionally to a a change in price.
Since the price elasticity is usually negative it is common to use the absolute value |EQd,P|.
Linear Demand Function and Elasticity In which point of a linear demand function the elasticity has the value −1?
elastic inelastic bc bc bc EQd,P = −1 EQd,P = −∞ EQd,P = 0 a 2b a 2 a P Qd Qd = a −bP EQd,P = dQd dP P Qd = −bP a −bP = −1 bP = a −bP P = a 2b Qd = a −b a 2b = a 2 Special cases . . .
Perfectly inelastic de-mand: EQd,P = 0 P Qd A change in price has no effect on the quantity demanded!
Perfectly elastic demand: EQd,P = −∞ P Qd A change in price has an infinitely large effect on the quantity de-manded!!
Determinants of price elasticity Determinants of price elasticity: ceteris paribus demand tends to be more elastic, . . .
the more and closer substitutes are available.
when the good is rather a luxury than a necessity.
the higher the proportion of income spent on the good.
the longer the time period under consideration.
demand for durable goods tends to be more elastic than demand for non-durables (consumers choose to hold on to the good instead of replacing it).
Price elasticities for cars Model Price Estimated EQd,P Mazda 323 $ 5,039 −6.358 Nissan Sentra $ 5,661 −6.528 Ford Escort $ 5,663 −6.031 Honda Accord $ 9,292 −4.798 Ford Taurus $ 9,671 −4.220 Nissan Maxima $13,695 −4.845 Cadillac Sevifle $24,544 −3.973 Lexus LS400 $27,544 −3.085 BMW 735i $37,490 −3.515 Source: Table V in S. Berry, Levinsohn, and A. Pakes, “Automobile Prices in Market Equilibrium”, Econometrica 63 (July 1995): 841-890. [aus: D. Besanko & D. Braeutigam, Microeconomics (Wiley)] ⇒probably cheaper cars are perceived more as substitutes than luxury cars.
Price Elasticity Attention: Even if the demand for the entire product is rather inelastic the elasticity for the individual producer might be quite large.
For example, the demand for eggs or potatoes is rather inelastic, but the elasticity for the eggs or potatoes of an individual farmer might be close to infinity!
Price Elasticity of Demand and Total Revenue Elasticity & Total Revenue Elastic Demand: P Qd bc bc Total revenue decreases when price increases!
Inelastic Demand: P Qd bc bc Total revenue increases when price increases!
Total revenue is P × Qd (i.e. the hatched area) Elasticity & Total Revenue elastic inelastic E = −1 E = −∞ E = 0 a 2 P Qd bc bc If demand is ineleas-tic price and total revenue move in the same direction!
Elasticity & Total Revenue elastic inelastic E = −1 E = −∞ E = 0 a 2 P Qd bc bc If demand is elastic price and total rev-enue move in opposite direction!
Elasticity & Total Revenue With Calculus: TR = P × Q(P) d(TR) dP = d[P × Q(P)] dP = Q + P dQ dP = Q 1 + dQ dP P Q = Q(1 + EQd,P) = Q(1 −|EQd,P|) Elasticity & Total Revenue d(TR) dP = Q(1 −|EQd,P|) d(TR) dP = 0 for |EQd,P| = 1 ⇒Max.!
d(TR) dP < 0 for |EQd,P| > 1 ⇒elastic d(TR) dP > 0 for |EQd,P| < 1 ⇒inelastic Other Elasticities Income Elasticity Demand function: Qd = Qd(P, M, PS, PC, . . .) Income Elasticity of Demand: EQd,M = Percentage change in quantity demanded Percentage change in income = dQd dM Qd M = dQd dM M Qd ≈%∆Qd %∆M The Income Elasticity of Demand shows the percentage change in quantity demanded if income ceteris paribus increases by one precent.
Income Elasticity of Demand Normal goods (necessities): 0 < EQd,M < 1: income elasticity is between 0 and 1.
Luxury or superior goods: EQd,M > 1: if income ceteris paribus increases by one percent the quantity demanded will increase by more than one percent! Example: lobster, . . .
Inferior goods: EQd,M < 0: the quantity demanded decreases if income increases! Example: second-hand clothes, rice, . . .
Income Elasticity of Demand Income elasticity of demand can be important for firms: Demand for luxuries increases more than proportional with income, markets for luxuries tend to grow more rapidly than markets for normal and inferior goods.
Firms can try to target marketing campaigns to consumer groups with higher income elasticity.
Developing countries are often specialized in primary production with low income elasticities.
Income Elasticity of Demand Estimates of the Income Elasticity of Demand for Selected Food Products Product Estimated EQd,M Product Estimated EQd,M Cream 1.72 Milk 0.50 Peaches 1.43 Butter 0.37 Apples 1.32 Potatoes 0.15 Fresh peas 1.05 Margarine −0.20 Oranges 0.83 Flour −0.36 Eggs 0.44 Source: Daniel B. Suits, “Agriculture”, in: The Structure of American Industry, W. Adams and J. Brock, eds.
(Englewood, Nj: Prentice Hall), 1995; H. S. Houthhakker and Lester D. Taylor, “Consumer Demand in the United States, 1929-1970” (Cambridge, MA: Harvard University Press), 1966.
taken from: D. Besanko & D. Braeutigam, Microeconomics (Wiley) Cross Price Elasticity Cross-price elasticity of demand: measures how demand for Good X varies with changes in the price of another Good Y .
Substitute goods have positive cross elasticity.
Complementary goods have negative cross elasticity.
Defines relevant market in which different products compete.
Cross Price Elasticity Demand function: Qd = Qd(P, M, PS, PC, . . .) Cross Price Elasticity of Demand: EQd,PS = Percentage change in demand for good A Percentage change in price for good B = dQdA dPB QdA PB = dQdA dPB PB QdA ≈%∆QdA %∆PB The Cross Price Elasticity of Demand shows the percentage change in the demand for a good, if the price of another good changes by one percent.
Cross Price Elasticity Substitutes: (e.g. Cafe and Tea) ⇒cross price elasticity is positive if tea becomes more expensive the demand for cafe increases.
EQd,PS = dQd dPS PS Qd > 0 Complementary Goods: (e.g. cafe and sugar) ⇒cross price elasticity is negative if cafe becomes more expensive the demand for sugar decreases.
EQd,PC = dQd dPC PC Qd < 0 Price and Cross Price Elasticities Demand for Price of Beef Price of Pork Price of Chicken Beef −0.65 0.01 0.20 Pork 0.25 −0.45 0.16 Chicken 0.12 0.20 −0.65 Source: Daniel B. Suits, “Agriculture”, in: The Structure of American Industry, W. Adams and J. Brock, eds.
(Englewood, Nj: Prentice Hall), 1995 entnommen aus: D. Besanko & D. Braeutigam, Microeconomics (Wiley) Price elasticities are on the main diagonale, offthe main diagonal are the cross price elasticities.
e.g.: −0.65 is the price elasticity of beef, 0, 01 is the cross price elasticity of the demand for beef with respect to the price of pork.
Price and Cross Price Elasticities Sometimes useful to judge whether markets are ‘related’.
Price of Price of Price of Price of Demand for Sentra Escort LS400 735i Sentra −6.528 0.078 0.000 0.000 Escort 0.454 −6.031 0.001 0.000 LS400 0.000 0.001 −3.085 0.093 735i 0.000 0.001 0.032 −3.515 Source: S. Berty Levinsohn, and A. Pakes, “Automobile Prices in Market Equilibrium”, Econometrica 63 (July 1995): 841-890.
entnommen aus: D. Besanko & D. Braeutigam, Microeconomics (Wiley) Diagonal elements: the price elasticity of demand Off-diagonal elements: the cross-price elasticity of demand.
Demand for Coca- and Pepsi Cola Econometrically estimated demand functions: Qdc = 26.17 −3.98Pc + 2.25Pp + 2.60Ac −0.62Ap + 0.99M + . . .
Qdp = 17.48 −5.48Pp + 1.40Pc + 2.83Ap −4.81Ac + 1.92M + . . .
Qdc quantity demanded of Coca-Cola (ten million cases) Qdp quantity demanded of Pepsi (ten million cases) Pc price of Coca-Cola (dollars per ten cases) Pp price of Pepsi (dollars per ten cases) Ac advertising expenditures on behalf of Coca-Cola Ap advertising expenditures on behalf of Pepsi M disposable income in the United States All prices expressed in 1986 U.S. dollars!
Elasticities: Coca Cola und Pepsi Cola By inserting the means (e.g. Pc = 12, 96, Pp = 8, 16; Ac = 5, 89; . . . ) one can calculate the elasticities in the mean: Price, Cross-Price, and Income Elasticities of Demand for Coca-Cola and Pepsi Elasticity Coca-Cola Pepsi Price elasticity of demand −1.47 −1.55 Cross-price elasticity of demand 0.52 0.64 Income elasticity of demand 0.58 1.38 Source: Gasmi, F., J.J. Laffont and Q. Vuong (1992): “Econometric Analysis of Collusive Behaviour in the Soft Drink Market”, Journal of Economics and Marketing Strategy, Vol. 1 entnommen aus: D. Besanko & D. Braeutigam, Microeconomics (Wiley) Advertising Elasticity of Demand Advertising elasticity of demand: the percentage change in quantity demanded of a good relative to the percentage change in advertising dollars spent on that good.
Marketing studies: e.g. Tellis, 1988; Sethuraman and Tellis, 1991; Hoch, et al, 1995 Advertising elasticities of demand tend to be much smaller than price elasticities of demand (by a factor 10-15).
Supply Elasticity Price elasticity of supply is the percentage change in quantity supplied resulting from a percent change in price.
ES,P = dS dP S P = dS dP P S ≥0 Supply is elastic when ES,P > 1 Supply is inelastic when ES,P < 1 Supply Elasticity Determinants of Elasticity of Supply: Time period: Supply is more elastic in the long run!
Ability of sellers to change the amount of the good they produce.
(Beach-front land is probably inelastic, while books, cars, or manufactured goods are rather elastic) Special cases . . .
Perfectly Inelastic Supply: ES,P = 0 P S An increase in price leaves the quantity supplied unchanged!
Perfectly Elastic Supply: ES,P = ∞ P Qs Below the price the quantity sup-plied is zero, above it is infinite!
Price Elasticity in Marketing Managerial Price Sensitivity Analysis The price elasticity of demand (in marketing literature often called price sensitivity) is one of the most important variables for managers.
A managerial analysis of price elasticity should be a written document that can be criticized and improved over time.
It should include some of the following questions: Reference Price Substitutes and Reference Price: Are there close substitutes to the product, and if so, are the buyers (or a segment of buyers) usually aware thereof when making a purchase? Can they compare prices?
Can buyers speed up or delay purchases based on expectations of future prices?
How difficult is it for buyers to compare offers of different suppliers?
Switching Cost & Expenditure Share To what extent have buyers already made investments (monetary and/or psychological) that they would need to incur again if they switched suppliers?
For how long are buyers presumably ‘locked’ by those expenditures?
How significant are buyers expenditures for the product?
For end consumers mainly the portion of income is important.
For business customers also the absolute price might be important.
Fairness Buyers are more sensitive to a product’s price when it is outside the range that they perceive as ‘fair’ or ‘reasonable’.
How does the current price compare with prices people have paid in the past?
What do buyers expect to pay for similar products?
Do customers perceive the product as ‘necessity’ or as a discretionary purchase?
Framing Effect Prospect Theory: (D. Kahneman & A. Tversky) Essential idea: people ‘frame’ purchasing decisions in their minds as a bundle of gains and losses.
Consumers tend to be more price sensitive when they perceive the price as a ‘loss’ rather than a foregone ‘gain’.
Additionally, they are more price sensitive when the price is paid separately rather than as part of a bundle.
Framing Effect Example: (Prospect Theory) Gas station A sells gasoline for $1.20 and gives a $0.10 per liter discount if the buyer pays with cash.
Gas station B sells gasoline for $1.10 and charges a $0.10 surcharge if the buyer pays with credit card.
Most people choose station A.
We’ll have to say a lot more about price elasticities when we study optimal pricing on imperfect markets. |
190238 | https://www.youtube.com/watch?v=jcFLhEBgi68 | Find the value of x and y
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okay so to answer this question right here where we've been asked to find the value of X and Y we we have a lot of methods on how we can answer this question but I'll just pick one for you so that you understand and you follow it okay this is called substitution method so if you can see nicely here we've got a equation one under equation two there are two equations so between the two equations pick one of the equations that you start working with so get to the first one which is uh 2 X Plus Y is equal to F4 so I've gotten the first one now after getting the first one you make one of the letters be subject of the formula so we have 2X plus y here whichever letter that you want to be subject to the formula you can make it so here I'll make y subject of the formula so I'll show y here to be 4 minus 2 X hopefully you've seen what has happened these two eggs cross the equal sign and this is what we we have after reaching this point we now go to equation two we know that this is equation one this is equation two in equation 2 we substitute where Y is here we substitute this which have three X minus 3 4 minus 2 there inside the brackets is equal to a 2. so hopefully you've seen the steps what I've done here is I've is that I've just gotten what is right here and substituted it there and this is the what we have the next thing here is to solve for x so how do we solve for x order of operation needs to be applied here so we have 3 x minus 12. we get rid of the bracket so negative three times the four it's a negative 12. plus 6 x negative and negative it's positive three times two it's a six and X there is equal to 2. okay so we group The like terms which have a 3 X plus six x is equal to 2 Plus 12. okay after grouping the like terms we now work out things we have here we have 9 x is equal to fourteen remember we're solving for x divide by nine divide by nine this side will cancel so our x value would just be equal to 14 over 9. like that so now uh we can also solve for for y we've we've solved for X so for y we come at this point where we made y subject to the formula so we shall say why is equal to 4 minus 2 x okay Y is equal to 4 minus 2 What is the value of x here it's 14 over 9. here we work out we're getting a 28 over 9 okay so the lowest common denominator here will be a nine nine uh nine into nine it's one one times z uh negative 28 is negative 28 here here is a one one and two nine it's nine nine times uh four it's thirty six okay so when we work out here to finish it up we find that the value of y will be equal to 36 minus twenty eight that's eight so the value of y is eight over nine thank you so much for watching this is Jacob hopefully you've learned one or two things out of this video bye |
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7: 1st Order RC/RL Circuit
Introduction to Circuit Analysis
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7.5: Transient Response of RL Circuits
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Contents
1. Home
2. Workbench
3. Introduction to Circuit Analysis
4. 7: 1st Order RC/RL Circuit
5. 7.5: Transient Response of RL Circuits
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Introduction to Circuit Analysis
Front Matter
1: Basic Concepts and Quantitites
2: Fundamental Laws
3: Nodal and Mesh Analysis, Dependent Sources
4: Analysis Theorems and Techniques
5: Advanced Topic- Operational Amplifiers
6: Capacitors and Inductors
7: 1st Order RC/RL Circuit
8: Parallel RLC Circuits
9: Series RLC Circuits
10: AC Signal Fundamentals
11: AC Power
12: Laplace Transform in Circuit Analysis
Back Matter
7.5: Transient Response of RL Circuits
Last updated Dec 21, 2023
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7.4: Initial and Steady-State Analysis of RL Circuits
7.6: Initial and Steady-State Analysis of RLC Circuits
Page ID 98488
James M. Fiore
Mohawk Valley Community College
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Example 9.5.1
2. Computer Simulation
1. Example 9.5.2
2. Example 9.5.3
Computer Simulation
References
The transient response of RL circuits is nearly the mirror image of that for RC circuits. To appreciate this, consider the circuit of Figure 9.5.1 .
Figure 9.5.1 : RL circuit for transient response analysis.
Again, the key to this analysis is to remember that inductor current cannot change instantaneously. When power is first applied, the circulating current must remain at zero. Therefore no voltage drop is produced across the resistor, and by KVL, the voltage across the inductor must equal the source, E. This establishes the initial rate of change of current via Equation 9.2.9 (di/dt=E/L) and is represented by the dashed red line in the graph of Figure 9.5.2 . As the current starts to increase, the voltage drop across the resistor begins to increase. This reduces the voltage available for the inductor, thus slowing the rate of change of current. This is depicted by the solid red curve on the graph. Meanwhile, the solid blue curve represents the decreasing inductor voltage. Thus, in the RL circuit, the inductor's voltage curve echoes the RC circuit's current curve (or resistor voltage curve), and the RL current curve echoes the RC circuit's capacitor voltage curve.
The curves presented in Figure 9.5.2 are identical to those presented in Chapter 8 when we discussed capacitors. They are reproduced here for your convenience.
Figure 9.5.2 : Normalized charge and discharge curves.
As noted before, the rate of current change versus time is equal to v/L, and therefore in this case, E/L. If the initial rate of change were to continue unabated, the maximum (steady-state) current, E/R, would be reached in L/R seconds 1. Therefore the time constant for an RL circuit is:
(7.5.1)τ=L R
Once again, five constants will achieve steady-state.
Following the prior work on capacitors, the relevant equations for the RL circuit can be shown to be 2:
(7.5.2)V L(t)=Eϵ−t τ
(7.5.3)V R(t)=E(1−ϵ−t τ)
(7.5.4)I(t)=E R(1−ϵ−t τ)
Where
V L(t) is the inductor voltage at time t,
V R(t) is the resistor voltage at time t,
I(t) is the current at time t,
E is the source voltage,
R is the series resistance,
t is the time of interest,
τ is the time constant,
ε (also written e) is the base of natural logarithms, approximately 2.718.
Time for an example.
Example 9.5.1
Given the circuit of Figure 9.5.3 , assume the switch is closed at time t=0. Determine the charging time constant, the amount of time after the switch is closed before the circuit reaches steady-state, and the inductor voltage and current at t=0, t=2 microseconds and t=1 millisecond. Assume the inductor is initially uncharged.
Figure 9.5.3 : Circuit for Example 9.5.1 .
First, the time constant:
τ=L R
τ=400μH 150Ω
τ≈2.667μs
Steady-state will be reached in five time constants, or approximately 13.33 microseconds. Thus we know that V L(0)=9 volts and V L(1ms)=0 volts. Because the inductor is an open initially, I L(0)=0 amps. At I L(1 ms), the circuit is in steady-state and the inductor acts like a short. Therefore, all of the 9 volt source drops across the 150 Ω resistor, for 60 mA.
To find V L(2 \mu s) we simply solve Equation 7.5.2.
V L(t)=Eϵ−t τ
V L(2μs)=9Vϵ−2μs 2.667μs
V L(2μs)≈4.251V
This value can also be determined graphically from Figure 9.5.2 . The time of 2 microseconds represents 75% of a time constant. Find this value on the horizontal axis and then track straight up to the solid blue curve that represents the charging inductor voltage. The point of intersection is right around 47% of the maximum value on the vertical axis. The maximum value here is the source voltage of 9 volts. Therefore the inductor will have reached approximately 47% of 9 volts, or just over 4.2 volts.
The current can be found in a similar manner using Equation 7.5.4.
I L(t)=E R(1−ϵ−t τ)
I L(2μs)=9V 150Ω(1−ϵ−2μs 2.667μs)
I L(2μs)=60mA(1−ϵ−0.75)
I L(2μs)=31.66mA
Computer Simulation
To verify our analysis, the circuit of Figure 9.5.3 is entered into a simulator, as shown in Figure 9.5.4 . In order to reflect the notion of a time-varying circuit with a switch, the 9 volt DC voltage source has been replaced with a rectangular pulse voltage source. This source starts at 0 volts and then immediately steps up to 9 volts. It stays at this level for 20 microseconds before dropping back to 0 volts.
Figure 9.5.4 : The circuit of Figure 9.5.3 in a simulator.
The results of a transient analysis are shown in Figure 9.5.5 . The waveform shown tracks the inductor's voltage at node 2 with respect to ground.
Figure 9.5.5 : Simulation results for the circuit of Figure 9.5.3 .
We can see that the voltage starts at 9 volts as expected. It then falls back to zero and is at steady-state in less than 15 microseconds, just as predicted. At 20 microseconds, the pulse source returns to zero volts. At this instant, the current through the inductor must still be the steady-state current of 60 milliamps. This current will still be flowing in a clockwise direction, thus it will produce a 9 volt drop across the 150 Ω resistor with a + to − polarity from left to right. This effectively places node 2 negative with respect to ground. The result is that the polarity of the inductor's voltage flips, with the inductor now acting as a short-lived source. We see this on the transient analysis as a negative 9 volt spike. The discharge time constant is identical to the charge constant, and thus we see the inductor's voltage fall back to zero in the same amount of time.
Example 9.5.2
Given the circuit of Figure 9.5.6 , find V L at t=1 microsecond after the circuit is energized. Assume the inductor is initially uncharged.
Figure 9.5.6 : Circuit for Example 9.5.2 .
First, the time constant:
τ=L R
τ=6mH 15kΩ
τ=400ns
Steady-state will be reached in five time constants, or 2 microseconds, at which point the inductor voltage will be zero as it will be behaving as a short. In contrast, as the inductor is initially an open (I L(0)=0), all of the current from the source will flow into the 15 k Ω resistor, producing 30 volts across this parallel network. Therefore, we can state that V L(0)=30 volts and V L(2 \mu s) = 0 volts, defining the extremes. In order to find V L(1 \mu s), we can use Equation 7.5.2.
V L(t)=Eϵ−t τ
V L(1μs)=30Vϵ−1μs 0.4μs
V L(1μs)≈2.463V
There are a few different ways to crosscheck this value. For starters, we can determine the inductor current using a slight modification of Equation 7.5.4 (the current source value is used in place of E/R as the equation effectively requires the maximum or steady-state current).
I L(t)=I(1−ϵ−t τ)
I L(1μs)=2mA(1−ϵ−1μs 0.4μs)
I L(1μs)=1.836mA
The inductor voltage of 2.463 volts would also have to appear across the parallel 15 k Ω resistor. This produces 2.463 V / 15 k Ω, or 0.164 mA of resistor current. By KCL, the remainder of the 2 mA source current must be flowing down through the inductor. This yields a net inductor current of 2 mA − 0.164 mA, or 1.836 mA, verifying our prior result.
Example 9.5.2 also reinforces the concept of the time constant being inversely proportional to the resistance, rather than directly proportional as in the RC case. In the circuit of Figure 9.5.6 , it should be obvious that the larger the resistance value, the larger the resulting initial-state voltage. From Equation 9.2.9 it can be seen that if the voltage across the inductor is increased, then the initial rate of change of current with respect to time will increase, and that implies a shorter time constant.
For more complex circuits, Thévenin's theorem may be used in order to determine the effective source voltage and charging resistance. As we saw with RC circuits, it's also possible that the discharge resistance may be considerably different from the charging resistance. In such a case, the charge and discharge curves can be highly asymmetric in both time and amplitude. Generally, the larger the discharge resistance is when compared to the charge resistance, the larger in voltage and the shorter in time the discharge spike will be (think in terms of the area under the curve staying constant).
Example 9.5.3
Assume the initial current through the inductor is zero in Figure 9.5.7 . Determine the time constant. Also, determine the inductor voltage and the voltage across the 6 k Ω resistor 200 nanoseconds after the switch is closed.
Figure 9.5.7 : Circuit for Example 9.5.3 .
This circuit is based on the circuit presented in Figure 9.3.2 as used in Example 9.3.1. In that analysis it was discovered that the steady-state voltage for the 6 k Ω and 2 k Ω resistors was 15 volts, the pair being in parallel. Further, the initial voltage across the 2 k Ω resistor and the inductor was 16.67 volts and for the 6 k Ω resistor, 0 volts.
For the Thévenin circuit, the open circuit voltage at the inductor would be the potential across the 2 k Ω resistor, which is obtained from a voltage divider between it and the 1 k Ω resistor, or 16.67 volts. The equivalent resistance is obtained by shorting the voltage source which leaves the 1 k Ω and 2 k Ω resistors in parallel, which is then in series with the 6 k Ω resistor, yielding approximately 6.667 k Ω. The equivalent is shown in Figure 9.5.8 .
Figure 9.5.8 : Thévenin equivalent of the circuit of Figure 9.5.7 driving the inductor.
Immediately, we can see that the initial-state voltage across the inductor is confirmed in the equivalent circuit. We can now determine the time constant.
τ=L R
τ=1mH 6.667kΩ
τ=150ns
Steady-state will be reached in 750 nanoseconds. To find the inductor voltage we can use Equation 7.5.2.
V L(t)=Eϵ−t τ
V L(200ns)=16.67Vϵ−200ns 150ns
V L(200ns)≈4.39V
Referring back to the original circuit, in order to determine the voltage across the 6 k Ω resistor we can find the current through it and use Ohm's law. The current would be the same as the inductor's current as the two are in series. Thus, Equation 7.5.4 would do the trick.
I L(t)=E R(1−ϵ−t τ)
I L(200ns)=16.67V 6.667kΩ(1−ϵ−200ns 150ns)
I L(200ns)=2.5mA(1−ϵ−1.333)
I L(200ns)≈1.841mA
And finally,
V 6k(200ns)=IR
V 6k(200ns)=1.841mA6kΩ
V 6k(200ns)≈11.05V
Further, the voltage across the 2 k Ω resistor must be the sum, or approximately 15.44 volts.
Computer Simulation
The results of Example 9.5.3 are crosschecked in a simulator. Once again the circuit is built using a pulse generator, as shown in Figure 9.5.9 .
Figure 9.5.9 : Circuit of Figure 9.5.7 in a simulator.
A transient analysis is run out to 1 microsecond which is modestly into steady-state. Node voltages 2 and 3 are plotted, as shown in Figure 9.5.10 . The initial voltage across the 2 k Ω resistor (node 2) is as predicted, approximately 16.7 volts, and falls to 15 volts at steady-state, approximately 750 nanoseconds later. The voltage across the 6 k Ω resistor (node 2) starts at zero volts and also winds up at 15 volts in steadystate, just as predicted. Further, note that the predicted voltages across the 2 k Ω and 6 k Ω resistors at 200 nanoseconds are verified.
Figure 9.5.10 : Transient analysis simulation of the circuit of Figure 9.5.7 .
The voltage across the inductor is node 2 minus node 3. This differential is plotted separately in Figure 9.5.11 . The expected initial, steady-state and 200 nanosecond voltages are as predicted.
Figure 9.5.11 : Simulation of the inductor voltage versus time for the circuit of Figure 9.5.7 .
One very important observation is that if an RL circuit is abruptly altered or opened, very large voltage spikes may occur. This is due to the fact that inductor current cannot change instantaneously. If the circuit is opened, the open represents a very large resistance. Ohm's law indicates that the inductor current times this very large resistance may produce a very large voltage across the new open. In fact, the potential may be sufficient to cause a spark or arc. Note that because the current cannot change instantaneously (both magnitude and direction), the inductor now behaves as a voltage source of very high magnitude and with reverse polarity. This phenomenon is used to create the ignition spark in internal combustion engines. In short, the ignition coil is charged, creating some current flow. The circuit is then interrupted leaving just the coil in series with the spark plug, the spark plug being little more than a precisely sized gap between two electrodes. This results in a large voltage being developed across the spark plug gap, typically in the neighborhood of 20,000 volts, which is sufficient to create a small arc (i.e., the spark) that then ignites the air-fuel mixture in the piston.
An example of producing a discharge voltage spike that is considerably larger than the source voltage can be illustrated with the circuit of Figure 9.5.12 . We shall assume that the inductor is initially uncharged when power is applied, and that the switch is in position 1. In this case, the circuit consists of just the 12 volt source, the 2.2 k Ω resistor, and the inductor. The circuit reaches steady-state in roughly 227 nanoseconds. At that point the inductor behaves as a short, leaving the full 12 volt source to drop across the 2.2 k Ω resistor. This produces a clockwise current of approximately 5.455 mA.
Figure 9.5.12 : Circuit illustrating large discharge voltage spike.
If we now move the switch to position 2, this current must be maintained because the current through the inductor cannot change instantaneously. The new discharge resistance is now the series combination of the two resistors, or 49.2 k Ω. Ohm's law and KVL dictate that the resulting inductor voltage must be 49.2 k Ω times 5.455 mA, or just beyond −268 volts. This potential is negative because the clockwise current is flowing up through the 47 k Ω resistor, producing a drop + to − from ground up. The increased resistance also shortens the time constant and now steady-state will be reached in only 10.1 nanoseconds. Thus, we see a much larger magnitude spike (over twenty times that of the source voltage) with a much shorter time duration (less than one-twentieth of the charge time).
Depending on the kind of switch that is used, things can be even more extreme than what has been just described. Switches come in two basic varieties: make-beforebreak and break-before-make. The former makes contact with the second position before it breaks contact with the first position, while the latter does the opposite. The behavior just described assumes that a make-before-break switch is being used. In contrast, if a break-before-make switch is used, we'd see a drastically different result.
Let us pick up back at steady-state with 5.455 mA flowing through the inductor. We now throw the switch to position 2. This new switch breaks contact with the wire leading back to voltage source prior to it making contact with the 47 k Ω resistor. For a short instant of time the switch is not making contact with anything and the resulting resistance in the loop is determined by the air gap between the switch contacts. Even if this was a mere 10 M Ω, the resulting potential would be over 50,000 volts. This will almost certainly create a spark and we will have inadvertently recreated the spark plug scenario. Hopefully, there are no combustible gases nearby.
References
1 Which is to say, E/L amps per second times L/R seconds yields E/R amps.
2 See also Appendix C.
This page titled 7.5: Transient Response of RL Circuits is shared under a Public Domain license and was authored, remixed, and/or curated by James M. Fiore.
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190242 | https://brainly.com/question/33890269 | [FREE] Two planes are equidistant from the center of a sphere and intersect the sphere. What is true of the - brainly.com
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Two planes are equidistant from the center of a sphere and intersect the sphere.
What is true of the circles? Are they lines in spherical geometry? Explain.
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When two planes are equidistant from the center of a sphere and intersect the sphere, they form circles on the surface of the sphere. These circles are not lines in spherical geometry, but rather curves that are parallel to each other and do not intersect.
Two planes that are equidistant from the center of a sphere and intersect the sphere will form circles on the surface of the sphere. These circles are not lines in spherical geometry.
In spherical geometry, a line is defined as the intersection of a plane with the sphere.
However, in this case, the planes are not intersecting the sphere at a single point, but instead intersecting it along a curve. This curve forms a circle on the surface of the sphere.
To understand this concept better, let's consider an example. Imagine a sphere representing the Earth and two planes that are equidistant from its center.
These planes could represent different latitudes on the Earth's surface. When these planes intersect the Earth, they will form circles that correspond to the latitudes. These circles are parallel to each other and do not meet.
In contrast, if we consider a line in spherical geometry, it would be a great circle on the surface of the sphere. A great circle is a circle that has the same center as the sphere itself and divides the sphere into two equal halves.
Examples of great circles on Earth are the equator and any line of longitude.
So, to summarize, when two planes are equidistant from the center of a sphere and intersect the sphere, they form circles on the surface of the sphere.
These circles are not lines in spherical geometry, but rather curves that are parallel to each other and do not intersect.
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When two planes equidistant from the center of a sphere intersect it, they form parallel circles on the sphere's surface, known as small circles. These small circles do not intersect and are different from straight lines in Euclidean geometry. In spherical geometry, such circles illustrate the concept of curves, similar to latitude lines on Earth.
Explanation
When two planes are equidistant from the center of a sphere and they intersect the sphere, they create circles on the surface of the sphere. These circles are called small circles if the planes do not pass through the center of the sphere, while circles formed by planes that pass through the center are called great circles.
Great Circles:
A great circle is formed when the plane goes through the center of the sphere.
An example of a great circle is the equator, which divides the sphere into two equal halves.
Small Circles:
If the planes are located at equal distances from the center but do not pass through it, they create small circles.
These small circles run parallel to each other and do not intersect.
In spherical geometry, the concept of a straight line differs from that in Euclidean geometry. Here, a 'line' is defined as the intersection of a plane with a sphere, resulting in a curve (a circle in this case). So, the circles created by these two planes are not straight lines; rather, they are curves that remain parallel and do not meet, similar to how lines of latitude work on Earth.
Additionally, observing Earth's latitude lines can assist in comprehending how these planes operate. The latitude lines are parallel and represent small circles when considering a plane intersecting the Earth.
To summarize, the circles formed by the intersection of two equidistant planes with a sphere are considered small circles in spherical geometry and are parallel, not straight lines.
Examples & Evidence
An example is the latitudinal lines on Earth, such as 30°N and 30°S, which are small circles formed by planes parallel to the equator and do not intersect with each other.
The understanding of great and small circles is established in geometry, particularly in spherical geometry, where the intersection of planes with a sphere can be explicitly defined. Many mathematical texts and educational resources confirm that great circles correspond to planes through the sphere's center, while equidistant planes create smaller circles.
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Community Answer thought provoking the postulates and theorems in this book represent euclidean geometry. in spherical geometry, all points are points on the surface of a sphere. a line is a circle on the sphere whose diameter is equal to the diameter of the sphere. explain how many right angles are formed by two perpendicular lines in spherical geometry.
Community Answer In spherical geometry, all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. A plane is the surface of the sphere. In spherical geometry, is it possible that two triangles are similar but not congruent? Explain your reasoning.
Community Answer 4.8 68 The center of a sphere is a line segment from the center point to the surface of the sphere. a fixed point equidistant from all points on the surface of the sphere. a three-dimensional circle in which all points are equidistant from a fixed point. the same as the base of the sphere.
Community Answer The diameter of a sphere is a chord that passes through the center of the sphere. a fixed point equidistant from all points on the surface of the sphere. a line segment from the center point to a point on the sphere. a three-dimensional circle in which all points are equidistant from a fixed point.
Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish?
Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit?
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Rationalize the denominator. 17−4 7
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Simplify by factoring. Assume that the variable in the radicand represents a positive real number and that the radicand does not involve negative quantities raised to even powers. 3 y 11
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190243 | https://www.isibang.ac.in/~d.yogesh/Course_Notes/DM1/Ch5.html | Discrete Mathematics - Lecture notes Some questions : Random spanning trees5.1 Erdös-Szekeres lemma and Dilworth’s theorem
Chapter 5 Extremal Combinatorics
The extremal refers to the fact that one is concerned with extremal (maximal or minimal) questions about graphs. A prototypical question is the maximal or minimal number of edges in a graph with a certain property. We have already answered such a question about maximal number of edges for a graph to be a tree.
A commonly used tool in extremal graph theory is the pigeonhole principle i.e., if a set constaining elements is partitioned into classes, then one of the classes must contain at least elements.
It is in answering a question in extremal graph theory, Erdös made powerful use of probabilistic ideas and this gave birth to what is now famously known as the probabilistic method. We shall see an illustration of this in the later section.
We shall begin with some combinatorial results such as Erdös-Szekeres lemma and Dilworth’s lemma before moving onto applications in Graph theory such as Turan’s theorem |
190244 | https://cds.cern.ch/record/1607220/files/epjc.74.2794.pdf | Eur. Phys. J. C (2014) 74:2794 DOI 10.1140/epjc/s10052-014-2794-6 Regular Article - Experimental Physics Measurement of negatively charged pion spectra in inelastic p+p interactions at plab = 20, 31, 40, 80 and 158GeV/c NA61/SHINE Collaboration N. Abgrall1, A. Aduszkiewicz2,a, Y. Ali3, T. Anticic4, N. Antoniou5, B. Baatar6, F. Bay7, A. Blondel1, J. Blumer8, M. Bogomilov9, A. Bravar1, J. Brzychczyk3, S. A.Bunyatov6, O. Busygina10, P. Christakoglou5, T. Czopowicz11, N. Davis5, S. Debieux1, H. Dembinski8, F. Diakonos5, S. Di Luise7, W. Dominik2, T. Drozhzhova12, J. Dumarchez13, K. Dynowski11, R. Engel8, A. Ereditato14, G. A. Feofilov12, Z. Fodor15, A. Fulop15, M. Ga´ zdzicki16,17, M. Golubeva10, K. Grebieszkow11, A. Grzeszczuk18, F. Guber10, A. Haesler1, T. Hasegawa19, M. Hierholzer14, R. Idczak20, S. Igolkin12, A. Ivashkin10, D. Jokovi´ c21, K. Kadija4, A. Kapoyannis5, E. Kaptur18, D. Kiełczewska2, M. Kirejczyk2, J. Kisiel18, T. Kiss15, S. Kleinfelder22, T. Kobayashi19, V. I. Kolesnikov6, D. Kolev9, V. P. Kondratiev12, A. Korzenev1, P. Kovesarki20, S. Kowalski18, A. Krasnoperov6, A. Kurepin10, D. Larsen18, A. László15, V. V. Lyubushkin6, M. Ma´ ckowiak-Pawłowska17, Z. Majka3, B. Maksiak11, A. I. Malakhov6, D. Mani´ c21, A. Marcinek3, V. Marin10, K. Marton15, H.-J. Mathes8, T. Matulewicz2, V. Matveev6,10, G. L. Melkumov6, St. Mrówczy´ nski16, S. Murphy1, T. Nakadaira19, M. Nirkko14, K. Nishikawa19, T. Palczewski23, G. Palla15, A. D. Panagiotou5, T. Paul24, C. Pistillo14, W. Peryt11,b, O. Petukhov10, R. Płaneta3, J. Pluta 11, B. A. Popov6,13, M. Posiadała2, S. Puławski18, J. Puzovi´ c21, W. Rauch25, M. Ravonel1, A. Redij14, R. Renfordt17, A. Robert13, D. Röhrich26, E. Rondio23, M. Roth8, A. Rubbia7, A. Rustamov17, M. Rybczy´ nski16, A. Sadovsky10, K. Sakashita19, M. Savi´ c21, K. Schmidt18, T. Sekiguchi19, P. Seyboth16, D. Sgalaberna7, M. Shibata19, R. Sipos15, E. Skrzypczak2, M. Słodkowski11, P. Staszel3, G. Stefanek16, J. Stepaniak23, H. Ströbele17, T. Šuša4, M. Szuba8, M. Tada19, V. Tereshchenko6, T. Tolyhi15, R. Tsenov9, L. Turko20, R. Ulrich8, M. Unger8, M. Vassiliou5, D. Veberiˇ c24, V. V. Vechernin12, G. Vesztergombi15, L. Vinogradov12, A. Wilczek18, Z. Włodarczyk16, A. Wojtaszek-Szwarc16, O. Wyszy´ nski3, L. Zambelli13, W. Zipper18 1 University of Geneva, Geneva, Switzerland 2 Faculty of Physics, University of Warsaw, Warsaw, Poland 3 Jagiellonian University, Cracow, Poland 4 Rudjer Boškovi´ c Institute, Zagreb, Croatia 5 University of Athens, Athens, Greece 6 Joint Institute for Nuclear Research, Dubna, Russia 7 ETH, Zurich, Switzerland 8 Karlsruhe Institute of Technology, Karlsruhe, Germany 9 Faculty of Physics, University of Sofia, Sofia, Bulgaria 10 Institute for Nuclear Research, Moscow, Russia 11 Warsaw University of Technology, Warsaw, Poland 12 St. Petersburg State University, St. Petersburg, Russia 13 LPNHE, University of Paris VI and VII, Paris, France 14 University of Bern, Bern, Switzerland 15 KFKI Research Institute for Particle and Nuclear Physics, Budapest, Hungary 16 Jan Kochanowski University in Kielce, Kielce, Poland 17 University of Frankfurt, Frankfurt, Germany 18 University of Silesia, Katowice, Poland 19 High Energy Accelerator Research Organization (KEK), Tsukuba, Ibaraki 305-0801, Japan 20 University of Wrocław, Wrocław, Poland 21 University of Belgrade, Belgrade, Serbia 22 University of California, Irvine, USA 23 National Centre for Nuclear Research, Warsaw, Poland 24 Laboratory of Astroparticle Physics, University Nova Gorica, Nova Gorica, Slovenia 2794 Page 2 of 22 Eur. Phys. J. C (2014) 74:2794 25 Fachhochschule Frankfurt, Frankfurt, Germany 26 University of Bergen, Bergen, Norway Received: 14 October 2013 / Accepted: 21 February 2014 © The Author(s) 2014. This article is published with open access at Springerlink.com Abstract We present experimental results on inclusive spectra and mean multiplicities of negatively charged pions produced in inelastic p+p interactions at incident projectile momenta of 20, 31, 40, 80 and 158GeV/c (√s = 6.3, 7.7, 8.8, 12.3 and 17.3GeV, respectively). The measurements were performed using the large acceptance NA61/SHINE hadron spectrometer at the CERN super proton synchrotron.
Two-dimensional spectra are determined in terms of rapid-ity and transverse momentum. Their properties such as the width of rapidity distributions and the inverse slope parame-ter of transverse mass spectra are extracted and their collision energy dependences are presented. The results on inelastic p+p interactions are compared with the corresponding data on central Pb+Pb collisions measured by the NA49 experi-ment at the CERN SPS. The results presented in this paper are part of the NA61/SHINE ion program devoted to the study of the properties of the onset of deconfinement and search for the critical point of strongly interacting matter. They are required for interpretation of results on nucleus–nucleus and proton–nucleus collisions.
1 Introduction This paper presents experimental results on inclusive spectra andmeanmultiplicitiesofnegativelychargedpionsproduced in inelastic p+p interactions at 20, 31, 40, 80 and 158GeV/c.
The measurements were performed by the multi-purpose NA61/SHINE—the SPS Heavy Ion and Neutrino Experi-ment —at the CERN Super Proton Synchrotron (SPS).
They are part of the NA61/SHINE ion program devoted to the study of the properties of the onset of deconfinement and search for the critical point of strongly interacting matter.
Within this program data on p+p, Be+Be and p+Pb colli-sions were recorded and data on Ar+Ca and Xe+La colli-sions will be registered within the coming years. The started two dimensional scan in collision energy and size of col-liding nuclei is mainly motivated by the observation of the onset of deconfinement [2,3] in central Pb+Pb collisions at about 30AGeV/c by the NA49 experiment at the CERN SPS. Recently the NA49 results were confirmed by the RHIC beam energy scan programme and their interpretation by the a e-mail: Antoni.Aduszkiewicz@fuw.edu.pl b Deceased onset of deconfinement is supported by the LHC results (see Ref. and references therein).
In addition to the ion programme, NA61/SHINE is con-ductingprecisehadronproductionmeasurementsforimprov-ing calculations of the initial neutrino flux in long-baseline neutrino oscillation experiments [5–8] as well as for more reliable simulations of cosmic-ray air showers [9,10].
An interpretation of the rich experimental results on nucleus–nucleus collisions relies to a large extent on a com-parison to the corresponding data on p+p and p+A inter-actions. However, the available data concern mainly basic features of unidentified charged hadron production and they are sparse. Many needed results on hadron spectra, fluctua-tions and correlations are missing. Detailed measurements of hadron spectra in a large acceptance in the beam momentum range covered by the data presented in this paper exist only for inelastic p+p interactions at 158GeV/c [11–13]. Thus the new high precision measurements of hadron production properties in p+p and p+A interactions are necessary and they are performed in parallel with the corresponding mea-surements in nucleus–nucleus collisions. Among the many different hadrons produced in high energy collisions pions are the lightest and by far the most abundant ones. Thus, data on pion production properties are crucial for constraining basic properties of models of strong interactions. In particu-lar, the most significant signals of the onset of deconfinement (the “kink” and “horn”) require precise measurements of the mean pion multiplicity at the same beam momenta per nucleon as the corresponding A+A data. Moreover, the NA61/SHINE data are taken with the same detector and the same acceptance.
In the CERN SPS beam momentum range of 10– 450GeV/c the mean multiplicity of negatively charged pions in inelastic p+p interactions increases from about 0.7 at 10GeV/c to about 3.5 at 450GeV/c . Among three charged states of pions the most reliable measurements in the largest phase–space are usually possible for π−mesons.
Neutral pions decay electromagnetically into two photons and thus measurements of their production properties require measurements of both photons and then extraction of the π0 signal from the two-photon mass spectra. Charged pions are easy to detect by ionisation detectors as they decay weakly with a long lifetime (cτ = 7.8 m). A significant fraction of positively charged hadrons are protons (25 %) and kaons (5 %) [11–13]. Therefore measurements of π+ mesons require their identification by measurements of the 123 Eur. Phys. J. C (2014) 74:2794 Page 3 of 22 2794 energy loss and/or time-of-flight (tof). This identification is not as important for π−mesons because the contamination of negatively charged particles by K−mesons and anti-protons is below 10 % [11–13] and can be subtracted reliably. The latter method is used in this paper and it allows to derive π−spectra in the broadest phase–space region in a uniform way. Results obtained using explicit pion identification are planned in future NA61/SHINE publications.
The paper is organised as follows. In Sect. 2 the NA61/ SHINE experimental set-up is described. Details on the beam, trigger and event selection are given in Sect. 3.
Datareconstruction,simulationanddetectorperformanceare described in Sect. 4. Analysis techniques and final results are presented in Sects. 5 and 6, respectively. These results are compared with the corresponding data on central Pb+Pb collisions in Sect. 7. A summary in Sect. 8 closes the paper.
The pion rapidity is calculated in the collision centre of mass system: y = atanh(βL), where βL = pL/E is the lon-gitudinal component of the velocity, pL and E are pion lon-gitudinal momentum and energy given in the collision centre of mass system. The transverse component of the momen-tum is denoted as pT and the transverse mass mT is defined as mT = m2 π + pT2, where mπ is the charged pion mass.
The nucleon mass and collision energy per nucleon pair in the centre of mass system are denoted as mN and √sNN, respectively.
2 The NA61/SHINE facility The NA61/SHINE experimental facility consists of a large acceptance hadron spectrometer located in the CERN North Area Hall 887 (EHN1) and the H2 beam-line to which beams accelerated in the CERN accelerator complex are delivered from the Super Proton Synchrotron. NA61/SHINE profits from the long development of the CERN proton and ion sources and the accelerator chain as well as the H2 beam line of the CERN North Area. The latter has recently been modified to also serve as a fragment separator as needed to produce the Be beams for NA61/SHINE. Numerous com-ponents of the NA61/SHINE set-up were inherited from its predecessor, the NA49 experiment .
The schematic layout of the NA61/SHINE detector is shown in Fig. 1.
A set of scintillation and Cherenkov counters as well as Beam Position Detectors (BPDs) upstream of the spectrome-ter provide timing reference, identification and position mea-surements of incoming beam particles. The trigger scintilla-tor counter S4 placed downstream of the target is used to select events with collisions in the target area. Details on this system are presented in Sect. 3.
The main tracking devices of the spectrometer are large volume Time Projection Chambers (TPCs). Two of them, the vertex TPCs (VTPC-1 and VTPC-2 in Fig. 1), are located in the magnetic fields of two super-conducting dipole magnets with a maximum combined bending power of 9 Tm which corresponds to about 1.5 and 1.1 T fields in the upstream and downstream magnets, respectively. This field configuration was used for data taking on p+p interactions at 158GeV/c.
In order to optimise the acceptance of the detector at lower collision momenta, the field in both magnets was lowered in proportion to the beam momentum.
Fig. 1 The schematic layout of the NA61/SHINE experiment at the CERN SPS (horizontal cut, not to scale). The beam and trigger detector configuration used for data taking on p+p interactions in 2009 is shown.
Alignment of the chosen coordinate system is shown in the figure; its origin lies in the middle of VTPC-2, on the beam axis. The nominal beam direction is along the z axis. The magnetic field bends charged particle trajectories in the x–z (horizontal) plane. The drift direction in the TPCs is along the y (vertical) axis 123 2794 Page 4 of 22 Eur. Phys. J. C (2014) 74:2794 Two large TPCs (MTPC-L and MTPC-R) are positioned downstream of the magnets symmetrically to the beam line.
The fifth small TPC (GAP-TPC) is placed between VTPC-1 and VTPC-2 directly on the beam line. It closes the gap between the beam axis and the sensitive volumes of the other TPCs.
The TPCs are filled with Ar:CO2 gas mixtures in propor-tions 90:10 for the VTPCs and the GAP-TPC, and 95:5 for the MTPCs.
The particle identification capability of the TPCs based on measurements of the specific energy loss, dE/dx, is aug-mented by tof measurements using Time-of-Flight (ToF) detectors. The high resolution forward calorimeter, the Pro-jectile Spectator Detector (PSD), measures energy flow around the beam direction, which in nucleus–nucleus col-lisions is primarily given by the projectile spectators.
NA61/SHINEusesvarioussolidnucleartargetsandaLHT (see Sect. 3 for details). The targets are positioned about 80 cm upstream of the sensitive volume of VTPC-1.
The results presented in this paper were obtained using information from the TPCs, the BPDs as well as from the beam and trigger counters.
3 Beams, target, triggers and data samples This section summarises basic information on the beams, target, triggers and recorded data samples which is relevant for the results presented in this paper.
Secondary beams of positively charged hadrons at 20, 31, 40, 80 and 158GeV/c are produced from 400GeV protons extracted from the SPS in a slow extraction mode with a flat-top of 10 seconds. The secondary beam momentum and intensity is adjusted by proper setting of the H2 beam-line magnet currents and collimators. The beam is transported along the H2 beam-line towards the experiment. The preci-sion of the setting of the beam magnet currents is approxi-mately 0.5 %. This was verified by a direct measurement of the beam momentum at 31GeV/c by bending the incoming beam particles into the TPCs with the maximum magnetic field . The selected beam properties are given in Table 1.
The set-up of beam detectors is illustrated in the inset on Fig. 1. Protons from the secondary hadron beam are iden-tified by two Cherenkov counters, a CEDAR (either CEDAR-W or CEDAR-N) and a threshold counter (THC).
The CEDAR counter, using a coincidence of six out of the eight photo-multipliers placed radially along the Cherenkov ring, provides positive identification of protons, while the THC, operated at pressure lower than the proton threshold, is used in anti-coincidence in the trigger logic. Due to their lim-ited range of operation two different CEDAR counters were used, namely for beams at 20, 31, and 40GeV/c the CEDAR-W counter and for beams at 80 and 158GeV/c the CEDAR-N Table 1 Basic beam properties and numbers of events recorded for p+p interactions at 20, 31, 40, 80 and 158GeV/c. The first column gives the beam momentum. The second and third columns list typical numbers of beam particles at NA61/SHINE per spill (about 10 s) and the fraction of protons in the beam, respectively pbeam (GeV/c) Particles per spill (k) Proton fraction (%) 20 1000 12 31 1000 14 40 1200 14 80 460 28 158 250 58 counter. The threshold counter was used for all beam ener-gies. A selection based on signals from the Cherenkov coun-ters allowed to identify beam protons with a purity of about 99 %. A consistent value for the purity was found by bend-ing the beam into the TPCs with the full magnetic field and using the dE/dx identification method . The fraction of protons in the beams is given in Table 1.
Two scintillation counters, S1 and S2, provide beam def-inition, together with the three veto counters V0, V1 and V1p with a 1 cm diameter hole, which are defining the beam before the target. The S1 counter provides also the timing (start time for all counters). Beam protons are then selected by the coincidence S1∧S2∧V0∧V1∧V1p∧CEDAR∧THC.
Trajectories of individual beam particles were measured in a telescope of BPDs placed along the beam line (BPD-1/2/3 in Fig. 1). These counters are small (4.8×4.8 cm2) proportional chambers with cathode strip readout, providing a resolution of about 100 μm in two orthogonal directions, see Ref. for more details. The beam profile and divergence obtained from the BPD measurements are presented in Fig. 2. Due to properties of the H2 beam line both the beam width and divergence at the NA61/SHINE target increase with decreas-ing beam momentum.
For data taking on p+p interactions a LHT of 20.29 cm length (2.8 % interaction length) and 3 cm diameter placed 88.4 cm upstream of VTPC-1 was used. The Liquid Hydro-gen Target facility (LHT) filled the target cell with para-hydrogen obtained in a closed-loop liquefaction system which was operated at 75 mbar overpressure with respect to the atmosphere. At the atmospheric pressure of 965 mbar the liquid hydrogen density is ρLH = 0.07 g/cm3. The boiling rate in the liquid hydrogen was not monitored dur-ing the data taking and thus the liquid hydrogen density is known only approximately. It has however no impact on the results presented in this paper as they are deter-mined from particle yields per selected event and thus they are independent of the target density. Data taking with inserted and removed liquid hydrogen in the LHT was alternated in order to calculate a data-based correc-123 Eur. Phys. J. C (2014) 74:2794 Page 5 of 22 2794 Fig. 2 Top The beam spot as measured by BPD-3 after the V1 cut described in the text for 20GeV/c (left) and 158GeV/c (right) beams. Bottom The beam divergence in x and y for 20GeV/c (left) and 158GeV/c (right) beams. All distributions were arbitrarily scaled to the full colour scale. Widths of the distributions are given in the legend x [cm] -0.5 0 0.5 y [cm] -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 20 GeV/c = 0.55 cm y FWHM = 0.55 cm x FWHM x [cm] -0.5 0 0.5 entries [arb. units] 0 1 2 3 4 5 6 7 8 9 10 158 GeV/c = 0.25 cm y FWHM = 0.25 cm x FWHM div x [mrad] -1 0 1 div y [mrad] -1.5 -1 -0.5 0 0.5 1 1.5 20 GeV/c = 0.90 mrad y FWHM = 0.80 mrad x FWHM div x [mrad] -1 0 1 entries [arb. units] 0 1 2 3 4 5 6 7 8 9 10 = 0.10 mrad y FWHM = 0.20 mrad x FWHM 158 GeV/c tion for interactions with the material surrounding the liquid hydrogen.
Interactions in the target are selected by the trigger system by an anti-coincidence of the incoming beam protons with a small, 2 cm diameter, scintillation counter (S4) placed on the beam trajectory between the two vertex magnets (see Fig. 1).
This minimum bias trigger is based on the disappearance of the incident proton. In addition, unbiased proton beam events were recorded with a frequency typically by a factor of 10 lower than the frequency of interaction events.
4 Data processing, simulation and detector performance Detector parameters were optimised by a data-based calibra-tion procedure which also took into account their time depen-dences. Small adjustments were determined in consecutive steps for: (i) detector geometry, TPC drift velocities and distortions due to the magnetic field inhomogeneities in the cor-ners of VTPCs, (ii) magnetic field setting, (iii) specific energy loss measurements, (iv) tof measurements.
Each step involved reconstruction of the data required to opti-mise a given set of calibration constants and time depen-dent corrections followed by verification procedures. Details of the procedure and quality assessment are presented in Ref. . The resulting performance in the measurements of quantities relevant for this paper is discussed below.
The main steps of the data reconstruction procedure are: (i) cluster finding in the TPC raw data, calculation of the cluster centre-of-gravity and total charge, (ii) reconstruction of local track segments in each TPC sep-arately, (iii) matching of track segments into global tracks, (iv) track fitting through the magnetic field and determi-nation of track parameters at the first measured TPC cluster, (v) determination of the interaction vertex using the beam trajectory (x and y coordinates) fitted in the BPDs and the trajectories of tracks reconstructed in the TPCs (z coordinate), (vi) refitting the particle trajectory using the interaction ver-tex as an additional point and determining the particle momentum at the interaction vertex, (vii) matching of ToF hits with the TPC tracks.
Anexampleof areconstructedp+pinteractionat 40GeV/c is shown in Fig. 3. Long tracks of one negatively charged and two positively charged particles are seen. All particles leave signals in the ToF detectors.
A simulation of the NA61/SHINE detector response is used to correct the reconstructed data. Several MC mod-els were compared with the NA61/SHINE results on p+p, p+C and π+C interactions: FLUKA2008, URQMD1.3.1, 123 2794 Page 6 of 22 Eur. Phys. J. C (2014) 74:2794 Fig. 3 An example of a p+p interaction at 40GeV/c measured in the NA61/SHINE detector. The measured points (green) are used to fit tracks (red lines) to the interaction point. The grey dots show the noise clusters. Due to the central gap of the VTPCs only a small part of the trajectory of the negatively charged particle is seen in VTPC-1 VENUS4.12, EPOS1.99, GHEISHA2002, QGSJetII-3 and Sibyll2.1 [17,22–24]. Based on these comparisons and tak-ing into account continuous support and documentation from the developers the EPOS model was selected for the MC simulation. The simulation consists of the following steps (see Ref. for more details): (i) generation of inelastic p+p interactions using the EPOS model , (ii) propagation of outgoing particles through the detector material using the GEANT 3.21 package which takes into account the magnetic field as well as rele-vant physics processes, such as particle interactions and decays, (iii) simulation of the detector response using dedicated NA61/SHINE packages which introduce distortions corresponding to all corrections applied to the real data, (iv) simulation of the interaction trigger selection by check-ing whether a charged particle hits the S4 counter, see Sect. 3, (v) storage of the simulated events in a file which has the same format as the raw data, (vi) reconstruction of the simulated events with the same reconstruction chain as used for the real data and (vii) matching of the reconstructed tracks to the simulated ones based on the cluster positions.
It should be underlined that only inelastic p+p interac-tions in the hydrogen in the target cell were simulated and reconstructed. Thus the Monte Carlo based corrections (see Sect. 5) can be applied only for inelastic events. The con-tribution of elastic events is removed by the event selection cuts (see Sect. 5.1), whereas the contribution of off-target interactions is subtracted based on the data (see Sect. 5.4).
Spectra of π−mesons presented in this paper were derived from spectra of all negatively charged hadrons corrected for a small contamination of mostly K−mesons and anti-protons.
The typical acceptance in rapidity and transverse momen-tum is presented in Fig. 4 for p+p interactions at 20 and 158GeV/c. This figure also shows acceptance regions for methods based on explicit pion identification using dE/dx and tof measurements. They are limited due to the geomet-rical acceptance of the ToF detectors, the finite resolution of the dE/dx measurements and limited data statistics.
The quality of measurements was studied by reconstruct-ing masses of K0 S particles from their V0 decay topology. As an example the invariant mass distributions of K0 S candidates found in p+p interactions at 20 and 158GeV/c are plotted in Fig. 5. The differences between the measured peak positions and the literature value of the K0 S mass are smaller than 1 MeV/c2. The width of the distributions, related to the detector resolution, is about 25 % smaller for the Monte Carlo than for the data. This implies that statistical and/or systematic uncer-tainties of track parameters reconstructed from the data are 123 Eur. Phys. J. C (2014) 74:2794 Page 7 of 22 2794 Fig. 4 Typical acceptance regions for π−meson spectra in p+p interactions at 20GeV/c (left) and 158GeV/c (right) for different analysis methods: the method used in this paper which does not require an explicit pion identification, the method which identifies pions via their energy loss (dE/dx) and, in addition, their time-of-flight (tof) y 0 1 2 3 4 [GeV/c] T p 0 0.2 0.4 0.6 0.8 1 1.2 1.4 this paper dE/dx ToF 20 GeV/c y 0 1 2 3 4 [GeV/c] T p 0 0.2 0.4 0.6 0.8 1 1.2 1.4 158 GeV/c counts 5 10 15 20 25 30 35 40 45 data MC 20 GeV/c ] 2 ) [GeV/c -π + π ( inv m 0.45 0.5 0.55 counts 0 500 1000 1500 2000 2500 158 GeV/c 0 Fig. 5 Invariant mass distribution of reconstructed K0 S candidates in p+p interactions at 20 (top) and 158GeV/c (bottom) for the measured data and EPOS model based Monte Carlo simulations. The MC plot was normalised to the peak height of the data. The K0 S candidates were selected within 0 < y < −1 and 0 < pT < 0.5 GeV/c for 20GeV/c and −1 < y < 0 and 0 < pT < 0.5 GeV/c for 158GeV/c. The distribution was fitted with the sum of a Lorentzian function (signal) and a second order polynomial (background) somewhat underestimated in the simulation. Systematic bias due to this imperfectness was estimated by varying the selec-tion cuts and was found to be below 2 % (see Sect. 5.7.2).
The track reconstruction efficiency and the resolution of kinematic quantities were calculated by matching recon-structed tracks to their generated partners. In only 0.1–0.2 % of cases a single generated track is matched to more than one reconstructed partner, typically due to failure of match-ing reconstructed track segments. This effect is taken into account in the correction described in Sect. 5.6. As exam-ples, the reconstruction efficiency as a function of rapidity and transverse momentum for negatively charged pions pro-duced in p+p interactions at 20 and 158GeV/c is shown in Fig. 6. The resolution of rapidity and transverse momentum measurements is illustrated in Fig. 7. The resolution was cal-culated as the FWHM of the distribution of the difference between the generated and reconstructed y and pT. These results were obtained for negatively charged pions passing the track selection criteria described in Sect. 5.2. Resolution of the transverse momentum is worse at low beam momenta.
This is caused by the lower magnetic field and by the fact that the same rapidity region in the centre of mass frame corresponds to lower momenta in the laboratory frame.
Figures 8 and 9 show further examples of the comparison between data and simulation. Distributions of the z coordi-nate of the fitted vertex are presented in Fig. 8. Distributions of the distance between the track trajectory extrapolated to the z coordinate of the vertex and the vertex in the x–y plane (bx and by impact parameters) are given in Fig. 9. Differ-ences visible in the tails of distributions are partially due to imperfect simulation of the detector response and, in case of the impact parameter, partially due to the contribution of background tracks from off-time beam particles which are not included in the simulation. The difference is smaller for events selected using more restrictive cuts on the off-time beam particles. A possible small bias due to these effects was estimated by varying the impact parameter cuts and was found to be below 1 %.
5 Analysis technique This section presents the procedures used for data analysis consisting of the following steps: (i) applying event and track selection criteria, 123 2794 Page 8 of 22 Eur. Phys. J. C (2014) 74:2794 y 0 1 2 3 [GeV/c] T p 0 0.5 1 1.5 98 98 97 97 98 98 96 98 98 98 98 99 99 99 99 99 99 98 99 99 98 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 98 99 99 99 99 99 99 99 99 99 100 20 GeV/c efficiency [%] 95 96 97 98 99 100 y 1 2 3 4 100 100 97 98 99 98 98 98 97 99 100 99 98 99 99 99 99 98 99 100 100 99 99 99 99 99 99 96 99 100 99 99 99 99 99 99 99 99 100 99 99 99 99 99 99 98 100 99 99 99 99 99 99 99 100 99 99 99 99 99 99 99 96 100 99 99 99 99 99 99 97 97 99 99 99 99 99 100 98 158 GeV/c 0 Fig. 6 Reconstruction efficiency of negatively charged pions produced in p+p interactions at 20 (left) and 158GeV/c (right) as a function of rapidity and transverse momentum. It was calculated by dividing the number of tracks passing the track selection cuts specified in Sect. 5.2 by the number of the generated tracks. The selection criteria include the requirement of at least 90 % reconstruction efficiency [GeV/c] T p 0.5 1 1.5 16 16 14 15 16 17 13 14 13 13 12 15 18 15 8 8 11 9 10 12 22 20 6 10 8 8 11 28 5 8 8 7 7 17 5 7 6 6 8 5 6 6 5 7 4 5 5 4 4 4 20 GeV/c y 0 1 2 3 [GeV/c] T p 0 0.5 1 3 4 3 4 4 6 5 5 5 5 5 6 8 7 4 10 10 7 7 8 18 13 14 14 11 9 10 24 23 19 13 11 15 26 30 26 20 16 10 40 30 20 16 19 58 39 29 16 54 24 20 GeV/c 3 10 × (y resolution) 0 5 10 15 20 25 30 35 40 45 50 1 2 8 8 9 10 10 14 16 18 16 12 12 13 13 14 24 28 28 24 12 13 12 12 20 36 33 35 12 13 11 12 28 36 33 36 10 10 9 12 34 33 44 10 8 9 16 33 30 31 9 8 8 21 33 35 8 6 6 24 30 41 6 5 6 26 30 158 GeV/c resolution [MeV/c] T p 0 5 10 15 20 25 30 35 40 45 y 1 2 3 4 2 3 3 3 3 5 7 6 6 3 3 4 3 4 7 9 8 6 4 4 4 4 6 12 11 14 5 5 4 5 11 15 13 12 7 6 5 7 19 18 23 8 6 6 11 22 20 21 10 8 7 18 26 32 12 8 8 26 34 47 16 10 10 38 57 158 GeV/c 0 0 Fig. 7 Resolution of rapidity (top, scaled by 103) and transverse momentum (bottom) measurements for negatively charged pions pro-duced in p+p interactions at 20 (left) and 158GeV/c (right) as a function of pion rapidity and transverse momentum. The results are obtained using the track selection cuts specified in Sect. 5.2 (ii) determination of spectra of negatively charged hadrons using the selected events and tracks, (iii) evaluation of corrections to the spectra based on exper-imental data and simulations, (iv) calculation of the corrected spectra.
Corrections for the following biases were evaluated and applied: (i) geometrical acceptance, (ii) contribution of off-target interactions, (iii) contribution of particles other than negatively charged pions produced in inelastic p+p interactions, (iv) losses of inelastic p+p interactions as well as of nega-tively charged pions produced in accepted interactions due to the trigger and the event and track selection cri-teria employed in the analysis.
These steps are described in the successive subsections.
The final results refer to π−mesons produced in inelas-tic p+p interactions by strong interaction processes and in electromagnetic decays of produced hadrons. Such pions are referred to as primary π−. The term primary will be used in the above meaning also for other particles.
The analysis was performed independently in (y, pT) and (y, mT) bins. The bin sizes were selected taking into account 123 Eur. Phys. J. C (2014) 74:2794 Page 9 of 22 2794 events/dz [1/cm] 10 2 10 3 10 4 10 5 10 data MC 20 GeV/c fitted vertex z [cm] -700 -650 -600 -550 -500 -450 events/dz [1/cm] 10 2 10 3 10 4 10 158 GeV/c 1 Fig. 8 Distribution of fitted vertex z coordinate for p+p interactions at 20 (top) and 158GeV/c (bottom). The black line shows the data after target removed subtraction (see Sect. 5.4). The filled area shows the distribution for the reconstructed Monte Carlo simulation. This distri-bution was normalised to the total integral of the data plot. The dashed vertical lines show the z vertex selection range Table 2 Number of events recorded with the interaction trigger (all) and selected for the analysis (selected) pbeam (GeV/c) Target inserted Target removed All (k) Selected (k) All (k) Selected (k) 20 1324 233 123 4 31 3145 843 332 15 40 5239 1578 529 44 80 4038 1543 429 54 158 3502 1650 427 51 the statistical uncertainties as well as the resolution of the momentum reconstruction. Corrections as well as statistical and systematic uncertainties were calculated for each bin.
5.1 Event selection criteria This section presents the event selection criteria. The num-ber of events selected by the trigger (see Sect. 3) and used in the analysis is shown in Table 2. The fraction of events selected for the analysis increases with the interaction energy, mostly due to lower beam intensity (see Table 1) and resulting smaller off-time particle contamination, and smaller fraction of the low multiplicity events for which no tracks are found within the acceptance.
Fig. 9 Distribution of the impact parameter in the x (left) and y (right) coordinate for p+p interactions at 20 (top) and 158GeV/c (bottom). The black line shows the data after target removed subtraction (see Sect. 5.4). The filled area shows the reconstructed Monte Carlo simulation. The dashed vertical lines show the accepted range (see Sect. 5.2). The dotted blue line in the 20GeV/c plots show the distribution obtained using the alternative event selection.
Namely only events with no off-time beam particles within the time window of ±6 μs around the trigger particle time were accepted [cm] x b -4 -2 0 2 4 db) [1/cm] ⋅ tracks/(events -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 1 10 20 GeV/c [cm] y b -4 -2 0 2 4 data (alternative) data MC [cm] x b -4 -2 0 2 4 db) [1/cm] ⋅ tracks/(events -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 1 10 158 GeV/c [cm] y b -4 -2 0 2 4 data MC 123 2794 Page 10 of 22 Eur. Phys. J. C (2014) 74:2794 Table 3 Momentum thresholds used to reject elastic interactions (cut (v)) Beam momentum (GeV/c) 20 31 40 80 158 Threshold momentum (GeV/c) 17 28 35 74 – The following event selection criteria were applied to the events recorded with the interaction trigger: (i) no off-time beam particle is detected within ±2 μs around the trigger particle, (ii) the beam particle trajectory is measured in at least one of BPD-1 or BPD-2 and in the BPD-3 detector posi-tioned just in front of the LHT, (iii) there is at least one track reconstructed in the TPCs and fitted to the interaction vertex, (iv) the vertex z position (fitted using the beam and TPC tracks) is not farther away than 40 cm from the centre of the LHT, (v) events with a single, well measured positively charged track with absolute momentum close to the beam momentum are rejected. The momentum thresholds are listed in Table 3.
The off-line (listed above) and on-line (the interaction trig-ger condition, see Sect. 3) event cuts select a large fraction of well measured (cuts (i) and (ii)) inelastic (cut (iii)) p+p interactions. The cut (iii) removes part of elastic interac-tions. However in some elastic events at beam momenta up to 80GeV/c the beam particle is deflected enough to be mea-sured in the detector. This is demonstrated in the momentum distributions shown in Fig. 10. Such events are removed by cut (v).
Moreover cut (iv) significantly suppresses interactions outside the hydrogen in the target cell. The corrections for the contribution of interactions outside the hydrogen in the target cell and the loss of inelastic events are presented in Sects. 5.4 and 5.6.
5.2 Track selection criteria In order to select well-measured tracks of primary nega-tively charged hadrons as well as to reduce the contamination of tracks from secondary interactions, weak decays and off-time interactions the following track selection criteria were applied: (i) the track momentum fit at the interaction vertex should have converged, (ii) the fitted track charge is negative, (iii) the fitted track momentum component px is negative.
This selection minimises the angle between the track trajectory and the TPC pad direction for the chosen magnetic field direction. This reduces statistical and systematic uncertainties of the cluster position, energy deposit and track parameters, (iv) the total number of reconstructed points on the track should be greater than 30, (v) the sum of the number of reconstructed points in VTPC-1 and VTPC-2 should be greater than 15 or the num-ber of reconstructed points in the GAP-TPC should be greater than 4, (vi) the distance between the track extrapolated to the inter-action plane and the interaction point (impact parame-ter) should be smaller than 4 cm in the horizontal (bend-ing) plane and 2 cm in the vertical (drift) plane, (vii) the track should be measured in a high (≥90 %) TPC acceptance region (see Sect. 5.3), (viii) tracks with dE/dx and total momentum values char-acteristic for electrons are rejected. The electron con-tribution to particles with momenta above 20GeV/c is corrected using the simulation. The electron selection procedure is visualised in Fig. 11.
p [GeV/c] 16 18 20 22 24 tracks/events 0 0.002 0.004 0.006 0.008 20 GeV/c p [GeV/c] 35 40 45 40 GeV/c p [GeV/c] 100 120 140 160 158 GeV/c Fig. 10 Momentum distributions at 20 (left), 40 (middle) and 158GeV/c (right) of the positively charged tracks in events passing selection cuts (i)–(iv), containing a single track, which is positively charged and measured in the GAP TPC and MTPC. The distributions were normalised to all events. The vertical dashed lines at 20 and 40GeV/c show the momentum threshold used to remove elastic events (cut (v)) 123 Eur. Phys. J. C (2014) 74:2794 Page 11 of 22 2794 [p/(1 GeV/c)] 10 log -1 -0.5 0 0.5 1 1.5 dE/dx [MIP] 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 1 10 2 10 3 10 -e -h dE/dx [MIP] 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 entries 1 10 2 10 3 10 -h -e total 0.79<p<1.00 [GeV/c] Fig. 11 Left Distribution of particle energy loss as a function of the logarithm of total momentum, for negatively charged particles produced in p+p interactions at 40GeV/c. The black contour shows the electron selection region. Right dE/dx distribution in the momentum range indi-catedinthefigureandmarkedwithverticaldashedlinesintheleftpanel.
Electrons and negatively charged hadrons show separated peaks The spectra of negatively charged particles after track and event selections were obtained in 2-dimensional bins of (y, pT) and (y, mT). The spectra were evaluated in the centre-of-mass frame after rotation of the z axis into the proton beam direction measured event-by-event by the BPDs.
5.3 Correction for detector acceptance The detection and reconstruction inefficiencies are corrected using the simulation described in Sect. 4. However, in order to limit the impact of possible inaccuracies of this simula-tion, only regions are accepted where the reconstruction effi-ciency (defined as the ratio of the number of reconstructed and matched MC π−tracks passing the track selection crite-ria to the number of generated π−) equals at least 90 %. These regions were identified using a separate, statistically indepen-dent simulation in three-dimensional bins of y, pT or mT and the azimuthal angle φ (5◦bin width). The resulting accep-tance maps are shown in Fig. 12. The acceptance calculated in the y < 0 region, not used for the final results, is shown also for comparison. We chose an upper limit of 1.5GeV/c for the transverse momentum spectra, because beyond the admixture of background tracks reaches a level which cannot be handled by the correction procedures used in this paper.
Future publications will be devoted to the high pT region.
Sinceneithertargetnorbeamarepolarized,wecanassume a uniform distribution of particles in φ. The data falling into the accepted bins is summed over φ bins and the (y, pT/mT) bin content is multiplied by a correction factor to compen-sate for the rejected φ ranges. The acceptance correction also compensates for the px < 0 selection (see Sect. 5.2, point (5.2)).
Even a small deviation of the beam direction from the nominal axis (z) results in a non-negligible bias in the recon-structed transverse momentum. The beam direction is mea-sured in the BPDs, and the particle momenta are recalculated to the frame connected with the beam direction. However, the detector acceptance depends on the momentum in the detec-tor frame. Therefore the acceptance selection is done in the detector frame, and the acceptance correction is applied as a weight to each track. The weights are used to obtain parti-cle spectra corrected for the detector acceptance in the frame connected with the beam direction.
5.4 Correction for off-target interactions The spectra were derived for events with liquid hydrogen in (I) and removed (R) from the LHT. The latter data set rep-resents interactions outside the liquid hydrogen (interactions with materials downstream and upstream of the liquid hydro-gen). The differential inclusive yield of negatively charged particles per event in interactions of beam protons with the liquid hydrogen inside the LHT (nT[h−]) is calculated as: nT[h−] = 1 1 −ε · nI[h−] −ε · nR[h−] , (1) where: (i) nI[h−] and nR[h−] is the number of tracks in a given bin per event selected for the analysis (see Sect. 5.2) for the data with the liquid hydrogen inserted and removed, respectively, (ii) ε is the ratio of the interaction probabilities for the removed and inserted target operation.
ε was derived based on the distribution of the fitted z coor-dinate of the interaction vertex. All vertices far away from the target originate from interactions with the beam-line and 123 2794 Page 12 of 22 Eur. Phys. J. C (2014) 74:2794 y -1 0 1 2 3 4 [GeV/c] T p 0 0.2 0.4 0.6 0.8 1 1.2 1.4 bins accepted [%] φ fraction of 0 10 20 30 40 50 60 70 80 90 100 20 GeV/c y -1 0 1 2 3 4 [GeV/c] T p 0 0.2 0.4 0.6 0.8 1 1.2 1.4 bins accepted [%] φ fraction of 0 10 20 30 40 50 60 70 80 90 100 158 GeV/c Fig. 12 Detector acceptance at 20 (left) and 158GeV/c (right). The acceptance is calculated as fraction of (y, pT, φ) bins accepted for given y and pT for tracks with px < 0 selected for this analysis Fig. 13 Distribution of fitted vertex z coordinate for data on p+p interactions at 40GeV/c.
The distribution for the data recorded with the removed liquid hydrogen was multiplied by a factor of NI[z > −450 cm]/NR[z > −450 cm] fitted vertex z [cm] -700 -650 -600 -550 -500 -450 -400 -350 -300 events 2 10 3 10 4 10 5 10 normalization region target inserted 3.69) × target removed ( Table 4 The ratio of the interaction probabilities, ε, for the removed and inserted target operation for data taking on p+p interactions at 20, 31, 40, 80 and 158GeV/c pbeam (GeV/c) ε (%) 20 8.0 ± 0.3 31 7.1 ± 0.1 40 10.4 ± 0.1 80 12.7 ± 0.1 158 12.6 ± 0.1 detector materials. Neglecting the beam attenuation in the target one gets: ϵ = NR NI · NI[z > −450 cm] NR[z > −450 cm] , (2) where N[z > −450 cm] is the number of events with fitted vertex z > −450 cm. Examples of distributions of z of the fitted vertex for events recorded with the liquid hydrogen inserted and removed are shown in Fig. 13. Values of ϵ are listed in Table 4.
The correction for the off-target interactions changes the yields obtained from the target inserted data by less than ±5 %, except in the regions where the statistical uncertainty is high.
5.5 The correction for contamination of primary π−mesons More than 90 % of primary negatively charged particles pro-duced in p+p interactions in the SPS energy range are π− mesons [11–13]. Thus π−meson spectra can be obtained by subtracting the estimated non-pion contribution from the spectra of negatively charged particles and additional particle identification is not required.
The simulation described in Sect. 4 was used to calculate corrections for the admixture of particles other than primary π−mesons to the reconstructed negatively charged particles.
The dominating contributions are primary K−and ¯ p, and sec-ondary π−from weak decays of and K0 S (feed-down) and from secondary interactions, incorrectly fitted to the primary vertex.
123 Eur. Phys. J. C (2014) 74:2794 Page 13 of 22 2794 The EPOS spectra were adjusted based on the existing data [24,29]. Preliminary NA61/SHINE results were used to scale double differential spectra of K−, and ¯ p . EPOS spectra of π−were replaced by the preliminary NA61/ SHINE results normalised to the multiplicity from the world data compilation . Spectra of and K0 S were scaled by a constant factor derived at each energy using the world data compilation of total multiplicities. The impact of the adjustments on the final spectra ranges from −2 % to +5 % in most regions, except of the low pT region at the low beam momenta, where it reaches +20 %.
As it was found in the yields of K−and ¯ p are strongly correlated with the π−yield. Thus the correction for the con-tribution of primary hadrons is performed via the multiplica-tive factor cK. On the contrary the contribution due to weak decays and secondary interactions is mostly located in the low pT region, and it is weakly correlated with the primary pion yield in this region. Thus this feed-down contribution is corrected for using the additive correction cV. The total correction is applied in as: nprim[π−] = nT[h−] −cV · cK, (3) where cV = n[π− ] + n[π− K0 S] + n[other] MC sel , (4) cK = n[π−] n[K−] + n[¯ p] + n[π−] MC sel .
(5) The spectrum of a particle x is denoted as n[x] whereas n[other] stands for all primary and secondary particles other than K−, ¯ p, π−and feed-down from and K0 S. The spec-trum n[other] of all other particles originates mostly from secondary interactions with >90 % occurring in the hydro-gen target. This contribution was taken from the simulations without an additional adjustment. The superscript MC marks adjusted EPOS spectra. The subscript sel indicates that the event and track selection criteria were applied and then the correction for the detector acceptance was performed; the reconstructed tracks were identified by matching.
5.6 Correction for event as well as track losses and migration The multiplicative correction closs for losses of inelastic events as well as losses and bin-to-bin migration of primary π−mesons emitted within the acceptance is calculated using the Monte Carlo simulation as: closs = n[π−]MC gen / n[π−]MC sel , (6) where the subscript gen indicates the generated spectrum of primary π−mesons binned according to the generated momentum vector. Then the final, corrected π−meson spec-trum in inelastic p+p interactions is calculated as n[π−] = closs · nprim[π−] .
(7) The dominating effects contributing to the closs correction are – losses of inelastic events due to the trigger and off-line event selection, – the pion migration between analysis bins, – the pion reconstruction inefficiency.
5.7 Statistical and systematic uncertainties 5.7.1 Statistical uncertainties Statistical errors receive contributions from the finite statis-tics of both the data as well as the simulated events used to obtain the correction factors. The dominating contribution is the uncertainty of the data which is calculated assuming a Poisson probability distribution for the number of entries in a bin. The Monte Carlo statistics was higher than the data statistics. Also the uncertainties of the Monte Carlo correc-tions are significantly smaller than the uncertainties of the number of entries in bins.
5.7.2 Systematic uncertainties Systematic errors presented in this paper were calculated tak-ing into account contributions from the following effects.
(i) Possible biases due to event and track cuts which are not corrected for. These are: – a possible bias due to the dE/dx cut applied to remove electrons, – a possible bias related to the removal of events with off-time beam particles close in time to the trigger particle.
The magnitude σi of possible biases was estimated by varying values of the corresponding cuts. The dE/dx cut was changed by ±0.01 dE/dx units (where 1 cor-responds to a minimum ionising particle, and 0.04 is a typical width of a single particle distribution), and the off-time interactions cut was varied from a ±1 μs to a ±3 μs time window. The assigned systematic uncer-tainty was calculated as the maximum of the absolute differences between the results obtained for both cut values. This contribution is drawn with a long-dashed red line in Fig. 14.
123 2794 Page 14 of 22 Eur. Phys. J. C (2014) 74:2794 Fig. 14 Statistical and systematic uncertainties in selected bins of pT for 20 (left) and 158GeV/c (right) p+p data.
The shaded band shows the statistical uncertainty. The coloured thin lines show the contributions to the systematic uncertainty listed in Sect. 5.7.2.
The thick black lines show the total systematic uncertainty, which was calculated by adding the contributions in quadrature (continuous line) or linearly (dashed/dotted line, shown for comparison) y 0 1 2 3 4 relative uncertainty [%] 0 5 10 15 20 25 30 <0.05 [GeV/c] T 0.00<p 20 GeV/c y 0 1 2 3 4 relative uncertainty [%] 0 5 10 15 20 25 30 Total (linear sum) Total (square sum) i σ ii σ iii σ <0.05 [GeV/c] T 0.00<p 158 GeV/c y 0 1 2 3 4 relative uncertainty [%] 0 5 10 15 20 25 30 <0.25 [GeV/c] T 0.20<p 20 GeV/c y 0 1 2 3 4 relative uncertainty [%] 0 5 10 15 20 25 30 <0.25 [GeV/c] T 0.20<p 158 GeV/c y 0 1 2 3 4 relative uncertainty [%] 0 5 10 15 20 25 30 <0.70 [GeV/c] T 0.60<p 20 GeV/c y 0 1 2 3 4 relative uncertainty [%] 0 5 10 15 20 25 30 <0.70 [GeV/c] T 0.60<p 158 GeV/c (ii) Uncertainty of the correction for contamination of the primary π−mesons. The systematic uncertainty σii of this correction was assumed as 20 % (for 40, 80 and 158GeV/c) and 40 % (for 20 and 31GeV/c) of the absolute value of the correction. At the low beam momenta there was less data available to adjust the sim-ulated spectra, which was the reason to increase the uncertainty. This contribution is drawn with a dashed-dotted blue line in Fig. 14. The absolute correction is small thus the related systematic uncertainty is small also.
(iii) Uncertainty of the correction for the event losses. The uncertainty was estimated using 20 % of the correc-tion magnitude and a comparison with the correction calculated using the VENUS [34,35] model: σiii = 0.2 · 1 −cEPOS loss + cEPOS loss −cVENUS loss .
(8) 123 Eur. Phys. J. C (2014) 74:2794 Page 15 of 22 2794 y -2 0 2 dy dn 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 NA61/SHINE (alternative) NA61/SHINE NA49 Fig. 15 Rapidity distribution of π−mesons produced in inelastic p+p interactions at 158GeV/c. The big blue points show the results obtained with an alternative method: without vertex fit requirement and rejection of events with a single very high momentum positively charged track.
The results of NA61/SHINE (this paper, red dots) are compared with the NA49 measurements (black squares). The open symbols show points reflected with respect to mid-rapidity. A single NA61/SHINE point measured at y < 0 is also shown for comparison. The shaded band shows the NA61/SHINE systematic uncertainty This contribution is drawn with a short-dashed green line in Fig. 14.
(iv) Uncertainty related to the track selection method. It was estimated by varying the track selection cuts: removing the impact parameter cut and decreasing the minimum number of required points to 25 (total) and 10 (in VTPCs) and by checking symmetries with respect to y = 0 and pT = 0. The potential bias is below 2 % and the corresponding contribution was neglected.
The total systematic uncertainty was calculated by adding in quadrature the contributions σsys = σ 2 i + σ 2 ii + σ 2 iii. This uncertainty is listed in the tables including numerical values and it is visualised by a shaded band around the data points in plots presenting the results. Statistical and systematic uncer-tainties in selected example regions are shown in Fig. 14.
Systematic biases in different bins are correlated, whereas statistical fluctuations are almost independent.
Figure 15 presents a comparison of the rapidity spec-trum of π−mesons produced in inelastic p+p interactions at 158GeV/c (for details see Sect. 6) from the present analy-sis with the corresponding spectrum measured by NA49 .
Statistical andsystematicuncertainties of theNA49spectrum are not explicitly given but the published information implies that the systematic uncertainty dominates and amounts to several %. The results agree within the systematic uncertain-ties of the NA61/SHINE spectra.
y -2 0 2 dy dn 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 NA61/SHINE (31 GeV/c) MIRABELLE (32 GeV/c) Fig. 16 Rapidity distribution of π−mesons produced in inelastic p+p interactions. The NA61/SHINE results at 31GeV/c (blue points) are compared with the MIRABELLE measurement (parametrised by the black line) at 32GeV/c. The shaded band shows the NA61/SHINE sys-tematic uncertainty The analysis method of p+p interactions at 158GeV/c per-formed by NA49 differed from the one used in this paper.
In particular, pions were identified by dE/dx measurement and the NA49 event selection criteria did not include the selection according to the fitted z coordinate of the interac-tion vertex and the rejection of elastic interactions. Namely, all events passing the trigger selection and off-line quality cuts were used for the analysis. For comparison, this event selectionprocedurewas appliedtotheNA61/SHINEdata. As a result 20 % more events were accepted. Approximately half of them were unwanted elastic and off-target interactions and half were wanted inelastic interactions. Then the corrections corresponding to the changed selection criteria were applied (the contribution of elastic events was subtracted using the estimate from Ref. ). The fully corrected rapidity spec-trum obtained using this alternative analysis is also shown in Fig. 15. The differences between the results for the standard and alternative methods are below 0.5 % at y < 2 and below 2 % at higher y.
Figure 16 shows a comparison of the rapidity distribution at 31GeV/c with the MIRABELLE results at 32GeV/c [36, 37]. A parametrisation of the distribution and the total π− multiplicity are provided. As the parametrisation appears to be incorrectly normalised, we normalised it to the total multi-plicity. The results agree within the NA61/SHINE systematic uncertainties.
The spectra measured in p+p interactions should obey reflection symmetry with respect to mid-rapidity. As the NA61/SHINE acceptance extends somewhat below mid-rapidity a check of the reflection symmetry can be performed and used to validate the measurements. It was verified that the yields measured for y < 0 agree with those measured for y > 0 in the reflected acceptance within 1.5 %. A similar 123 2794 Page 16 of 22 Eur. Phys. J. C (2014) 74:2794 agreement was also found at lower beam momenta. The mea-surements above mid-rapidity are taken as the final results.
Nevertheless, for comparison the points at y < 0 were added in Figs. 15 and 20 in the regions where the pT acceptance extends to zero.
6 Results This section presents results on inclusive π−meson spectra in inelastic p+p interactions at beam momenta of 20, 31, 40, 80 and 158GeV/c. The spectra refer to pions produced by strong interaction processes and in electromagnetic decays of produced hadrons.
Numerical results corresponding to the plotted spectra as well as their statistical and systematic uncertainties are given in Ref. .
6.1 Double differential spectra The double differential inclusive spectra of π−mesons in rapidity and transverse momentum produced in inelastic p+p interactions at 20, 31, 40, 80 and 158GeV/c are shown in Fig. 17. The transverse momentum distributions at 20 and 158GeV/c are plotted in Fig. 18. Here d2n dy dpT or d2n dy dmT were calculated by dividing the fully corrected bin contents n[π−] (see Sect. 5) by the bin size. The spectra in (y, mT) are not shown here but they are given in the compilation of the numerical values .
6.2 Transverse mass spectra The transverse mass spectra at mid-rapidity (0 < y < 0.2) are shown in Fig. 19 (left). A function dn dmT = A · mT · exp −mT T (9) was fitted in the range 0.2 < mT −mπ < 0.7 GeV/c2 and is indicated by lines in Fig. 19 (left). The fitted parameters were the normalisation A and the inverse slope T . They minimise the χ2 function which was calculated using statistical errors only. In the χ2 calculation a measured bin content (dn/dmT) was compared with the integral of the fitted function in a bin divided by the bin width.
Similar fits were performed to spectra in other rapidity bins containing data in the fit range. The rapidity depen-dence of the fitted inverse slope parameter T is presented in Fig. 19 (right). The T parameter decreases significantly when going from mid-rapidity to the projectile rapidity (ybeam = 1.877, 2.094, 2.223, 2.569 and 2.909 at 20, 31, 40, 80 and 158GeV/c, respectively).
6.3 Rapidity spectra The rapidity spectra are shown in Fig. 20 (left). They were obtained by summing the measured mT spectra and using the exponential function Eq. (9). The function was fitted in the range ending at the maximum measured mT, and starting 0.9GeV/c2 below (note this is a different fit from the one shown in Fig. 19). The correction is typically below 0.2 % and becomes significant (several %) only at y > 2.4. Half of the correction is added in quadrature to the systematic uncertainty in order to take into account a potential imper-fectness of the exponential extrapolation. The pion yield increases with increasing collision energy at all measured rapidities.
The rapidity spectra are parametrised by the sum of two Gaussian functions symmetrically displaced with respect to mid-rapidity: dn dy = ⟨π−⟩(y0, σ0) 2σ0 √ 2π · · exp −(y −y0)2 2σ 2 0 + exp −(y + y0)2 2σ 2 0 , (10) where y0 and σ0 are fit parameters, and the total multiplic-ity ⟨π−⟩(y0, σ0) is calculated from the requirement that the integral over the measured spectrum equals the integral of the fitted function Eq. (10) in the range covered by the measure-ments. The χ2 function was minimised in a similar way as in case of the mT spectra, namely using the integral of the func-tion in a given bin. The numerical values of the fitted param-eters as well as the r.m.s. width σ = y2 0 + σ2 0 are given in Table 5.
6.4 Mean multiplicities Mean multiplicities of π−mesons, ⟨π−⟩, produced in inelas-tic p+p interactions at 20, 31, 40, 80 and 158GeV/c were calculated as the integral of the fitted function Eq. (10). The extrapolation into the unmeasured region at large y con-tributes about 1 %. Half of it is added in quadrature to the systematic uncertainty.
The dependence of the produced average ⟨π−⟩multiplic-ity per inelastic p+p collision on the Fermi’s energy mea-sure , F ≡ (√sNN −2mN)3 √sNN 1/4 (11) is plotted in Fig. 20 (right). The results of NA61/SHINE are in agreement with a compilation of the world data [11,32].
123 Eur. Phys. J. C (2014) 74:2794 Page 17 of 22 2794 Fig. 17 Double differential spectra d2n/(dy dpT)[(GeV/c)−1] of π−mesons produced in inelastic p+p interactions at 20, 31, 40, 80 and 158GeV/c y 0 1 2 3 4 [GeV/c] T p 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 20 GeV/c y 0 1 2 3 4 [GeV/c] T p 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 31 GeV/c y 0 1 2 3 4 [GeV/c] T p 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 40 GeV/c y 0 1 2 3 4 [GeV/c] T p 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 80 GeV/c y 0 1 2 3 4 [GeV/c] T p 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 158 GeV/c 7 Comparison with central Pb+Pb collisions InthissectiontheNA61/SHINEresultsoninelasticp+pinter-actions are compared with the corresponding data on central Pb+Pb collisions published previously by NA49 [2,3]. Pion production properties which are different and similar in p+p interactions and central Pb+Pb collisions are identified. For completeness selected plots include the compilation of the world data on inelastic p+p interactions [11,32], as well as results on central Au+Au collisions from AGS [40,41] and RHIC [42–46], as processed in Ref. .
Figure 21 shows the ratio of transverse mass spectra of π−mesons produced at mid-rapidity (0 < y < 0.2) in central Pb+Pb collisions and p+p interactions at the same collision energy per nucleon. The spectra were normalised to unity before dividing. First, one observes that the ratio is not constant implying that the spectral shapes are differ-ent in p+p interactions and central Pb+Pb collisions. Sec-ond, it is seen that the ratio depends weakly, if at all, on collision energy. The ratio is higher than unity in the left (mT−mπ < 0.1 GeV/c2) and right (mT−mπ > 0.5 GeV/c2) 123 2794 Page 18 of 22 Eur. Phys. J. C (2014) 74:2794 Fig. 18 Transverse momentum spectra of π−mesons produced in inelastic p+p interactions at 20 (left) and 158GeV/c (right) in various rapidity ranges. The legend provides the centres of the rapidity bins, ybin and the scaling factor c used to separate the spectra visually [GeV/c] T p 0 0.5 1 1.5 2 2.5 ] -1 [(GeV/c) dy T dp n 2 d -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 1 20 GeV/c [GeV/c] T p 0 0.5 1 1.5 2 2.5 ] -1 [(GeV/c) dy T dp n 2 d 158 GeV/c -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 1 ] 2 [GeV/c π - m T m 0 0.5 1 ] -2 ) 2 [(GeV/c T dy dm n 2 d T m 1 -3 10 -2 10 -1 10 1 10 2 10 16 × 8 × 4 × 2 × 158 GeV/c 80 GeV/c 40 GeaV/c 31 GeV/c 20 GeV/c beam y/y 0 0.2 0.4 0.6 0.8 1 ] 2 T [MeV/c 80 90 100 110 120 130 140 150 160 170 158 GeV/c 80 GeV/c 40 GeV/c 31 GeV/c 20 GeV/c Fig. 19 Left Transverse mass spectra at mid-rapidity (0 < y < 0.2).
The fitted exponential function Eq. (9) is indicated by solid lines in the fit range 0.2 < mT −mπ < 0.7 GeV/c2 and dashed lines outside the fit range. The data points for different beam momenta were scaled for better readability. Right The inverse slope parameter T of the transverse mass spectra as a function of rapidity divided by the projectile rapidity.
The fit range is 0.2 < mT −mπ < 0.7 GeV/c2. The results refer to π−mesons produced in inelastic p+p interactions at 20, 31, 40, 80 and 158GeV/c 123 Eur. Phys. J. C (2014) 74:2794 Page 19 of 22 2794 y -2 0 2 dy dn 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 158 GeV/c 80 GeV/c 40 GeV/c 31 GeV/c 20 GeV/c ] 1/2 F [GeV 0 1 2 3 4 5 〉 -π 〈 0 0.5 1 1.5 2 2.5 3 NA61/SHINE world data Fig. 20 Left Rapidity spectra obtained from sums of the measured and extrapolated mT spectra. Closed symbols indicate measured points, open points are reflected with respect to mid-rapidity. The measured points at y < 0 are shown for systematic comparison only. The plotted statistical errors are smaller than the symbol size. The systematic uncer-tainties are indicated by the coloured bands. The lines indicate fits of the sum of two symmetrically displaced Gaussian functions (see Eq. (10)) to the spectra. The results refer to π−mesons produced in inelastic p+p interactions at 20, 31, 40, 80 and 158GeV/c. Right Dependence of the mean total multiplicity of π−mesons produced in inelastic p+p interactions on Fermi’s energy measure F (see Eq. (11)). The results of NA61/SHINE are indicated by filled circles and the compilation of the world data [32,11] by open circles. The plotted statistical errors are smaller than the symbol size. The systematic uncertainties are indicated by the coloured band Table 5 Numerical values of the parameters fitted to rapidity (see Eq. (10)) and transverse mass (see Eq. (9)) spectra of π−mesons produced in inelastic p+p interactions at 20, 31, 40, 80 and 158GeV/c. In case of the rapidity fit parameters ⟨π−⟩, σ, σ0 and y0, the systematic uncertainty dominates. The uncertainties written in the table are the quadrature sum of the statistical and systematic uncertainties. All uncertainties are given numerically in . For T and ⟨mT⟩the statistical uncertainty is written first and the systematic one second pbeam(GeV/c) ⟨π−⟩ σ σ0 y0 T (y = 0) (MeV/c2) ⟨mT⟩(y = 0) −mπ (MeV/c2) 20 1.047 ± 0.051 0.981 ± 0.017 0.921 ± 0.118 0.337 ± 0.406 149.1 ± 5.0 ± 4.8 237.8 ± 6.4 ± 2.3 31 1.312 ± 0.069 1.031 ± 0.016 0.875 ± 0.050 0.545 ± 0.055 153.3 ± 2.2 ± 1.2 246.1 ± 2.7 ± 0.9 40 1.478 ± 0.051 1.069 ± 0.014 0.882 ± 0.045 0.604 ± 0.044 157.7 ± 1.7 ± 2.1 247.3 ± 2.0 ± 0.9 80 1.938 ± 0.080 1.189 ± 0.026 0.937 ± 0.019 0.733 ± 0.010 159.9 ± 1.5 ± 4.1 253.5 ± 1.9 ± 1.1 158 2.444 ± 0.130 1.325 ± 0.042 1.007 ± 0.051 0.860 ± 0.021 159.3 ± 1.3 ± 2.6 253.6 ± 1.6 ± 1.4 parts of the mT range. It is below unity in the central region 0.1 < mT −mπ < 0.5 GeV/c2.
The inverse slope parameter T of transverse mass spectra fitted in the range 0.2 < mT −mπ < 0.7 GeV/c2 is plotted versus the collision energy in Fig. 22 (left). The T parameter is larger by about 10–20MeV/c2 in central Pb+Pb collisions than in p+p interactions.
The transverse mass spectra measured by NA61/SHINE and NA49 allow a reliable calculation of mean transverse mass. A small correction to the measured value for the high mT region not covered by the measurements was applied based on the exponential extrapolation of the tail of the dis-tributions. Half of the correction was added to the systematic uncertainty on ⟨mT⟩. In spite of the different shapes of the mT spectra the mean transverse mass calculated for p+p inter-actions and central Pb+Pb collisions is similar, see Fig. 22 (right). This is because the differences shift the mean mT in opposite directions for different regions of mT and as a result leave it almost unchanged. Thus the mean transverse mass appears to be insensitive to the apparent changes of the pion production properties observed between p+p interactions and central Pb+Pb collisions.
Figure 23 (left) presents the ratio of the normalised π− rapidity spectra produced in central Pb+Pb and inelastic p+p interactions at the same collision energy per nucleon. The spectra are plotted versus the rapidity scaled by the beam rapidity. Only weak, if any, energy dependence of the ratio is observed. Moreover, the ratio is close to unity in the central rapidity region (y/ybeam < 0.6), whereas it is higher closer to beam rapidity (y/ybeam > 0.6).
Consequently the r.m.s. width σ of rapidity distributions of π−mesons produced in p+p interactions is smaller than 123 2794 Page 20 of 22 Eur. Phys. J. C (2014) 74:2794 ] 2 [GeV/c π - m T m 0 0.2 0.4 0.6 0.8 1 (Pb+Pb)/(p+p) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 GeV/c A 158 GeV/c A 80 GeV/c A 40 GeV/c A 31 GeV/c A 20 Fig. 21 The ratio of the normalised transverse mass spectra of π− mesons at mid-rapidity produced in central Pb+Pb collisions and inelas-tic p+p interactions at the same collision energy per nucleon. The coloured bands represent the systematic uncertainty of the p+p data the width in central Pb+Pb collisions. This is seen in Fig. 23 (right) where the energy dependence of σ is plotted. Addi-tionally, p+p data from [11,36,37,47] are shown; they agree with the NA61/SHINE results.
Note, that when interpreting differences between results obtained for inelastic p+p interactions and central Pb+Pb collisions the isospin effects should be taken into account.
This concerns both the spectra as well as the total multiplic-ities .
In order to reduce their influence the mean multiplicity of pions is obtained from a sum of mean multiplicities of neg-atively and positively charged pions using the phenomeno-logical formula : ⟨π⟩= 3 2 ⟨π+⟩+ ⟨π−⟩ .
(12) The results divided by the mean number of wounded nucle-ons (NW = 2 for p+p) are shown in Fig. 24 as a func-tion of the Fermi energy measure F. The value of ⟨π+⟩for the NA61/SHINE results on inelastic p+p interactions was estimated from the measured ⟨π−⟩multiplicity assuming ⟨π+⟩= ⟨π−⟩+2/3. This assumption is based on the compi-lation of the world data presented in Ref. and the model presented therein. At beam momenta lower than 40AGeV/c the ⟨π⟩/⟨NW⟩ratio is higher in p+p interactions than in cen-tral Pb+Pb collisions. The opposite relation holds for beam momentahigherthan40AGeV/c.Theenergydependencefor inelastic p+p interactions crosses the one for central Pb+Pb (Au+Au) collisions at about 40AGeV/c.
8 Summary We presented experimental results on inclusive spectra and mean multiplicities of negatively charged pions produced in inelastic p+p interactions at 20, 31, 40, 80 and 158GeV/c.
Two dimensional spectra in transverse momentum and rapid-ity and parameters characterizing them were given. The results agree with existing sparse measurements, extend their range, accuracy and depth of detail.
The results on inelastic p+p interactions were compared with the corresponding data on central Pb+Pb collisions obtained by NA49. The spectra in p+p interactions are nar-rower both in rapidity and in transverse mass, which might be [GeV] NN s 6 7 8 9 10 20 ] 2 T [MeV/c 140 150 160 170 180 190 200 p+p (SPS - NA61/SHINE) Pb+Pb (SPS) [GeV] NN s 1 10 2 10 ] 2 [GeV/c π - m 〉 T m 〈 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 p+p (SPS - NA61/SHINE) Au+Au (AGS) Pb+Pb (SPS) Au+Au (RHIC) Fig. 22 Left Inverse slope parameter T of the transverse mass spectra at mid-rapidity (0 < y < 0.2) plotted against the collision energy per nucleon. The parameter T was fitted in the range 0.2 < mT −mπ < 0.7 GeV/c2. The systematic uncertainty for the two lowest energy points for Pb+Pb, not given in is assumed to be the same as for the higher energies . Right Mean transverse mass ⟨mT⟩at mid-rapidity (0 < y < 0.2) versus the collision energy. The results on inelastic p+p interactions are compared with the corresponding data on central Pb+Pb (Au+Au) collisions 123 Eur. Phys. J. C (2014) 74:2794 Page 21 of 22 2794 beam y/y 0 0.5 1 1.5 (Pb+Pb)/(p+p) 0.6 0.8 1 1.2 1.4 GeV/c A 158 GeV/c A 80 GeV/c A 40 GeV/c A 31 GeV/c A 20 [GeV] NN s 6 7 8 9 10 20 σ 0.9 1 1.1 1.2 1.3 1.4 1.5 p+p (SPS - NA61/SHINE) p+p (world) Pb+Pb (SPS) Fig. 23 Left The ratio of normalised rapidity spectra of π−mesons produced in central Pb+Pb collisions and inelastic p+p interactions at the same collision energy per nucleon plotted versus the rapidity scaled by the beam rapidity. The coloured bands represent the NA61/SHINE systematic uncertainty. Right Energy dependence of the width of the rapidity distribution of π−mesons produced in p+p interactions and central Pb+Pb collisions. The systematic uncertainty for the Pb+Pb points is not given ] 1/2 F [GeV 0 1 2 3 4 5 〉 W N 〈 / 〉 π 〈 0 1 2 3 4 5 6 7 8 p+p (SPS - NA61/SHINE) p+p (world) Au+Au (AGS) Pb+Pb (SPS) Au+Au (RHIC) Fig. 24 Mean multiplicity of all pions per wounded nucleon produced in inelastic p+p interactions and central Pb+Pb (Au+Au) collisions. The vertical lines show the total uncertainty attributed to isospin effects. The mean pion multiplicity per wounded nucleon in p+p interactions increases more slowly with energy in the SPS range and crosses the correspond-ing dependence measured in the Pb+Pb collisions at about 40AGeV/c.
Acknowledgments ThisworkwassupportedbytheHungarianScien-tific Research Fund (Grants OTKA 68506 and 71989), the Polish Min-istry of Science and Higher Education (Grants 667/N-CERN/2010/0, NN 202 48 4339 and NN 202 23 1837), the National Science Cen-ter of Poland (Grant UMO-2012/04/M/ST2/00816), the Foundation for Polish Science—MPD program, co-financed by the European Union within the European Regional Development Fund, the Federal Agency of Education of the Ministry of Education and Science of the Rus-sian Federation (Grant RNP 2.2.2.2.1547), the Russian Academy of Science and the Russian Foundation for Basic Research (Grants 08-02-00018, 09-02-00664, and 12-02-91503-CERN), the Ministry of Educa-tion, Culture, Sports, Science and Technology, Japan, Grant-in-Aid for Scientific Research (Grants 18071005, 19034011, 19740162, 20740160 and20039012),theGermanResearchFoundation(GrantsGA1480/2-1, GA 1480/2-2), Bulgarian National Scientific Foundation (Grant DDVU 02/19/2010), Ministry of Education and Science of the Republic of Ser-bia (Grant OI171002), Swiss Nationalfonds Foundation (Grant 200020-117913/1) and ETH Research Grant TH-01 07-3. Finally, it is a pleasure to thank the European Organization for Nuclear Research for a strong support and hospitality and, in particular, the operating crews of the CERN SPS accelerator and beam lines who made the measurements possible.
Open Access This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited.
Funded by SCOAP3 / License Version CC BY 4.0.
References 1. N. Antoniou et al. (NA49-future Collaboration), CERN-SPSC-2006-034 (2006) 2. S.V. Afanasiev et al., (NA49 Collaboration). Phys. Rev. C 66, 054902 (2002). [arXiv:0205002 [nucl-ex]] 3. C. Alt et al., (NA49 Collaboration), Phys. Rev. C 77, 024903 (2008). [arXiv:0710.0118 [nucl-ex]] 4. A. Rustamov, Central Eur. J. Phys. 10, 1267 (2012). [arXiv:1201.
4520 [nucl-ex]] 5. Y. Itow et al., (T2K Collaboration), arXiv:0106019 [hep-ex] (2001) 6. K. Abe et al., (T2K Collaboration), Nucl. Instrum. Meth. A 659, 106–135 (2011) 7. N. Abgrall et al., (NA61/SHINE Collaboration), CERN-SPSC-2012-022; SPSC-P-330-ADD-6 8. K. Abe et al., (T2K Collaboration), Phys. Rev. D 87, 012001 (2013) 9. J. Abraham et al., (Pierre Auger Collaboration), Nucl. Instrum.
Meth. A 523, 50 (2004) 10. T. Antoni et al., (KASCADE Collaboration), Nucl. Instrum. Meth.
A 513, 490 (2003) 123 2794 Page 22 of 22 Eur. Phys. J. C (2014) 74:2794 11. C. Alt et al., (NA49 Collaboration), Eur. Phys. J. C 45, 343–381 (2006). [arXiv:0510009 [hep-ex]] 12. T. Anticic et al., (NA49 Collaboration), Eur. Phys. J. C 65, 9 (2010).
[arXiv:0904.2708 [hep-ex]] 13. T. Anticic et al., (NA49 Collaboration), Eur. Phys. J. C 68, 1 (2010).
[arXiv:1004.1889 [hep-ex]] 14. M. Gazdzicki, M. Gorenstein, P. Seyboth, Acta Phys. Polon. B 42, 307 (2011). [arXiv:1006.1765 [hep-ph]] 15. A.I. Golokhvastov, Phys. Atom. Nucl. 64, 1841 (2001). [Yad. Fiz.
64, 1924 (2001)] 16. S. Afanasiev et al., (NA49 Collaboration), Nucl. Instrum. Meth. A 430, 257 (1999) 17. N. Abgrall et al., (NA61/SHINE Collaboration), Phys. Rev. C 84, 034604 (2011). [arXiv:1102.0983 [hep-ex]] 18. C. Bovet, R. Maleyran, L. Piemontese et al., CERN-YELLOW-82-13 19. C. Strabel, Ph.D. Thesis, ETH, Zurich, Switzerland (2011). https:// edms.cern.ch/document/1136130/1 20. N. Abgrall et al., (NA61/SHINE Collaboration), J. Instrum. (sub-mitted). [arXiv:1401.4699 [physics.ins-det]] 21. N. Abgrall et al., (NA61/SHINE Collaboration), Report No.
CERN-SPSC-2008-018 (2008) 22. M. Unger, for the NA61/SHINE Collaboration, EPJ Web Conf. 52, 01009 (2013). [arXiv:1305.5281 [nucl-ex]] 23. M. Unger et al., (NA61/SHINE Collaboration), PoS ICHEP 2010, 449 (2010). arXiv:1012.2604 [nucl-ex] 24. A. Ilnicka, Master Thesis, University of Warsaw, Poland (2013) 25. K. Werner, Nucl. Phys. B 175(Proc. Suppl.), 81 (2008) 26. N. Abgrall, Ph.D. Thesis, University of Geneva, Geneva, Switzer-land (2011). 27. R. Brun, F. Carminati, GEANT Detector Description and Simula-tion Tool, CERN Program Library Long Writeup W5013 (1993).
28. J. Beringer, Particle Data Group, Phys. Rev. D 86, 010001 (2012) 29. A. Aduszkiewicz, Ph.D. Thesis in preparation, University of War-saw, Poland (2014) 30. S. Puławski, PoS, CPOD 2013, 056 (2013). archive/conferences/185/056/CPOD%202013_056.pdf 31. A. Aduszkiewicz, EuNPC 2012 (2012). indico/contributionDisplay.py?contribId=268&sessionId=2& confId=0 32. M. Gazdzicki, D. Roehrich, Z.Phys.C 65, 215 (1995) 33. M. Ga´ zdzicki, D. Röhrich, Z.Phys.C 71, 55 (1996) 34. K. Werner, Nucl. Phys. A 525, 501c (1991) 35. K. Werner, Phys. Rep. 232, 87 (1993) 36. E.E. Zabrodin et al., Phys. Rev. D 52, 1316 (1995) 37. E.E. Zabrodin, private communication.
38. 39. E. Fermi, Prog. Theor. Phys. 5, 570 (1950) 40. J. Klay et al., (E895 Collaboration), Phys. Rev. C 68, 054905 (2003) 41. L. Ahle et al., E802 Collaboration, Phys. Rev. C 57, 466 (1998) 42. B.B. Back et al., (PHOBOS Collaboration). [arXiv:0301017 [nucl-ex]] 43. K. Adcox et al., (PHENIX Collaboration), Phys. Rev. C 69, 024904 (2004) 44. S.S.Adleret al.,(PHENIXCollaboration),Phys.Rev.C 69,034909 (2004) 45. J. Adams et al., (STAR Collaboration), Phys. Rev. Lett. 92, 112301 (2004) 46. I.G. Bearden et al., (BRAHMS Collaboration), Phys. Rev. Lett. 94, 162301 (2005) 47. V.V. Ammosov et al., Nuovo Cimento 40, 237 (1977) 123 |
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in:Math exercises, Eureka Math/EngageNY exercises, 6th grade math exercises,
and4 more
6th grade (U.S.): Data and statistics
6th grade (Eureka Math/EngageNY): Module 6: Statistics
High school statistics and probability exercises
High school statistics and probability: Displaying and describing data
Reading histograms
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| Reading histograms |
| |
| Description |
| Exercise Name: | Reading histograms |
| Math Missions: | 6th grade (U.S.) Math Mission, High school statistics and probability Math Mission |
| Types of Problems: | 1 |
The Reading histograms exercise appears under the 6th grade (U.S.) Math Mission and High school statistics and probability Math Mission. This exercise analyzes a histogram created from a set of data.
Types of Problems[]
There is one type of problem in this exercise:
Answer the question based on the histogram: This problem has a histogram that models some data and a question related to the histogram. The user is asked to correctly answer the question based on the histogram.
Answer the question based on the histogram
Strategies[]
Knowledge of statistical vocabulary and the method to read and create frequency tables are encouraged to ensure success on this exercise.
The height of a bar represents the number of times the values occur in that category.
Possible question include:
How many total pieces of data?
How many pieces of data in certain categories?
How many different categories?
Real-life Applications[]
Data and statistics appear in news reports and in the media every day.
Many of the problems in this exercise could be viewed as real-life applications.
Statistics can be seen more frequently than calculus in every day life.
Categories
Categories:
Math exercises
Eureka Math/EngageNY exercises
6th grade math exercises
6th grade (U.S.): Data and statistics
6th grade (Eureka Math/EngageNY): Module 6: Statistics
High school statistics and probability exercises
High school statistics and probability: Displaying and describing data
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190246 | https://www.acxesspring.com/spring-force-constant-calculator.html?srsltid=AfmBOorm2zhqaH9Z3N520dBj0qY0tdQ_j_HJAVONoavUSwjUlFfqfstb | Spring Force Constant Calculator
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Spring Force Constant Calculator
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Use our spring force constant calculator to find the force in a spring. To calculate spring force constant enter the inputs and click calculate. Outputs include compression spring force, force on a spring, spring force constant, and spring rates. This is also a spring rates calculator that calculates spring rates. The spring constant is also known as your spring rate or compression spring rate. How to measure spring rate: enter your inputs into our spring force calculator and review your answers. Outputs include spring ratings and spring constant. Spring ratings are a constant, example: If your spring rate is 10 lbsf/in then it will take you 10 lbsf to force to compress the spring one inch of distance. To determine a spring load or load of a spring: rate times distance traveled equal = spring load or load of a spring. Our spring force calculator generates spring blueprints, you can email your spring blueprints once your spring design is finished. You can also save your spring blueprints as a pdf.
Spring Calculator Instructions
COMPRESSION
EXTENSION
TORSION
Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
Choose Your Unit of Measure:
English Metric
Wire Diameter:
Optional
min
to
max
inchesmm
[x] Outer Diameter Inner Diameter
Outer Diameter:
Mandatory
min
to
max
inchesmm
Inner Diameter:
Mandatory
min
to
max
inchesmm
Free Length:
Mandatory
min
to
max
inchesmm
Total Coils:
Optional
min
to
max
Material Type:
Optional
Required Field Calculate Reset Form
Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
Choose Your Unit of Measure:
English Metric
Wire Diameter:
Optional
min
to
max
inchesmm
Outer Diameter:
Mandatory
min
to
max
inchesmm
Length Inside Hook:
Mandatory
min
to
max
inchesmm
Material Type:
Optional
Required Field Calculate Reset Form
Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
Choose Your Unit of Measure:
English Metric
Wire Diameter:
Optional
min
to
max
inchesmm
[x] Outer Diameter Inner Diameter
Outer Diameter:
Mandatory
min
to
max
inchesmm
Inner Diameter:
Mandatory
min
to
max
inchesmm
Leg Length:?Both legs are equal in length
Optional
min
to
max
inchesmm
Free Position or Leg Position in degrees:?Free Position or Leg Position in degrees
Optional
degrees
Total Coils:
Mandatory
min
to
max
Material Type:
Optional
Optional
[x] Right Hand (RH)
[x] Left Hand (LH)
Required Field Calculate Reset
For instructions on how to use our Free Online Spring Calculator visit our pages:Compression Spring Calculator InstructionsExtension Spring Calculator InstructionsTorsion Spring Calculator Instructions
Spring Force Constant Calculator: How to Determine Spring Rates with Confidence
Spring Calculator Instructions
COMPRESSION
EXTENSION
TORSION
Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
Choose Your Unit of Measure:
English Metric
Wire Diameter:
Optional
min
to
max
inchesmm
[x] Outer Diameter Inner Diameter
Outer Diameter:
Mandatory
min
to
max
inchesmm
Inner Diameter:
Mandatory
min
to
max
inchesmm
Free Length:
Mandatory
min
to
max
inchesmm
Total Coils:
Optional
min
to
max
Material Type:
Optional
All
Required Field Calculate Reset Form
Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
Choose Your Unit of Measure:
English Metric
Wire Diameter:
Optional
min
to
max
inchesmm
Outer Diameter:
Mandatory
min
to
max
inchesmm
Length Inside Hook:
Mandatory
min
to
max
inchesmm
Material Type:
Optional
All
Required Field Calculate Reset Form
Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
Choose Your Unit of Measure:
English Metric
Wire Diameter:
Optional
min
to
max
inchesmm
[x] Outer Diameter Inner Diameter
Outer Diameter:
Mandatory
min
to
max
inchesmm
Inner Diameter:
Mandatory
min
to
max
inchesmm
Leg Length:?Both legs are equal in length
Optional
min
to
max
inchesmm
Free Position or Leg Position in degrees:?Free Position or Leg Position in degrees
Optional
Ÿ Ý 0 degrees
degrees
Total Coils:
Mandatory
min
to
max
Material Type:
Optional
All
Optional
[x] Right Hand (RH)
[x] Left Hand (LH)
Required Field Calculate Reset
Table of Contents
What is a Spring Force Constant?
Why is the Spring Constant Important in Design?
How Do You Calculate the Spring Constant of a Compression Spring?
How Do You Calculate an Extension Spring's Constant?
How Can You Adjust a Spring’s Constant?
How Can You Calculate a Spring Constant Online?
Wrapping It All Up: Why Spring Constants Matter
What is a Spring Force Constant?
The spring force constant, or spring rate, measures how much force is needed to compress, extend, or twist a spring by a certain amount. It’s commonly written as "k" and is central to Hooke’s Law: F = k × x, where F is force, x is deflection, and k is the spring constant.
For compression or extension springs, spring rate is measured in lbf/in or N/mm. For torsion springs, it’s torque per angle, like in-lb per degree or N·m per radian. A higher k value means a stiffer spring that resists movement more.
Why is the Spring Constant Important in Design?
The spring constant is a key factor in performance. If the spring is too soft, it won’t support the load properly. If it’s too stiff, it may damage parts, reduce motion, or make the system difficult to actuate.
For example, a car suspension uses springs with a specific k to balance comfort and control. A pen uses a soft spring that’s easy to compress repeatedly without excessive force.
With the spring rate, you can calculate the load at a given deflection using Hooke’s Law. It’s a reliable formula that helps you design with precision and avoid costly trial and error.
How Do You Calculate the Spring Constant of a Compression Spring?
To calculate a compression spring’s constant, you can measure it manually or apply a proven formula based on the spring’s geometry and material properties. This constant, also known as spring rate, defines how much force is needed to compress the spring a certain distance and is usually expressed in lbf/in or N/mm.
Experimentally: This method is useful for existing springs. Apply a known load and measure the resulting deflection. For example, if you apply a 50 lb force and the spring compresses exactly 1 inch, the spring rate is 50 lbf/in. This hands-on method provides a quick approximation when lab tools or detailed specs aren’t available.
Using the formula: For a more precise and predictive calculation, especially during spring design, use the following formula:
k = (G × d^4) / (8 × D^3 × N)
Where:
G = shear modulus of the material (psi or N/mm²)
d = wire diameter (in or mm)
D = mean coil diameter (in or mm)
N = number of active coils
This formula is ideal when designing custom springs or verifying spring behavior before production. It helps engineers factor in how material stiffness (G), wire thickness, and coil geometry affect performance.
Design Note: Music wire (ASTM A228) is commonly used in compression springs and has a shear modulus around 11,492,970.929 psi. The higher the modulus and wire diameter, the stiffer the spring. Reducing coil diameter or the number of active coils also increases stiffness.
Example: Let’s say you have Acxess Spring’s part number PC054-484-12000-MW-1934-C-N-IN, a compression spring made of music wire (G ≈ 11,492,970.929 psi), with:
Wire diameter (d) = 0.054 in
Mean coil diameter (D) = 0.43 in
Active coils (N) = 10
Plugging into the formula:
k = (11,492,970.929 × 0.054⁴) / (8 × 0.43³ × 10)
k ≈ (11,492,970.929 × 0.0000085) / (8 × 0.079507 × 10)
k ≈ 97.6902529 / 6.36056 = 15.35875031 lbf/in
This means it takes around 15.36 pounds of force to compress the spring by one inch.
How Do You Calculate an Extension Spring's Constant?
Extension springs follow the same formula as compression springs. However, they also have initial tension—a preload force that must be overcome before stretching begins. This initial tension allows the spring to maintain its coiled position without any external force and adds to the total force the spring provides when stretched.
To account for initial tension in your force calculations:
Spring force (F) = k × extension (x) + initial tension
This equation tells us that even before the spring starts to stretch, a certain amount of force is already present due to initial tension. Once the applied load exceeds this value, the spring begins to extend, and the force increases linearly with the stretch distance.
Design Insight: Initial tension is determined during the manufacturing process and depends on how tightly the coils are wound. It's especially useful in applications like screen doors or trampolines, where a firm closure or retraction force is needed from the start.
Example: Imagine an extension spring, like Acxess Spring’s part number PE063-750-10937-MW-2000-CO-N-IN, with a spring rate of 6.382 lbf/in and an initial tension of 1.511 lbf. If the spring is stretched 1.5 inch:
Spring force = (6.382 × 1.5) + 1.511 = 9.573 + 1.511 = 11.084 lbf
So, a 1.5 inches stretch would produce 11.08687 pounds of force.
How Can You Adjust a Spring’s Constant?
You can increase or decrease the spring constant by modifying specific elements of the spring’s design. These variables give engineers and designers flexibility to tailor a spring’s behavior to meet application requirements. Here’s how each factor plays a role:
Material: The type of material used directly affects the modulus of elasticity or shear modulus. Materials like music wire offer higher stiffness compared to softer metals like phosphor bronze. Choosing a higher modulus material increases the spring rate.
Wire Diameter (d): Spring rate increases significantly with wire diameter, since it's raised to the fourth power in the formula. For example, going from a 0.1 in to a 0.12 in wire can yield a dramatic rise in spring stiffness.
Coil Diameter (D): A smaller mean coil diameter leads to a stiffer spring, as it reduces the leverage acting on the coils. Reducing the diameter tightens the spring, making it resist force more strongly.
Number of Active Coils (N): The more coils a spring has, the more it distributes force, which lowers stiffness. Reducing the number of active coils will increase the spring rate, while adding coils will soften the spring.
Initial Tension (Extension Springs): For extension springs, initial tension is the preload force built into the spring. Increasing this tension allows the spring to resist initial stretch, acting stiffer in early movement. It's adjusted during manufacturing by the way the coils are wound.
Example Adjustment Scenario: Suppose you designed a spring with a spring rate of 20 lbf/in but need it closer to 30 lbf/in. You could:
Use a material with a higher shear modulus
Increase the wire diameter slightly (e.g., from 0.09 in to 0.1 in)
Decrease the coil diameter (e.g., from 0.6 in to 0.5 in)
Remove one active coil if feasible
Each of these changes will bring the spring rate closer to your target, allowing you to fine-tune your design based on mechanical constraints, available space, or cost.
How Can You Calculate a Spring Constant Online?
The easiest and most efficient way to calculate spring constants is by using our Spring Force Constant Calculator—a free tool designed to make your spring design process faster, simpler, and more accurate.
Try the calculator at the top of this article, or visit Spring Creator 5.0. The tool is browser-based, so there's no need to install anything.
Choose the spring type. Select from compression, extension, or torsion springs, depending on your application.
Input your specifications. Enter essential spring details such as wire diameter, outer or mean coil diameter, free length, number of active coils, and material type. These inputs are used to compute the spring rate accurately.
Click "Calculate." The calculator instantly processes your inputs using industry-standard formulas and displays the spring constant along with other vital data.
Review and refine. You'll get your spring rate in both imperial and metric units, along with information like maximum load, travel distance, and stress levels. You can make quick adjustments to the input and recalculate in real time.
The calculator also generates a detailed spring blueprint. This includes a 3D visual model of the spring, which you can rotate and inspect. You can download the blueprint as a PDF or send it to your inbox. This is perfect for record-keeping or sharing with suppliers and colleagues.
One of the most powerful features is the ability to explore alternate designs. If the calculated spring doesn’t fully meet your requirements, the tool will suggest nearby configurations with slightly higher or lower spring rates, helping you fine-tune your design effortlessly.
Whether you're a student learning about Hooke’s Law, an engineer refining a mechanical assembly, or a buyer preparing for production, our spring calculator online is your go-to resource. It removes the math hurdles and puts professional-grade spring engineering right at your fingertips.
Wrapping It All Up: Why Spring Constants Matter
Understanding the spring force constant is essential for any spring-based project. Whether you're designing a new product, troubleshooting an existing system, or optimizing performance, having a solid grasp of spring behavior will give you an edge. It empowers you to calculate forces accurately, prevent failures, and create systems that perform as expected.
For beginner engineers and students, this knowledge is a stepping stone into more advanced mechanical design. For buyers and DIY designers, it’s a way to confidently choose the right spring without relying on guesswork. With the help of modern tools like our online spring calculator, what once required complex math is now accessible in seconds.
Whether you’re working on compression springs for motion control, extension springs for return mechanisms, or torsion springs for torque applications, getting the spring constant right is key. Combine your understanding of spring rates with our free online calculator and spring blueprint tools, and you’ll be well on your way to smarter, more precise spring designs.
Don’t hesitate to reach out to our team at Acxess Spring for custom design help or to manufacture springs built to your exact specs. We’re here to help you turn great ideas into great products.
Key Takeaways
The spring force constant (k) shows how stiff a spring is.
Use Hooke’s Law to relate force, deflection, and rate.
Spring constants differ for compression, extension, and torsion springs.
Change the constant by adjusting material, wire size, coil size, or number of coils.
Use our free spring calculator online for quick results and professional blueprints.
Explore our full suite of spring design services and custom spring manufacturing at Acxesspring.com.
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190247 | https://kids.wordsmyth.net/we/school/ISBLibrary/?level=1&ac=823&rid=12637 | dry | definition for kids | Wordsmyth Word Explorer Children's Dictionary; WILD dictionary K-2 | Wordsmyth
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190248 | https://dev.to/jenshaw/reversing-an-integer-mathematically-fmg | Reversing an Integer Mathematically - DEV Community
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Jenny Shaw
Posted on Oct 9, 2019
1611
Reversing an Integer Mathematically
#javascript#beginners#todayilearned#problemsolving
The Problem
This an algorithm problem I've encountered a couple of times called Reverse the Integer.
```
Write a program or function called reverseInteger
to reverse the order of digits of a given integer.
Input: 12345
Output: 54321
```
At first glance, it appeared easy enough to figure out. One of the first problems I ever remember having to solve was Reverse the String, and so, because the two seemed similar enough, I figured I could use the same approach for both.
Reverse the String
Here's one solution to Reverse the String:
```
function reverseString(str) {
let reversedStr = ''; // initialize variable with empty string
for (let i = str.length - 1; i >= 0; i--) { // iterate backwards through each character of str (input)
reversedStr = reversedStr.concat(str[i]); // add each character to end of reversedStr
}
return reversedStr; // after completion of iterations, return reversedStr
}
reverseString('dog')
// returns 'god'
```
Here, I plan to return a variable called reversedStr at the end of my function. First, I initialize it as an empty string, and as I iterate backwards through each character of str, the original string input, I take that character to build up reversedStr using concatenation. Almost like .pop() and .push() in an array situation.
Reverse the Integer (like a string)
We could reverse integers using a similar algorithm, but there are a couple of caveats:
integers can't be iterated through, and
digits can't be concatenated.
If our input for reverseTheString() were an integer, the function would just spit back an empty string. Useless.
To resolve this, we'd have to first convert the integer input into a string before iteration and concatenation. And if we're expected to return an integer in the end, we'd also have to convert the string back into an integer before returning the value.
```
function reverseInteger(num) {
let numStr = num.toString(); // <-- convert integer to string
let reversedNumStr = '';
for (let i = numStr.length - 1; i >= 0; i--) {
reversedNumStr = reversedNumStr.concat(numStr[i]);
}
let reversedInt = Number(reversedNumStr); // <-- convert string back to integer
return reversedInt; // return a reversed integer
}
reverseInteger(12345)
// returns 54321
```
I've never been very enthusiastic about reversing an integer like a string for a few reasons.
Although this function certainly gets the job done for (most) integer inputs, I don't like having to go through the extra trouble of converting data types back and forth. I'd rather stick to just one data type.
Also, we're asked to reverse integers, yet we're largely manipulating strings, so this feels like a rather tangential approach, a little like a cheat. And I'm no cheater, so we're going to learn to do this right.
Reverse the Integer with Math
Let's approach this problem instead in a way where we can still cleanly 'pop' and 'push' digits, do it all mathematically, and completely avoid the need to convert our integer into a string and back.
(By the way, if you're worried about the math, don't be. We're sticking with basic arithmetic here. Elementary school level stuff. If you understand subtraction, multiplication, division, and place values, then you've got this, kid.)
Keep in mind that in this function, we'll be handling two variables. The first, num, is the input from which we'll be 'popping' digits until there are none left. The second, reversedInteger, will be our output. Here we'll be building up the reversed order of digits by 'pushing' on the 'popped' digits from num.
Step 1:
We'll start with the variable, reversedInteger, and initialize its value at 0.
```
function reverseIntegerWithMath(num) {
let reversedInteger = 0; // <-- initialize reversedInteger
}
```
Step 2:
We're going to start a while loop and continue it while num still has a value greater than 0. Every loop, we'll be chipping away one digit from num and using the digit to build reversedInteger.
```
function reverseIntegerWithMath(num) {
let reversedInteger = 0;
while (num > 0) { // <-- open while loop
}
}
```
Step 3:
At the beginning of each loop, we'll multiply reversedInteger by 10.
```
function reverseIntegerWithMath(num) {
let reversedInteger = 0;
while (num > 0) {
reversedInteger = 10; // <-- set up for proper place value
}
}
// Let's keep track of what num and reversedNumber look like
// starting from here...
// num => 1234
// reversedInteger => 0 10
// => 0
```
Step 4:
Now, let's take our num and divide by 10 using the modulo operator. This is to find a single-digit remainder which equals the current last digit of nums. We'll initialize a variable called rem at the top of our function, and tuck that value safely into it.
Then subtract rem from num and divide the result by 10. And now we're left with the same integer, but one digit less.
POP!
```
function reverseIntegerWithMath(num) {
let reversedInteger = 0;
let rem = 0; // <-- initialize remainder
while (num > 0) {
reversedInteger = 10;
rem = num % 10; // <-- remainder grabs last digit
num = (num - rem) / 10; // <-- eliminate zero in num
}
}
// rem => 1234 % 10
// => 4
// num => 1234 - rem
// => 1230 / 10
// => 123
// reversedInteger => 0
```
In case you're curious...
Why are we dividing and multiplying numbers by 10?
It's because we're replicating decimal place values where each place has a value of times ten from right to left.
Dividing by 10 eliminates the last zero in num, which then gives us access to the next digit that ends up in the ones place.
Multiplying reversedInteger by 10 makes room in the ones place where we can place the digit we popped off from num.
Step 5:
Next, it's time to "push" the "popped" digit from num by taking the remainder and adding it to reversedInteger.
PUSH!
```
function reverseIntegerWithMath(num) {
let reversedInteger = 0;
let rem = 0;
while (num > 0) {
reversedInteger = 10;
rem = num % 10;
num = (num - rem) / 10;
reversedInteger += rem; // <-- 'push' remainder onto end of reversedInteger
}
}
// rem => 4
// num => 123
// reversedInteger => 0 + 4
// => 4
```
Step 6:
We've completed one cycle of this process. Repeat until num's value dwindles to 0 and there are no more digits to 'pop' or 'push'.
After the reversal of digits is complete, we can finally return reversedInteger.
```
function reverseIntegerWithMath(num) {
let reversedInteger = 0;
let rem = 0;
while (num > 0) {
reversedInteger = 10;
rem = num % 10;
num = (num - rem) / 10;
reversedInteger += rem;
}
return reversedInteger; // <-- done!
}
// if you want to see what happens in the next loop
// num => 123 - 3 (step 4)
// => 120 / 10
// => 12 [pops the 3 from original integer]
// rem => 123 % 10 (step 3)
// => 3
// reversedInteger => 4 10 (step 2)
// => 40 + 3 (step 5)
// => 43 [pushes the 3 onto reversedInteger]
```
This is a pretty simple and neat trick in numeric manipulation and a much-improved approach to the reverseInteger problem. I'm always looking for other creative ways to solve simple problems like this, so if you've got clever ones to share, drop them in the comments!
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Steve Geluso
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•Oct 9 '19
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Yessss! This is definitely a problem people should be exposed to. It's too common that people sneak out a solution via Strings. Excellent explanation!
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Augusto Hernández
Augusto Hernández
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underrated!
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190249 | https://www.youtube.com/watch?v=k54pJCF7PN0 | Converting Repeating Decimals to Fractions | Math with Mr. J
Math with Mr. J
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Welcome to Converting Repeating Decimals to Fractions with Mr. J! Need help with writing repeating decimals as fractions? You're in the right place!
Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with how to convert repeating decimals to fractions. Mr. J will go through examples of repeating decimals and explain how to convert them to fractions.
✅ Chapters and Timestamps
00:00 - Intro
00:35 - Section 1 (only repeating digits)
15:31 - Section 2 (non-repeating and repeating digits)
About Math with Mr. J: This channel offers instructional videos that are directly aligned with math standards. Teachers, parents/guardians, and students from around the world have used this channel to help with math content in many different ways. All material is absolutely free. #MathWithMrJ
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Email: math5.mrj@gmail.com
Music:
Hopefully this video is what you're looking for when it comes to converting repeating decimals to fractions.
Have a great rest of your day and thanks again for watching! ✌️✌️✌️
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Transcript:
Intro [Music] welcome to math with Mr J in this video I'm going to cover how to convert repeating decimals to fractions we will take a look at four examples where we only have repeating digits to the right of the decimal and then we will take a look at two examples where we have non-re repeating and repeating digits to the right of the decimal now this process may seem confusing and overwhelming at first but after some practice and examples it will make a lot more sense let's jump into Section 1 (only repeating digits) number one where we have 0.7 repeating and our first step is going to be to write an equation x equals the repeating decimal so all we need to do for Step One X equals and then we have 777 and this continues on forever now let's move on to step two which we are going to work off of that original equation from step one we need to create two equations one with the repeating digit or digits just to the left of the decimal and then another equation with the repeating digit or digits just to the right of the decimal and we can multiply by powers of 10 in order to do this so 10 100 1,000 so on and so forth so let's start by getting the repeating digit 7 just to the left of the decimal so we can do this by multiplying the right side of the equation by 10 that shifts our digits one place to the left and we can think of this as just moving the decimal one place to the right so we are multiplying this side of the equation by 10 whatever we do to one side of an equation we must do to the other in order to keep it equivalent this gives us 10 X = 7.7 repeating now that we have the repeating digit seven just to the left of the decimal we need an equation where the repeating digit 7 is just to the right of the decimal and we add actually already have this equation written x equal 0.7 repeating so x equals 0.7 repeating so here is our equation with the repeating digit just to the left of the decimal and here is our equation with the repeating digit just to the right of the decimal now we can move on to step three where we subtract these equations to remove the repeating digit or digits so let's subtract these equations 10 x - x gives us 9 x equals and then on the right side of the equations the repeating sevens if we subtract these they cancel out they equals z so now we just have 7 - 0 which is 7 so the whole point here was to get rid of the repeating digit seven and now we have an equation to give us our fraction and now we can move on to step four we need to solve this equation for x and then simplify if possible so let's isolate that variable of X we have 9x = 7 so let's divide both sides by 9 so on the left side of the equation we just have X now and then on the right side we have 7 9ths which is in simplest form the only common factor between s and 9 is 1 so 0.7 repeating equals 7 9ths as a fraction so just remember we're looking to remove the repeating digit or digits and and we need to create an equation that we can solve for x now you can always double check to see if your fraction is correct by dividing the numerator of your fraction by the denominator so either using a calculator or you can do it by hand either way you should get the original repeating decimal and one more thing I do want to mention before moving on to number two all of those steps like I mentioned earlier may seem confusing long and overwhelming at first but I like to simplify them and think of this as Left Right subtract solve so get the repeating digit or digits to the left to the right subtract and solve let's move on to number two where we have 0.36 repeating let's start by writing an equation where x equals the repeating decimal so 0 36 36 and this continues on forever now for number two we have two repeating digits so we need to multiply by 100 in order to get those repeating digits just to the left of the decimal so we need to shift those digits two places to the left and we can think of this as moving the decimal twice to the right so we're multiplying this side by 100 whatever we do to one side of an equation we must do to the other in order to keep it balanced and equivalent so multiply the left side x by 100 as well that gives us 100 x equal 36 36 repeating now we need those repeating digits just to the right of the decimal and we already have that equation x equals 0.36 repeating so left right and now we subtract to remove the repeating digits so let's subtract 100x - x gives us 99x equal the repeating digits cancel out those equal 0 when we subtract so we have 36 - 0 which gives us 36 now we have an equation where we need to solve for x so 99x = 36 we need to isolate that variable of X by dividing both sides by 99 so on the left side x is now isolated so we have x equals and then 36 99 now 36 99s can be simplified we have a greatest common factor of nine that we can divide both the numerator and denominator by 36 / 9 is 4 99 / 9 is 11 so we get 4 11ths the only common factor between four and 11 is 1 so this is now simplified so 0 36 repeating equals 4 11ths as a fraction let's move on to numbers three and four so here are numbers three and four let's jump into number three where we have 2.4 repeating now for this one we have a whole number in front two but let's forget about that whole number until the end we're going to convert the repeating decimal to a fraction and then again worry about the whole number at the end so let's first write an equation where x equals the repeating decimal so x equals 0.4 repeating now here we have one repeating digit four so we need to multiply the right side of the equation by 10 in order to get that digit just to the left of the decimal whatever we do to one side of an equation we must do to the other so multiply the left side by 10 as well that gives us 10 x equal 4.4 repeating now we need our equation with the repeating digit just to the right of the decimal that's going to be x equals 0.4 repeating and now we can subtract 10 x - x gives us 9x equals our repeating fours cancel each other out we we subtract those they equal 0 so now we have 4 - 0 which gives us 4 and now we have our equation where we need to solve for x so 9x = 4 let's isolate that variable of X by dividing both sides by 9 on the left side of the equation x is now isolated so we have x equals and then on the right side of the equation we have 4 9 the only common factor between 4 and 9 is 1 so 4 9ths is in simplest form so4 repeating equals 4 9ths as a fraction so going back up to our original number here we have two we need to remember our whole number and 4 9ths so 2.4 repeating equals 2 and 4 9th in fractional form form we have a mixed number here because again we need to remember that whole number of Two And like I mentioned earlier you can always check here so do 4 / 9 either with a calculator or by hand and you will get 0.4 repeating and lastly let's move on to number four where we have 0.150 repeating let's start by writing an equation here where x equals the repeating decimal so 0.150 150 and this continues on now here we have three repeating digits just to the right of the decimal so we need to multiply by 1,00 that shifts the digits three places to the left and we can just think through this by moving the decimal three times to the right whatever we do to one side of the equation we must do to the other so multiply the left side by 1,000 as well that gives us 1,000 x equal 1505 repeating now we need the repeating digits just to the right of the decimal and we already have that x equal 0 0.150 150 and that continues on now that we have our two equations we can subtract 1X - x gives us [Music] 999x equals and then on the right side our repeating digits cancel each other out we subtract them and that equals zero so we are left with 150 - 0 and that gives us 150 now we have our equation that we can solve for x so 999x equal 150 let's isolate our variable of X by dividing both sides by 999 on the left side x is now isolated and then on the right side we have 150 999 so 150 over 999 now this can be simplified we have a greatest common factor of three that we can divide the numerator and denominator by 150 / 3 gives us 50 999 ID 3 gives us 333 so we end up with 50300 33 the only common factor between 50 and 333 is 1 so this is now simplified so 0.150 repeating equals 50 over 333 50333 now you may have noticed as we were working through these we ended up with something over 9 or 99 or 999 this is going to be the case for whenever we have only repeating digits to the right of the decimal so for example 0.5 repeating will be written as 5 over9 as a fraction 0.87 repeating will be written as 87 over 99 as a fraction and then something to the thousandth like 0.337 repeating will be written at as 337 over 999 as a fraction so that's something to keep in mind from here on out but I wanted to go through the how and why that all works out now with that being said don't just put any repeating decimal over 9ines that will not work when we have digits that do not repeat to the right of the decimal mixed with digits that do repeat to the right of the decimal now let's move on to those Section 2 (non-repeating and repeating digits) examples here are our examples with non-re repeating and repeating digits to the right of the decimal let's jump into number one where we have 0.2 and then 47 repeating so the two in the 10th place is a non-re repeating digit and then the four and the 7even do repeat now we're going to use the same steps we used in part one so for step one we need to write an equation where x equals the repeating decimal so x equals 0.2 4747 and this continues on forever now we can move on to step two and we'll use that equation from step one when we work through step two so we need to create two equations one with the repeating digit or digits just to the left of the decimal and then the other equation we need the repeating digit or digits just to the right of the decimal and we're going to multiply by powers of 10 in order to do this let's start by getting the repeating digits 4 and 7 just to the left of the decimal it looks like we need to shift the digits three places to the left in order to do this and we can think through this by moving the decimal three times to the right so we're multiplying that side of the equation by 1,000 whatever we do to one side of an equation we must do to the other so we need to multiply the left side by 1,000 as well that gives us 1,000 x equal 247 47 repeating now we need an equation with those repeating digits just to the right of the decimal so we need to shift the digits one place to the left and we can think of this as moving the decimal once to the right so the decimal will be right here so we have those repeating digits four and seven again just to the right right of the decimal so we multipli that right side of the equation by 10 whatever we do to one side of an equation we must do to the other so we need to multiply the left side by 10 as well that gives us 10 x equal 2.47 repeating so here we have the repeating digit just to the left of the decimal and here we have the repeating digits just to the right of the decimal that brings us to step three where we can subtract these equations to remove the repeating digits four and 7 so let's subtract here and we'll start with 1X - 10 x that gives us 990 X and then then on the right side the repeating digits 047 cancel each other out 47 repeating minus 47 repeating those are going to equal zero again they cancel out so we are getting rid of the repeating digits 7 - 2 is 5 then we have four and 2 so we have 2045 now we have an equation that we can solve for x 9 0x = 245 so moving on to step four we need to solve for x and then simplify if possible let's start by isolating the variable of X so we need to divide both sides by 990 so on the left side x is now isolated and then on the right side we have 245 900 90ths 245 over 990 now we can look to simplify if possible we do have a greatest common factor of five between 245 and 990 so let's divide the numerator by five and the denominator by 5 245 ID 5 gives us 49 and then 990 / 5 gives us 198 the only common factor between 49 and 198 is 1 so this is now in simplest form so going back up to the original decimal here 0.2 and then 47 repeating equals 4998 as a fraction 49 over 198 and remember we can always double check this by dividing the numerator by the denominator either with a calculator or by hand either way you should get the original repeating decimal once you divide let's move on to number two where we have 9.1 and then six repeating so here we have a whole number nine but let's forget about that whole number for now we will worry about that at the end let's focus on the repeating decimal so x equals again don't worry about the nine we'll worry about that at the end so we'll put 0.1 and then the six repeats now we need to create two equations one with the repeating digit of six just to the left of the decimal and then one with the repeating digit of six just to the right of the decimal so let's start by getting that repeating digit of six just to the left so we need to shift the digits two places to the left and we can think through this by moving the decimal twice to the right so we're multiplying by 100 so multiply both sides of the equation by 100 that gives us 100x equal 16.6 repeating now we need an equation with the repeating six just to the right of the decimal so let's move the digits one place to the left and we can think through that by moving the decimal once to the right so we're multiplying by 10 multiply both sides of the equation by 10 that gives us 10 x = 1 six repeating so we have the repeating digit just to the left of the decimal and the repeating digit just to the right of the decimal and now we are able to subtract and get rid of those repeating digits 100x - 10 x gives us 90x equals and then on the right side the repeating sixes if we subtract those they cancel out they equals zero so we end up with 16 - 1 which gives us 15 so now we have the equation 90x = 15 so let's solve for x and we can simplify if possible in order to isolate that variable of X we need to divide both sides by 90 so on the left side of the equation x is now isol ol ated and then on the right side of the equation we have 15 90th or 15 over 90 and that fraction equals 0.1 and then 6 repeating there is a greatest common factor of 15 between 15 and 90 so we can simplify here let's divide the numerator and denominator by 15 15 / 15 gives us 1 and then 90 / 15 gives us six the only common factor between 1 and six is one so this is now in simplest form 1 16 so going back up to the original number don't forget this whole number of nine we have 9 and 6 so again don't forget the whole number there 9.1 and then 6 repeating equals 9 and 1 16 so there you have it there's how to convert repeating decimals to fractions I hope that helped thanks so much for watching until next time peace |
190250 | https://brainly.com/question/12126792 | [FREE] In a trapezoid with bases of lengths a and b , a line parallel to the bases is drawn through the - brainly.com
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In a trapezoid with bases of lengths a and b, a line parallel to the bases is drawn through the intersection point of the diagonals. Find the length of the segment that is cut from that line by the legs of the trapezoid.
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a+b 2 ab.
Explanation
Consider the trapezoid ABCD. In this trapezoid BC=a and AD=b.
Since triangles BOC and AOD are somilar, then
OC A O=OB D O=BC A D=a b.
Triangles OAE and CAB are similar, then
BC EO=\dfar A O A C=a+b b.
This means that
EO=b+a ba.
Similarly, from similar triangles FDO and CDB:
OF=b+a ba.
Thus,
EF=a+b 2 ab.
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The length of the segment cut from the line drawn through the intersection of the diagonals in a trapezoid with bases a and b is given by the formula EF=a+b 2 ab. This is derived from the properties of similar triangles in the trapezoid. Thus, this length depends on the lengths of the trapezoid's bases.
Explanation
To find the length of the segment cut by the legs of the trapezoid from a line parallel to the bases that passes through the intersection point of the diagonals, we can apply some properties of trapezoids and similar triangles. Let's denote the trapezoid as ABCD, where AB is one base with length a, and CD is the other base with length b. We will follow these steps:
Intersecting Diagonals: In trapezoid ABCD, the diagonals AC and BD intersect at point O. A line parallel to the bases AB and CD, through point O, will divide the trapezoid into two smaller trapezoids (let's call the top one AOD and the bottom one BOC).
Use of Similar Triangles: The triangles AOB and COD formed by the intersection of the diagonals and the segments of the bases are similar due to the angle-angle similarity criterion. This means that we can set up a ratio based on the lengths of these bases.
Finding the Length of the Segment: The length of segment EF, where E and F are the points where the line drawn through O intersects the legs AD and BC respectively, can be computed using the following relation, which arises from the properties of similar triangles. We have:
EF=a+b 2 ab
This formula shows us that the length of the segment is influenced by the lengths of the bases of the trapezoid. To derive this, note that similar triangles’ properties give us proportional segments, leading us to the conclusion above.
Examples & Evidence
For instance, if you have a trapezoid with bases of lengths 4 units and 6 units, you can find the length of the segment EF by substituting the values into the formula: EF=4+6 2×4×6=10 48=4.8 units. This example illustrates how to apply the formula in practice.
The relationship derived is consistent with geometric properties of trapezoids and has been confirmed through various mathematical proofs involving the properties of similar triangles and trapezoidal geometry.
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Final answer:
In a trapezoid with bases of lengths 'a' and 'b', the length of the segment that is cut from a line parallel to the bases and passing through the intersection point of the diagonals can be found using similar triangles.
Explanation:
In a trapezoid with bases of lengths a and b, a line parallel to the bases is drawn through the intersection point of the diagonals. The length of the segment that is cut from that line by the legs of the trapezoid can be found using similar triangles.
Let's call the length of the segment that is cut from the line by the legs of the trapezoid as x. Using similar triangles, we can set up the following proportion:
x/a = (a+b)/b
Cross-multiplying, we get:
x = (a(a+b))/b
Therefore, the length of the segment cut by the legs of the trapezoid is (a(a+b))/b.
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190251 | https://www.nature.com/articles/s41467-025-58558-z.pdf | Article Bromide-mediated membraneless electrosynthesis of ethylene carbonate from CO2 and ethylene Menglu Cai 1,2,3, Siyun Dai 1,3, Jun Xuan2,3 & Yiming Mo 1,3 Cyclic carbonates, such as ethylene carbonate, are crucial in various applica-tions, including lithium-ion batteries and polymers. Traditional production routes for ethylene carbonate rely on high-temperature thermocatalytic pro-cesses that use fossil-fuel-derived epoxides and carbon dioxide (CO2). Herein, we report a bromide-mediated membraneless electrosynthesis strategy for direction conversion of ethylene and CO2 into ethylene carbonate. This method leverages electrolyte engineering to modulate the kinetics of solution chemistry to proceed at rates that match the high-current bromide electro-oxidation, and cathode protection with chromium hydroxide film to suppress the parasitic bromine reduction reaction. These enable the system to operate at 10–250 mA/cm2 current density with 47–78% Faraday efficiency towards ethylene carbonate. The system’s practicality is underscored by achieving an ethylene carbonate product concentration of 0.86 M and maintaining stability for over 500 hours. Furthermore, we demonstrate the integration of this process with CO2 electroreduction to ethylene, enabling a cascade ethylene carbonate electrosynthesis using only CO2 and water as feedstocks. A com-prehensive techno-economic analysis confirms the strong economic potential of this method for future applications.
As the world is moving towards a sustainable future powered by the renewable electricity, the surge in the use of lithium-ion batteries as the electricity storage media requires green and cost-effective routes to manufacture their required raw materials1,2.
Cyclic carbonates, such as ethylene carbonate (EC) and propylene carbonate (PC), are the most widely used electrolytes in lithium-ion batteries3,4. The global production of EC, valued at $0.94 billion in 2023, is anticipated to nearly double to $1.49 billion by 20285. However, existing industrial routes for EC production involve high-temperature (100–200 °C) and high-pressure (3.5–10 MPa) thermocatalytic reaction between epoxides and carbon dioxide (CO2) (Fig. 1a)6,7. The corresponding feedstock ethylene oxide is derived from fossil fuels through the energy-intensive aerobic silver-catalyzed epoxidation process6,8. In light of the surging market demand for cyclic carbonates along with the desired greener production, the development of sustainable and mild alternative synthesis methods is imperative.
Electrosynthesis driven by renewable electricity offers a sustain-able route to access value-added chemicals9,10. Significant advances have been made to achieve electrocatalytic upgradation of gaseous feedstocks, such as the electrochemical conversion of CO2 to C2+ hydrocarbons11,12 and the ammonia synthesis via nitrogen electroreduction13. The selective electrochemical transformation of olefins to the corresponding epoxides, the intermediates for cyclic carbonates, has recently been advanced using various oxidative or reductive mechanisms, such as homogeneous halide-mediated8,14 or heterogeneous Ag-catalyzed oxidative pathway using water as the oxygen source15 and reductive pathway via gaseous dioxygen reduction-derived H2O2 to prepare epoxides16. However, the epoxides are still one step away from cyclic carbonates.
Received: 23 March 2024 Accepted: 24 March 2025 Check for updates 1College of Chemical and Biological Engineering, Zhejiang University, Hangzhou, China. 2Department of Chemistry, Zhejiang University, Hangzhou, China.
3ZJU-Hangzhou Global Scientific and Technological Innovation Center, Zhejiang University, Hangzhou, Zhejiang, China.
e-mail: yimingmo@zju.edu.cn Nature Communications| (2025) 16:3285 1 1234567890():,; 1234567890():,; The direct electrosynthesis of cyclic carbonates still represents a straightforward approach to meet the requirement for sustainable synthesis. Dimethyl carbonate was electrosynthesized using CO2 and methanol via a redox-neutral palladium-catalyzed pathway, achieving 60% Faraday efficiency (FE) at 12 mA/cm2 current density17. Using widely available diols and CO2 as the starting material for cyclic car-bonate synthesis could proceed with electrogenerated base18. How-ever, a stoichiometric amount of expensive tetraethylammonium iodide and iodomethane (Supplementary Table 1) were required as the supporting electrolyte and esterification reagent, respectively, and yields for various cyclic carbonates were generally lower than 60%.
Similarly, with the help of electrogenerated base from cathodic hydrogen evolution, cyclic carbonates can be synthesized by reacting CO2 with the corresponding epoxides at ambient conditions19,20.
Nonetheless, to balance the charges used for cathodic base generation necessitates the anodic metal dissolution as the sacrificial electron donor. Alternatively, an oxidative pathway using iodide as the med-iator can successfully convert aromatic and aliphatic olefins into the iodohydrin intermediates, which further react with CO2 to obtain corresponding cyclic carbonates21. Its compatibility with gaseous ole-fins, such as ethylene and propylene, is yet to be validated (Entry 4 in Supplementary Table 2). Nam et al. reported a succinimide- and bromide-mediated cascade strategy for EC preparation, which was electrochemically driven for the initial bromine and base generation22.
This strategy involved four separate reaction steps for the final EC synthesis, highlighting the need for a streamlined one-pot EC elec-trosynthesis method.
To pursue a practical cyclic carbonate electrosynthesis route under industrially relevant current densities (Supplementary Note 6), we here developed an electrochemical bromide-mediated membra-neless synthesis strategy to convert readily available ethylene and CO2 to EC under ambient conditions (Fig. 1b). Mechanistically, the anodi-cally generated bromine/hypobromite reacts with the dissolved ethy-lene via halohydrin pathway to obtain 2-bromoethanol (2-BrEtOH) (Eq.
(3)), which further undergoes intermolecular cyclization with CO2 to form EC and regenerate bromide (Eq. (4)). The overall electrochemical conversion only requires ethylene, CO2, and water as feedstocks to co-generate value-added EC and hydrogen (Eq. (6)).
Anode: 2Br 2e ! Br2 E° = 1:065 V vs: SHE ð1Þ Br2 + H2O"HOBr + HBr ð2Þ C2H4 + HOBr ! HOCH2CH2Br ð3Þ HOCH2CH2Br + CO2 + OH ! C3H4O3 + Br + H2O ð4Þ Cathode: 2H2O + 2e ! H2 + 2OH E° = 0:826 V vs:SHE ð5Þ Overall: C2H4 + CO2 + H2O ! C3H4O3 + H2 ΔG° = 287:2kJ=mol ð6Þ The key to this efficient one-pot reaction design is modulating the reaction kinetics through electrolyte engineering, such that solution chemistry steps (Eqs. (2)–(4)) can proceed at high rates to match the high-current electrode reaction (Eqs. (1) and (5)). This direct EC elec-trosynthesis method is capable of operating at 10–250 mA/cm2 current density with 47–78% FE towards EC. Furthermore, the maximum con-centration of the produced EC can reach 0.86 M (i.e., 7.6 wt%), thus significantly reducing the downstream separation cost. In addition, the electrosynthesis process can run stably for ~500 h without noticeable performance degradation. This bromide-mediated direct electro-synthesis approach can extend the starting olefin beyond ethylene to other olefins, including propylene and allylbenzene, to synthesize corresponding cyclic carbonates with high efficiency. Pursuing a carbon-neutral chemical production, we further developed a cascade system by integrating electrochemical CO2-to-ethylene conversion and ethylene-to-EC electrosynthesis, thus achieving to use CO2 and water as the only feedstocks for EC production. The techno-economic analysis of the developed electrochemical CO2-to-EC process con-firmed its potential for future practical application.
Results Reaction design for electrosynthesis of ethylene carbonate The overall reaction of using ethylene and CO2 as the feedstocks to synthesize EC is an oxidative process that requires anode as the elec-tron acceptor. Existing heterogenous activation strategies to achieve partial oxidation of alkene at anode are limited by their current density and selectivity23–25. In addition, the ethylene is potentially subject to overoxidation to aldehydes or even CO2 as the complete oxidation products26. The state-of-the-art electrochemical heterogeneous olefin epoxidation achieved propylene conversion to propylene oxide with 66% FE at 50 mA/cm2 current density on palladium-platinum alloy electrocatalyst27. However, FE for ethylene oxide electrosynthesis from ethylene reduced to 25% in this system due to the low electrophilicity of ethylene.
Here, we took an alternative approach that implemented a redox mediator to facilitate the selective and high-rate electron transfer between ethylene and anode, preventing the overoxidation of direct electrooxidation approach8. Conventionally, cyclic carbonate can be b A n o d e Br-Br2 -e Step 1 Step 2 Step 3 Br HO C2H4 H2O OH-CO2 Br-Br-Electrode reaction Solution chemistry reaction + + HER O O O a O O O Fig. 1 | Routes for ethylene carbonate production. a Existing industrial thermocatalytic route for ethylene carbonate (EC) production. b The proposed bromide-mediated membraneless direct EC electrosynthesis strategy using ethylene and CO2 as the feedstocks.
Article Nature Communications| (2025) 16:3285 2 synthesized from halohydrin and CO2 through intermolecular cyclization28,29, and the corresponding halohydrin can be prepared by the reaction between olefin and halogen or hypohalite30. However, this chemical route is seldom implemented in industrial production due to the consumption of stoichiometric amount of hazardous halogens.
Inspired by the recent success of halide-mediated epoxide electrosynthesis8, we postulated that direct electrosynthesis of EC can be also accomplished using a halide-mediated halohydrin pathway (Fig. 1b). However, the major challenge associated with this triple-step reaction in one-pot, which includes anodic halide oxidation, halohy-drin formation, and intermolecular cyclization with CO2, is the orchestration of their reaction kinetics in order to achieve a high productivity of EC (i.e., large current density with good FE).
Electrolyte engineering for one-pot ethylene carbonate electrosynthesis We sought to modulate the triple-step reaction kinetics via electrolyte engineering, beginning with a detailed investigation and optimization of each individual step. First, with a brief survey of suitable halide mediator for this integrated EC electrosynthesis (Supplementary Fig. 1), we identified bromide (Br-) as the most effective mediator compared to chloride and iodide, achieving a 68% FE towards EC.
Compared to Cl-mediated system, hypobromite could react with ethylene to form 2-BrEtOH at a much higher rate than hypochlorite, and bromide was a better-leaving group during cycloaddition (Eq.
(4))31–33. For I-mediated system, no EC was generated, which was attributed to the sluggish disproportionation reaction of I2 into hypoiodite34.
To identify the optimal anode material for bromide oxidation reaction (BOR), various electrode materials, including IrO2 based dimensionally stable anodes (IrO2-DSA), platinum (Pt), carbon paper, and graphite, were investigated using linear scanning voltammetry (LSV) (Supplementary Fig. 2). IrO2-DSA electrode has the lowest onset potential and largest current density for BOR, and the competing oxygen evolution reaction (OER) is unfavored at the oxidation potential of bromide (Supplementary Fig. 3). In bulk electrolysis, IrO2-DSA also outperformed other electrode materials in terms of BOR FE (86%) and generated bromine concentration (21 mM) (Supplementary Fig. 4), and showed the highest Faraday efficiency for EC electro-synthesis (68% FE, Supplementary Fig. 5). Thus, IrO2-DSA was selected the optimal anode material for the following study.
We next focused on engineering the kinetics of solution chemis-tries to match with electrode reaction. The proton or hydroxide ions are involved in the formation of 2-BrEtOH (Eqs. (2) and (3)) and EC (Eq.
(4)), and thus, pH of the electrolyte would significantly impact the reaction kinetics (Fig. 2a). The ethylene to 2-BrEtOH formation rate decreased with increasing pH30,32, and pH above 10 would totally inhibit the reaction possibly due to the lack of active HOBr species at high pH35. On the other hand, the nucleophilic cycloaddition of CO2 and 2-BrEtOH to synthesize EC favors a basic electrolyte environment to the promote formation of the required alcoholate intermediate for the nucleophilic pathway28. Thus, a weakly alkaline environment (pH~8) would be ideal for both reactions (Eqs. (2)–(4)).
On the other hand, the nucleophilic CO2 cycloaddition to 2-BrEtOH (Eq. 4) was sluggish in aqueous electrolyte (Fig. 2b), thus limiting its successful coupling with 2-BrEtOH formation for one-pot a b c d Reaction time (h) After electrolysis Formation rate (mmol h-1) Fig. 2 | Investigation of the reaction kinetics through species balance and electrolyte engineering. Effects of (a) pH, (b) DMF content and (c) cations in solution on the formation of 2-BrEtOH and EC. For 2-BrEtOH formation, 0.5 M KBr, 0.2 M Cs2CO3, and 3 mM K2Cr2O7 in 30 mL solvent with 30 sccm C2H4 supply at 20 mA/cm2 current density for 30 min. IrO2-DSA and nickel foam (2 cm2) was used as anode and cathode, respectively. For EC formation, adding 0.12 mmol BrCH2CH2OH into 0.5 M KBr and 0.2 M Cs2CO3 in 3 mL CO2-saturated DMF/water electrolyte, and the solution was shaken at 25 °C for 30 min in CO2 atmosphere. For cation study, KBr and Cs2CO3 were replaced with corresponding cation bromides and carbonates. d Species concentration during electrolysis and when fully con-verted after stopping electrolysis. Reaction condition: 0.5 M KBr, 0.2 M Cs2CO3, and 3 mM K2Cr2O7 in 120 mL DMF/water electrolyte (30:70 vol%) with 30 sccm ethylene and 80 sccm CO2 supplying,and the reaction was performed under 20 mA/cm2 with 7488C charges using IrO2-DSA anode (8 cm2) and nickel foam cathode (8 cm2).
Article Nature Communications| (2025) 16:3285 3 EC electrosynthesis. To get around this kinetic barrier, aprotic polar organic solvent, such as dimethylformamide (DMF), was considered to expedite the CO2 cycloaddition reaction36. Compared to aqueous electrolyte, 30 vol% DMF improved cycloaddition reaction rate by 4.5 times. A higher DMF content (>30 vol%) would limit the solubility of inorganic electrolytes (Supplementary Fig. 7a), which was unamenable for EC electrosynthesis in the electrochemical flow cells. However, DMF content in the electrolyte had a negative effect on 2-BrEtOH formation rate due to the inhibition of bromine disproportionation to hypobromite (Fig. 2b). Thus, the DMF/H2O (30:70) mixture electrolyte solvent proved to be optimal after trading-off between reaction kinetics and electrolyte solubilities.
The alkali metal cations, including lithium (Li+), sodium (Na+), potassium (K+), and cesium (Cs+) were further investigated to fine tune the electrolyte environment (Fig. 2c and Supplementary Table 3). The type of cations had little effect on 2-BrEtOH formation rate, but sig-nificantly impacted EC formation rate (Fig. 2c). Due to the different nucleophilic strength during cycloaddition29,37, Cs+ showed the highest FE for one-pot EC electrosynthesis (Supplementary Table 3). However, considering the high price of CsBr (Supplementary Table 4), CsBr was replaced with the cheaper KBr, and this mixed cation electrolyte (KBr and Cs2CO3) gave similarly good performance as that of the Cs-only electrolyte (CsBr and Cs2CO3) (Supplementary Table 3).
The optimal electrolyte composition was 0.5 M KBr, 0.2 M Cs2CO3, and 3 mM K2Cr2O7 in 30:70 vol% DMF/water. With this opti-mized electrolyte, the concentrations of 2-BrEtOH and EC (Fig. 2d) gradually increased during the early stage of the electrolysis, and after 9-hour electrolysis, 2-BrEtOH reached a plateau concentration while EC continued to accumulate, indicating a characteristic series reaction pattern. Continuing supplying CO2 after stopping the electrolysis converted all remaining 2-BrEtOH to EC (Fig. 2d, Supplementary Fig. 10). In addition, an intermittent electrolysis experiment (Supple-mentary Fig. 11) revealed that the formation rates of 2-BrEtOH and EC were of similar magnitudes, confirming the desired outcome of elec-trolyte engineering on solution chemistry kinetics.
Reaction mechanism investigation Next, we focused on elucidating the mechanism for conversion of 2-BrEtOH to EC (Eq. 4) since the halohydrin formation reaction mechanism has been extensively studied in existing literatures8,32,38.
First, we sought to understand the reactivity of HCO3 −, CO3 2-, and dissolved CO2 with 2-BrEtOH. Without the dissolved CO2, HCO3 − showed higher reactivity compared to CO3 2- (Entries 1-2, Supplemen-tary Table 5). In addition, supplying the electrolyte with CO2 acceler-ated the formation of EC (Entries 3-4, Supplementary Table 5). Since the cathodic HER would generate OH−to react with CO2 to formHCO3 −, the concentrations of HCO3 −and CO2 should remain relatively unchanged during electrolysis. The pH of solution remained stable within 8.05-8.27 over the 13-hour electrolysis period (Supplementary Fig. 6). Thus, there are two plausible pathways occurring concurrently for the conversion of 2-BrEtOH to EC (Fig. 3a). (1) Reaction with HCO3 −: the oxide anion of HCO3 −undergoes nucleophilic addition to 2-BrEtOH with bromide as the leaving group, forming the corresponding bicar-bonate intermediate (I). The intermediate I then converts to EC via the intramolecular cyclization39,40. (2) Reaction with dissolved CO2: 2-BrEtOH first gets deprotonated in the alkaline electrolyte to form corresponding alcoholate (II), which attacks the dissolved CO2 mole-cule to form a carbonate intermediate (III). This intermediate goes through the nucleophilic cycloaddition to obtain EC28,37.
To further verify this reaction mechanism, we performed an 18O isotope experiment using H2 18O as the 18O source in the chemical conversion of 2-BrEtOH to EC (Supplementary Table 6). CsHC16O3 was first equilibrated in H2 18O/DMF to ensure complete oxygen atom exchange between CsHC16O3 and H2 18O41. Without supplying CO2, GC-MS analysis of the resulting EC revealed three mass spectra peaks at88, 90, and 92 m/z, corresponding to EC containing zero, single, and two 18O atoms (Fig. 3b and Entry 1 in Supplementary Table 6). This indi-cated that EC has one oxygen atom from 2-BrEtOH, and the other two are from HCO3 −, confirming the proposed HCO3 −reaction pathway. If supplying the electrolyte with C16O2 (Entry 2 in Supplementary Table 6), the abundance of 16O in the formed EC increased, suggesting participation of CO2 in the EC formation, consistent with CO2 reaction pathway.
Suppressing cathodic parasitic bromine reduction in membra-neless electrolysis The divided cell with a separating membrane provides an ion-selective transport or diffusion barrier for species in the electrolyte to avoid parasitic reaction of intermediates or products at the counter electrode42. But the associated cost, stability, and ohmic resistance issues of membranes diminish the advantages of electrosynthesis as a sustainable alternative for practical applications43. In the designed bromide-mediated electrosynthesis of EC, the anodically generated bromine is subject to the bromine reduction reaction (BRR) at the cathode, thus decreasing FE of the overall process. This parasitic BRR was confirmed by that FE for EC electrosynthesis in a divided with an anion-exchange membrane (82%) (Supplementary Fig. 12) was sig-nificantly higher than that in the undivided cell (55%) (Fig. 4a). Fur-thermore, reductive LSV scan of bromine-containing electrolyte showed a characteristic peak for BRR at −0.74 V vs. Ag/AgCl, which was Fig. 3 | Investigation of the reaction mechanism. a Proposed reaction mechanism of EC in membraneless electrochemical cell. b The 18O isotope experiments results.
Reaction condition: 0.5 M KBr and 0.2 M Cs2CO3 in 3 mL CO2-saturated DMF/ labeled H2 18O water electrolyte (30:70 vol%), and then 0.5 mmol of 2-BrEtOH was added and CO2 was bubbled during the reaction.
Article Nature Communications| (2025) 16:3285 4 thermodynamically more favorable than hydrogen evolution reaction (HER, −1.06 V vs. Ag/AgCl) (Fig. 4b).
Using a shielding layer on the electrode surface to impose mass transfer limitation on thermodynamically more favorable reaction is an emerging strategy to overcome parasitic reactions44,45. The inor-ganic passivating agents (e.g., K2Cr2O7, K2MoO4, and K2FeO4) are implemented in the chlor-alkali industry to improve FE towards sodium hypochlorite in the undivided cell46,47. Thus, we decided to investigate the effect of forming a similar shielding film on the cathode surface for suppressing BRR. A minor amount (3 mM) of K2Cr2O7, K2MoO4, or K2FeO4 was added to the electrolyte for in-situ reductive formation of non-soluble metal hydroxide film, which could poten-tially serve as a selective diffusion barrier for bromine or hypobromite (Fig. 4c). LSV study revealed that the addition of K2Cr2O7 and K2MoO4 could suppress the characteristic reduction peak of BRR (Fig. 4b), while K2FeO4 showed negligible impact on the behavior of BRR and HER. This BRR suppression effect was exemplified by the significantly improved FE for EC electrosynthesis in the membraneless electro-chemical cell (Fig. 4a). The FE of EC was improved to 77% and 67% for electrolyte with K2Cr2O7 and K2MoO4, respectively, close to that in divided flow cell. Consistent with the LSV analysis, K2FeO4 brought no benefits in suppressing BRR, but instead, slightly reduced FE to 50% possibly due to the instability of deposited Fe(OH)X in the bromine-containing electrolyte (Supplementary Fig. 14).
To probe the shielding mechanism, scanning electron microscopy (SEM) was used to study the morphology difference of the nickel plate cathodes with and without K2Cr2O7 additive. When conducting EC electrosynthesis in a membraneless cell without any passivating addi-tives, the originally smooth nickel cathode surface got severely cor-roded after the reaction (Fig. 4d and Supplementary Fig. 14b), which was attributed to the bromine corrosion of nickel cathode. Adding 3 mM K2Cr2O7 resulted in a dense deposition of Cr(OH)x film, which had a 2.8~3.3 μm thickness (Fig. 4d). Energy-dispersive X-ray spectro-scopy (EDS) confirmed the uniform dispersion of Cr on the nickel plate (Fig. 4d). X-ray photoelectron spectroscopy (XPS) was used to detect the Cr valence state (Fig. 4e). The peak at 587.2 eV and 577.5 eV cor-responded to the Cr 2p1/2 and Cr 2p3/2 of Cr(III) species and were attributed to Cr(OH)3. A minor amount of Cr(VI) and Cr(0) was also present, but Cr(III) species were dominant on the cathode surface, indicating Cr(OH)3 hydroxide film was the main coating component after the electrolysis48. This porous Cr(OH)3 shielding film served as a selective diffusion barrier to suppress the mass transport of negatively charged hypobromite anion via adverse potential gradient46,49,50, while water molecules are still able to access the cathode surface (Fig. 4c).
Towards practical electrosynthesis of the ethylene carbonate With the optimal electrode and electrolyte composition identified above for membraneless EC electrosynthesis, we next sought to explore its performance under practical current densities. The FE for EC reached 52% in a batch reactor at 20 mA/cm2 current density (Supplementary Fig. 18). To boost the mass transport between the bulk electrolyte and electrode surface, a circulating electrochemical flow reactor was employed (Supplementary Fig. 19). The improved mass transfer allowed the current density to increase up to 250 mA/cm2 5 um Without K2Cr2O7 With K2Cr2O7 5 um 5 um Cr 5 um Ni 20 um Ni plate 3.2 um 2.9um Cr(OH)3 OBr-H2O Cr(OH)3 passivation layer Cathode × H2 + OH-e a b c e Cr 2p1/2 Cr 2p3/2 d Fig. 4 | Suppression of parasitic bromine reduction reaction (BRR) on cathode in a membraneless electrochemical cell. a FE for EC electrosynthesis in divided and undivided cells. Reaction condition: 0.5 M KBr and 0.2 M Cs2CO3 in DMF-water (30:70 vol%) solution at 20 mA/cm2 current density with IrO2-DSA anode (2 cm2) and nickel foam cathode (2 cm2). For divided reaction, the reaction was performed in a flow-cell consisting of the IrO2-DSA anode (2.25 cm2), anion exchange mem-brane and nickel foam cathode (2.25 cm2) at 20 mA/cm2 current density. b The LSVs for cathodic BRR in presence of different additives (3 mM if added) performed in the CO2-saturated DMF-water (30:70 vol%) solution containing 0.2 M Cs2CO3 and 0.01 M Br2 (if added). Working electrode: IrO2-DSA. Counter electrode: Nickel foam.
Reference electrode: Ag/AgCl. c Schematics of BRR suppression mechanism.
d Surface morphology and the corresponding EDS of the used nickel cathodes with and without K2Cr2O7. e XPS spectra of the nickel plate cathode after being used for EC electrosynthesis in the electrolyte with K2Cr2O7.
Article Nature Communications| (2025) 16:3285 5 while still keeping a 47% FE (Fig. 5a). The EC productivity thus increased substantially, resulting in 8.6 times improvement to 4.4 mmol/h at 250 mA/cm2 over the 0.51 mmol/h productivity at 20 mA/cm2.
The product concentration in the reaction crude is one of the determining factors to the downstream separation cost. Our method could steadily increase the concentration of generated EC to 0.86 M (equivalent to 7.6 wt%) by extending the electrolysis time (Fig. 5b). This achieved concentration is significantly higher than that of the reported electrochemical EC synthesis method19,21,51–53. The stability of this electrochemical system was further investigated. Batches of fresh electrolytes were sequentially electrolyzed through the same flow reactor over the 550-hour stability test, during which the cell voltage remained stable at ~3.0 V and an accumulative 11.2 g EC was synthe-sized (Fig. 5c and Supplementary Figs. 20–22). While keeping the current density unchanged, we further scaled-up electrode area from 2 cm2 to 8 cm2, FE for EC in the large-scale flow cell was 74% similar to the small-scale, demonstrating the scaling potential of this strategy (Supplementary Figs. 23-24). Overall, our EC electrosynthesis strategy outperformed existing reported EC electrosynthesis methods in terms of system stability, EC concentration, current density, and productivity (Fig. 5d, Supplementary Fig. 25 and Supplementary Table 2)19–21,51–54.
Scope for electrosynthesis of cyclic carbonates This strategy can also be adopted to produce other valued cyclic carbonates in addition to EC. Propylene carbonate (PC), an important electrolyte for lithium-ion batteries, could be prepared using pro-pylene as the starting substrate. Performing the electrolysis at 20 mA/cm2, a 0.43 mmol/h productivity and 58% FE was achieved.
Increasing the current density to 100 mA/cm2 resulted in an improved PC productivity (1.7 mmol/h) with 45% FE (Fig. 5e, f). Going beyond gaseous olefins, the olefin substate with larger molecular weight, such as allyl benzene, can also undergo directly conversion to its corresponding cyclic carbonate with 82% yield (Fig. 5e). The demonstrated scope shows the potential of this bromide-mediated direct electrosynthesis method to be applied to other cyclic carbonates.
Techno-economic analysis and CO2 emission of EC electrosynthesis To evaluate the potential of this developed EC electrosynthesis method for future practical application, we conducted a techno-economic analysis (TEA) to assess the process economics (see detailed TEA calculation in Supplementary Note 1)8,55. Using the performance specifications of the developed EC electrosynthesis demonstrated above, the EC electrosynthesis from ethylene and CO2 is able to pro-duce EC at an estimated cost of US $1379.9/ton under the existing electricity cost of US $0.1/kWh8. Compared with the prices in the United States in March 202456,57, the production cost of EC electrolysis is lower than the $2305/ton market price of EC (Fig. 6a and Supple-mentary Note 1). The value increase of the developed EC electro-synthesis demonstrates the strong economic driving force to implement this methodology in practice. Sensitivity analysis reveals that the electricity cost is major influencing factor for the EC electro-synthesis cost (Supplementary Fig. 30 and Supplementary Table 8).
Compared with the existing EC production routes based on fossil-fuel-derived EO feedstock6,14, our electrosynthesis process could rea-lize a lower CO2 emission under the existing average grid CO2 emission b c d e f a CO2 O O O 20 mA cm-2 5 mA cm-2 CO2 O O O Fig. 5 | The performance of bromide-mediated cyclic carbonate electrosynth-esis in a membraneless flow cell. a The productivity and FE of EC electrosynthesis under different current densities. b The concentration profile of EC in extended electrolysis at 30 mA/cm2. c Stability study for EC electrosynthesis. Multiple bat-ches of fresh electrolyte were sequentially electrolyzed at 20 mA/cm2 for 13 h and continuing supply CO2 for 11 h to ensure full conversion of the remaining active bromine and bromoethanol. Reaction conditions for (a–c): electrolysis performed in a flow cell (IrO2-DSA anode and nickel foam cathode) using 0.5 M KBr and 0.2 M Cs2CO3 in DMF-water (30:70 vol%) solution with 30 sccm ethylene and 80 sccm CO2 supply. d Comparison of current density and productivity for EC electrosynthesis with reported literatures. e Mass spectra of propylene carbonate and 4-(phe-nylmethyl)−1,3-dioxolan-2-one. f The productivity and FE values of PC at different current densities. Reaction conditions for (e, f): For PC, the electrolysis condition was same as the EC production, except that the substrate changes from ethylene to propylene. For allybenzene, 1 mmol of substrate was dissolved in 20 mL DMF-water (3:7 volume ratio) containing 0.2 M Cs2CO3, 0.15 M KBr, and 1 mM K2Cr2O7. The flowrate for CO2 supply was 40 sccm.
Article Nature Communications| (2025) 16:3285 6 intensity (0.295 kg CO2/kWh) based on the European Union’s data (Fig. 6b and Supplementary Note 2)10. Furthermore, a negative CO2 emission (consuming 0.45 ton of CO2 per ton of EC produced) can be achieved when using renewable electricity as the energy source (Supplementary Note 3-4)58. This underscored the potential of the developed method in future shift towards carbon-neutral chemical production using renewable energy.
Cascade ethylene carbonate production from CO2 and water The current industrial route for EC uses the fossil fuel derived ethylene as the starting feedstock. To further reduce the carbon footprint of the overall EC production process, we sought to develop an integrated electrochemical system, relying on CO2 and H2O as the only feedstocks to synthesize EC. Our integrated system involves a two-step cascade electrochemical conversion. In the initial stage, within an electro-chemical flow reactor with gas diffusion electrode (GDE), CO2 was reduced to ethylene, which was then channeled into the subsequent EC electrosynthesis stage. This system provides a route to use renewable electricity for converting CO2 and H2O into a valuable commodity chemical.
In detail, for CO2 electroreduction, sheet-shaped copper fluoride hydroxide (Cu(OH)F) nanoparticles were used as the electrocatalyst for selective conversion of CO2 into ethylene (Supplementary Figs. 26–28)59. When operating at 100 mA/cm2, a 63 ± 4% C2H4 FE was achieved. To obtain sufficient supply of ethylene for the second EC electrosynthesis step, an electrolyzer with 5 cm2 electrode area was constructed to convert 5 sccm CO2 inlet steam into an outlet gas mixture containing 0.016 mmol/min ethylene. This outlet stream was directly sparged, without intermediate purification, into a batch cell for the subsequent EC electrosynthesis (Fig. 6c). Operating the EC electrosynthesis at a 6 mA/cm2 current density, 37% FE was achieved.
The lower FE compared to previous single-step EC electrosynthesis was primarily due to the low ethylene flowrate generated by 5 cm2 electrolyzer for CO2 electroreduction, which was confirmed by directly sparging the electrolyte with 0.09 mmol/min ethylene mixed in 5 sccm CO2 leading to an increased FE of 46% (Supplementary Fig. 29). If perform CO2 to ethylene conversion at large scale, distillation60,61 and other emerging methods like physisorption60,62,63 and membrane separation64 can beused for ethylene separation, which is beneficial for improving the downstream EC electrosynthesis FE and current density.
For the integrated cascade EC synthesis using CO2 and water as the feedstock, the production cost increased to US $2326/ton due to the high cost of C2H4 electrosynthesis using existing technology and electricity cost (US $0.1/kWh), which was con-sistent with the reported analysis (Supplementary Note 7)10.
Considering that the electricity cost continues to decrease along with future development of the CO2-to-C2H4 electrosynthesis process, the production cost of the integrated cascade CO2-EC system is projected to drop to US $1275/ton under US $0.04/kWh electricity cost (Supplementary Fig.
31 and Supplementary Note 8)10. When using renewable electricity to power the cascade EC electrosynthesis, the CO2 emission for electricity generation is 0.01 t/MWh58, based on this parameter the CO2 emission was only −1.39 tonCO2/tonEC, lower than the −0.46 tonCO2/tonEC CO2 emission when using fossil-fuel-derived ethylene in the single-step EC electrosynthesis method above (Supplementary Fig. 32 and Supplementary Note 4, 10).
Discussion In summary, we have demonstrated an electrochemical bromide-mediated EC synthesis directly from ethylene and CO2 in a membra-neless flow cell. Using an IrO2-DSA as the anode and nickel foam as the a b c CO2-to-ethylene reactor Ethylene-to-EC reactor Renewable electricity Existing grid electricity Thermocatalytic process Market price Electrosynthesis process Fig. 6 | Techno-economic analysis and CO2 emission of EC electrosynthesis, and cascade electrochemical CO2-to-EC synthesis process. a Techno-economic ana-lysis of single-step ethylene-to-EC electrosynthesis. b The conparison of the CO2 emission for EC synthesis from ethylene using different production methods.
c Schematics of the two-stage cascade CO2-to-EC system, including electroreduc-tion of CO2 to ethylene and bromide-mediated EC electrosynthesis using CO2 and generated ethylene.
Article Nature Communications| (2025) 16:3285 7 cathode, 47–78% FE towards EC was achieved at 10–250 mA/cm2 cur-rent density with 0.27–4.4 mmol/h productivity. Prolonging the elec-trolysis time, the EC concentration could reach to 0.86 M.
Furthermore, this system has shown an excellent stability, maintaining stable for over 500 h with negligible performance degradation. This method could extend to other olefin substrates to synthesize corre-sponding valued cyclic carbonates. We also developed a cascade sys-tem for EC production, which coupled EC electrosynthesis with CO2 electroreduction to ethylene, achieving to use CO2 and water as the only feedstocks for EC synthesis. The corresponding techno-economic analysis and CO2 emission assessment illustrated its potential for a cost-effective and environmentally sustainable pathway to EC synthesis.
Methods Materials and chemicals The IrO2-DSA electrode (thickness 1 mm, 15 mm × 25 mm) used for anodic bromide oxidation was purchased from Shengbailong and the content of IrO2 is around 10–15 g/m2. The nickel foam (thickness 1 mm, 15 mm × 25 mm) electrode used as cathode for C2H4-EC production.
The conductive polypropylene (PP) plate (thickness 1 mm) was pur-chased from Nake. Carbon gas-diffusion electrode (YLS-30T) for CO2 electroreduction, and anion exchange membrane (FAB-PK-130) used in CO2 electroreduction were purchased from Suzhou Sinero Technol-ogy. The Ag/AgCl (φ = 4 mm) was purchased from Xianfengyiqi with 3 M KCl as the electrolyte. The Nafion D520 polymer binder solution (5 wt%) was purchased from SCI Materials Hub. Chemicals used, such as KBr (AR 99%), Cs2CO3 (AR 98%), K2Cr2O7(AR 99.8%), DMF (AR 99.5%), CsBr (AR 99%), CsHCO3(99.9%), Cu(NO3)2·2H2O (AR 99.0%) and NH4HF2(AR ≥98%), were purchased from Sinopharm Chemical Reagent and Macklin, and used without further purification. CO2 (99.99%) was purchased from Hangzhou Mingxinghuagong and C2H4 (99.9%) was purchased from Hangzhou Jingongwuzi.
Characterization and product analysis The morphologies of the electrode were investigated through SEM (Thermo Fisher Scientific, Scios2 Hivac) equipped with an Oxford Xplore 50 EDX detector. The X-ray diffraction (XRD) was performed on an X-ray diffractometer (Bruker, D2 PHASER). The surface chemical composition of prepared samples was obtained by X-ray photoelec-tron spectroscopy (XPS) on a Thermo Fisher Scientific K-Alpha spec-trometer using Al Kα radiation (1486.6 eV). C 1s peak (BE = 284.8 eV) was used as the reference to calibrate the XPS data.
The linear sweep voltammetry (LSV) was conducted on a three-electrode setup with Ag/AgCl (3M KCl) as the reference electrode at a scan rate of 20 mV/s (CHI1140C potentiostat). Unless otherwise spe-cified, the test is a static test at room temperature, and the test room maintains at about 25 °C all year round.
The product yield was determined by gas chromatography-mass spectrometry (GC-MS, Agilent 8890-5977B) using naphthalene as the internal standard or high-performance liquid chromatography (HPLC, Agilent 1260) with a refractive index detector (RID) using n-butyl alcohol as the internal standard. Before GC-MS analysis, EC was extracted three times using ether-dichloroethylene (1:5) from the reaction solvent to avoid injecting nonvolatile salts into GC-MS system.
1H NMR spectra were recorded on a Bruker 600 MHz NMR spectro-meter in DMSO-d6 with mesitylene as the internal standard.
Procedure for electrosynthesis of cyclic carbonates For bromide-mediated EC electrosynthesis experiment, we used a membraneless flow cell as the reactor (Supplementary Fig. 19), which was equipped with a porous IrO2-DSA anode (2 cm2 in size and 1 mm thick) and a nickel foam cathode (2 cm2 in size and 1 mm thick). The flow rate of CO2 and C2H4 was controlled by mass flow controllers (CS200-A, Beijing Sevenstar Flow), and was set to 80 sccm and 30 sccm, respectively. 0.2 M Cs2CO3, 0.5 M KBr and 3 mM K2Cr2O7 dissolved in 30 mL DMF-water blended solution (3:7 volume ratio) was used as the electrolyte for EC electrosynthesis. Before electrolysis, the solution was saturated by CO2, and then circulated through the flow cell with a 10 mL/min flowrate using a peristaltic pump (L100-1S−2, Longer). A constant current (10–250 mA/cm2) was applied for a designated reaction time at ambient temperature, and continuing supply CO2 after stopping the electrolysis to consume all remaining bromine and 2-BrEtOH to EC. The product samples were analyzed by GC-MS with naphthalene as the internal standard. The 2-BrEtOH was analyzed by HPLC with n-butyl alcohol as the internal standard using RID. The electrosynthesis setup and operating parameters of propy-lene carbonate were the same as that of EC electrosynthesis, except for the ethylene feedstock was replaced with propylene.
For the electrochemical conversion of liquid olefins, 1 mmol of substrate was dissolved in 20 mL DMF-water electrolyte (3:7 volume ratio) composed of 0.2 M Cs2CO3, 0.15 M KBr, and 1 mM K2Cr2O7. The flowrate for CO2 supply was 40 sccm, and electrolyte was circulated at 10 mL/min. The electrolysis was carried out at ambient temperature until 3 F/mol electrons were supplied. The reaction crude was washed with saturated NaCl aqueous solution, and then extracted with ethyl acetate for three times. The organic layers were combined and then dried with anhydrous Na2SO4. Unless otherwise stated in the figure legend, all data are from one run. The yield was calculated based on the 1H NMR with mesitylene as the internal standard.
Unless otherwise state in the figure legend, the FE value towards cyclic carbonate was calculated from one experiment as follows: FE = nNF Q ð7Þ where n is the number of electrons transferred, N is the molar quantity of cyclic carbonate synthesized, F is Faraday constant, and Q is the total charged passed.
Preparation of Cu(OH)F-GDE catalyst The copper-based catalyst for CO2 reduction was synthesized using a hydrothermal method according to the literature59,65. 4 mmol of Cu(NO3)2·2H2O was added into 100 mL DMF, and then, 4 mmol NH4HF2 was added followed by 30-min vigorous stirring. The mixture was trans-ferred to a 300 mL Teflon-lined autoclave, and kept under 160 °C for 4 h.
After cooling to room temperature, the generated solid nanoparticles were first centrifuged to remove liquid, and sequentially, rinsed with deionized water and ethanol. The cleaned nanoparticles was then dried under vacuum at 60 °C overnight to obtain the desired Cu(OH)F catalyst.
To prepare the electrode for CO2 reduction to ethylene, 14 mg of the prepared Cu(OH)F catalyst and 20 μL Nafion D520 solution (5 wt%) were added to 1.6 mL isopropanol-tetrahydrofuran solution (1:1 volume ratio). Sonicating the mixture for 30 min to fully disperse Cu(OH)F catalyst particles. The prepared catalyst ink was air-brushed on a gas diffusion electrode (YLS-30T, 3 cm × 3 cm) to prepare the Cu(OH)F-GDE catalyst. In order to further increase the hydrophobicity of the catalyst, 200 μL of 1H,1H,2H,2H-perfluorooctyltrichlorosilane solution (5 wt% in toluene) was painted on the carbon fiber side (i.e., the side without brushed catalyst). Finally, the Cu(OH)F-GDE catalyst was vacuum dried at 30 °C before use.
Procedure for electroreduction of CO2 to ethylene For electroreduction of CO2 to ethylene, Cu(OH)F-GDE catalyst was used as the cathode (5 cm2) and IrO2-DSA was used as the anode (5 cm2). The catholyte was 150 mL of 2.0 M KOH aqueous solution and anolyte was 150 mL of 1.0 M KOH aqueous solution. Two chamber was separated by an anion exchange membrane (FAB-PK-130), which was activated in 0.5 M K2CO3 for 24 h before use. During the electrolysis, 5 sccm of humidified CO2 controlled by mass flow controllers (CS200-Article Nature Communications| (2025) 16:3285 8 A, Beijing Sevenstar Flow) continuously flowed into the cathodic gas chamber, and the catholyte and anolyte were both circulated at 5 mL/ min flowrate. For the ethylene product analysis, collecting the 0.6 mL gas mixture at the outlet of the cathodic gas chamber using a gas-tight syringe, which was then injected into a gas chromatography system with a thermal conductivity detector (TCD) and flame ionization detector (FID). Each sample was analyzed three times and the averaged result was reported. The FE was calculated as in Eq. (8): FE = naFVgasmol%C2H4 QVm ð8Þ where na is the number of electrons transferred for 1 mol C2H4 pro-duction, Vgas is the flow rate of CO2, mol%C2H4 is the ethylene molar percent in the reactor outlet gas mixture measured via GC, Q is the applied current per unit time, and Vm is the unit molar volume of CO2.
Cascade electrosynthesis of EC using CO2 and water For cascade electrosynthesis of EC using CO2 and water, the CO2 electroreduction settings are the same to the aforementioned part.
The outlet stream was directly sparged into a batch cell for the sub-sequent EC electrosynthesis without intermediate purification. 0.2 M Cs2CO3, 0.5 M KBr and 3 mM K2Cr2O7 dissolved in 3 mL DMF-water blended solution (3:7 volume ratio) was used as the electrolyte in a batch cell and the electrolysis was conducted under 6 mA/cm2 using IrO2-DSA anode materials (size 0.5 cm2, 1 mm thick) and nickel foam cathode (0.5 cm2, 1 mm thick) for 12 h.
Data availability The datasets generated and analyzed in this study are included in the main text and supplementary information. Source data are provided with this paper.
References 1.
Chen, Y. et al. A review of lithium-ion battery safety concerns: The issues, strategies, and testing standards. J. Energy Chem. 59, 83–99 (2021).
2.
Khan, F. M. N. U., Rasul, M. G., Sayem, A.S. M. & Mandal, N. K. Design and optimization of lithium-ion battery as an efficient energy sto-rage device for electric vehicles: A comprehensive review. J. Energy Storage 71, 108033 (2023).
3.
Monta, M. & Matsuda, Y. Ethylene carbonate-based organic elec-trolytes for high performance aluminium electrolytic capacitors. J.
Power Sources 60, 179–183 (1996).
4.
Bhuyan, Md. S. A. & Shin, H. Green Recovery of Cathode Active Materials from Li-Ion Battery Electrode Scraps Using Propylene Carbonate: A Novel Approach for Direct Recycling. ACS Sustain.
Chem. Eng. 11, 10677–10687 (2023).
5.
Ethylene Carbonate Market Size.
com/industry-reports/ethylene-carbonate-market. (Accessed 2023-12-18).
6.
Gu, X., Zhang, X., Zhang, X. & Deng, C. Simulation and assessment of manufacturing ethylene carbonate from ethylene oxide in mul-tiple process routes. Chin. J. Chem. Eng. 31, 135–144 (2021).
7.
Pescarmona, P. P. & Taherimehr, M. Challenges in the catalytic synthesis of cyclic and polymeric carbonates from epoxides and CO2. Catal. Sci. Technol. 2, 2169–2187 (2012).
8.
Leow, W. R. et al. Chloride-mediated selective electrosynthesis of ethylene and propylene oxides at high current density. Science 368, 1228–1233 (2020).
9.
Service, R. F. Renewable bonds. Science 365, 1236–1239 (2019).
10.
De Luna, P. et al. What would it take for renewably powered elec-trosynthesis to displace petrochemical processes? Science 364, eaav3506 (2019).
11.
Zheng, T. et al. Upcycling CO2 into energy-rich long-chain com-pounds via electrochemical and metabolic engineering. Nat. Catal.
5, 388–396 (2022).
12.
Lee, M. G. Selective synthesis of butane from carbon monoxide using cascade electrolysis and thermocatalysis at ambient con-ditions.Nat. Catal. 6, 310–318 (2023).
13.
Li, K. et al. Enhancement of lithium-mediated ammonia synthesis by addition of oxygen. Science 374, 1593–1597 (2021).
14.
Li, Y. et al. Redox-mediated electrosynthesis of ethylene oxide from CO2 and water. Nat. Catal. 5, 185–192 (2022).
15.
Ke, J. et al. Facet-dependent electrooxidation of propylene into propylene oxide over Ag3PO4 crystals. Nat. Commun. 13, 932 (2022).
16.
Ho, K.-P., Chan, T. H. & Wong, K.-Y. An environmentally benign catalytic system for alkene epoxidation with hydrogen peroxide electrogenerated in situ. Green Chem. 8, 900 (2006).
17.
Lee, K. M. et al. Redox-neutral electrochemical conversion of CO2 to dimethyl carbonate. Nat. Energy 6, 733–741 (2021).
18.
Chiarotto, I., Mattiello, L. & Feroci, M. The Electrogenerated Cya-nomethyl Anion: An Old Base Still Smart. Acc. Chem. Res. 52, 3297–3308 (2019).
19.
Niu, D. et al. Synthesis of cyclic carbonates from epoxides and CO2 in acetonitrile via the synergistic action of BMIMBr and electrogenerated magnesium. Chin. J. Catal. 37, 1076–1080 (2016).
20. Buckley, B. R., Patel, A. P. & Wijayantha, K. G. U. Electrosynthesis of cyclic carbonates from epoxides and atmospheric pressure carbon dioxide. Chem. Commun. 47, 11888–11890 (2011).
21.
Gao, X. et al. Efficient conversion of CO2 with olefins into cyclic carbonates via a synergistic action of I2 and base elec-trochemically generated in situ. Electrochem. Commun. 34, 242–245 (2013).
22. Jang, J. H. et al. Electrochemically initiated synthesis of ethylene carbonate from CO2. Nat. Synth. 3, 846–857 (2024).
23. Dorchies, F. et al. Controlling the Hydrophilicity of the Electro-chemical Interface to Modulate the Oxygen-Atom Transfer in Electrocatalytic Epoxidation Reactions. Chem. Soc. 144, 22734–22746 (2022).
24. Jin, K. et al. Epoxidation of Cyclooctene Using Water as the Oxygen Atom Source at Manganese Oxide Electrocatalysts. J. Am. Chem.
Soc. 141, 6413–6418 (2019).
25. Tayebi, M. et al. Photoelectrochemical Epoxidation of Cyclohexene on an α-Fe2O3 Photoanode Using Water as the Oxygen Source. ACS Appl. Mater. Interfaces 15, 20053–20063 (2023).
26. Miller, J. H., Joshi, A., Li, X. & Bhan, A. Catalytic degradation of ethylene oxide over Ag/α-Al2O3. J. Catal. 389, 714–720 (2020).
27.
Chung, M. et al. Direct propylene epoxidation via water activation over Pd-Pt electrocatalysts. Science 383, 49–55 (2024).
28. Reithofer, M. R., Sum, Y. N. & Zhang, Y. Synthesis of cyclic carbo-nates with carbon dioxide and cesium carbonate. Green Chem. 15, 2086–2090 (2013).
29. Ochoa-Gómez, J. R. et al. Synthesis of glycerol carbonate from 3-chloro-1,2-propanediol and carbon dioxide using triethylamine as both solvent and CO2 fixation–activation agent. Chem. Eng. J. 175, 505–511 (2011).
30. Huang, L. et al. Ethylene Electrooxidation to 2-Chloroethanol in Acidic Seawater with Natural Chloride Participation. J. Am. Chem.
Soc. 145, 15565–15571 (2023).
31.
Skaff, O., Pattison, D. I. & Davies, M. J. Kinetics of Hypobromous Acid-Mediated Oxidation of Lipid Components and Antioxidants.
Chem. Res. Toxicol. 20, 1980–1988 (2007).
32. Wang, Q. et al. Electrocatalytic CO2 Upgrading to Triethanolamine by Bromine-assisted C2H4 Oxidation. Angew. Chem. Int. Ed. 62, e202212733 (2023).
Article Nature Communications| (2025) 16:3285 9 33. Ayers, P. W., Anderson, J. S. M., Rodriguez, J. I. & Jawed, Z. Indices for predicting the quality of leaving groups. Phys. Chem. Chem.
Phys. 7, 1918–1925 (2005).
34. Chi, M. et al. Spatial decoupling of bromide-mediated process boosts propylene oxide electrosynthesis. Nat. Commun. 15, 3646 (2024).
35. Grgur, B. N. Electrochemical Oxidation of Bromides on DSA/RuO2 Anode in the Semi-Industrial Batch Reactor for On-Site Water Dis-infection. J. Electrochem. Soc. 166, E50–E61 (2019).
36. Kozak, J. A. et al. Bromine-Catalyzed Conversion of CO2 and Epoxides to Cyclic Carbonates under Continuous Flow Conditions.
J. Am. Chem. Soc. 135, 18497–18501 (2013).
37. Hirose, T. et al. Economical synthesis of cyclic carbonates from carbon dioxide and halohydrins using K2CO3. RSC Adv. 6, 69040–69044 (2016).
38. Xue, W. et al. Bromine‐Enhanced Generation and Epoxidation of Ethylene in Tandem CO2 Electrolysis Towards Ethylene Oxide.
Angew. Chem. Int. Ed. 62, e202311570 (2023).
39. Roshan, K. R. et al. A computational study of the mechanistic insights into base catalysed synthesis of cyclic carbonates from CO2: bicarbonate anion as an active species. Catal. Sci. Technol. 6, 3997–4004 (2016).
40. Bobbink, F. D., Gruszka, W., Hulla, M., Das, S. & Dyson, P. J. Synthesis of cyclic carbonates from diols and CO2 catalyzed by carbenes.
Chem. Commun. 52, 10787–10790 (2016).
41.
Fortier, S. M. An on-line experimental/analytical method for mea-suring the kinetics of oxygen isotope exchange between CO2 and saline/hypersaline salt solutions at low (25–50 °C) temperatures.
Chem. Geol. 116, 155–162 (1994).
42. Hashemi, S. M. H. et al. A versatile and membrane-less electro-chemical reactor for the electrolysis of water and brine. Energy Environ. Sci. 12, 1592–1604 (2019).
43. Esposito, D. V. Membraneless Electrolyzers for Low-Cost Hydrogen Production in a Renewable Energy Future. Joule 1, 651–658 (2017).
44. Li, X.-Y. et al. Mechanism of Cations Suppressing Proton Diffusion Kinetics for Electrocatalysis. Angew. Chem. Int. Ed. 62, e202218669 (2023).
45. Zhang, X. et al. Electrochemical oxygen reduction to hydrogen peroxide at practical rates in strong acidic media. Nat. Commun. 13, 2880 (2022).
46. Endrődi, B., Simic, N., Wildlock, M. & Cornell, A. A review of chro-mium(VI) use in chlorate electrolysis: Functions, challenges and suggested alternatives. Electrochimica Acta 234, 108–122 (2017).
47. Lindbergh, G. & Simonsson, D. Inhibition of cathode reactions in sodium hydroxide solution containing chromate. Electrochimica Acta 36, 1985–1994 (1991).
48. Biesinger, M. C. et al. Resolving surface chemical states in XPS analysis of first row transition metals, oxides and hydroxides: Cr, Mn, Fe, Co and. Ni. Appl. Surf. Sci. 257, 2717–2730 (2011).
49. Wagner, C. The Cathodic Reduction of Anions and the Anodic Oxidation of Cations. J. Electrochem. Soc. 101, 181 (1954).
50. Tilak, B. V., Viswanathan, K. & Rader, C. G. On the Mechanism of Sodium Chlorate Formation. J. Electrochem. Soc. 128, 1228 (1981).
51.
Gallardo-Fuentes, S. et al. On the mechanism of CO2 electro-cycloaddition to propylene oxides. J. CO2 Util. 16, 114–120 (2016).
52. Xiao, Y., Chen, B.-L., Yang, H.-P., Wang, H. & Lu, J.-X. Electrosynth-esis of enantiomerically pure cyclic carbonates from CO2 and chiral epoxides. Electrochem. Commun. 43, 71–74 (2014).
53. Iglesias, D. et al. Multi-step oxidative carboxylation of olefins with carbon dioxide by combining electrochemical and 3D-printed flow reactors. Green Chem. 25, 9934–9940 (2023).
54. Wu, L., Yang, H., Wang, H. & Lu, J. Electrosynthesis of cyclic car-bonates from CO2 and epoxides on a reusable copper nanoparticle cathode. RSC Adv. 5, 23189–23192 (2015).
55. Fan, L. et al. Selective production of ethylene glycol at high rate via cascade catalysis. Nat. Catal. 6, 585–595 (2023).
56. Ethylene price trend and forecast.
procurementresource.com/resource-center/ethylene-price-trends. (Accessed 2024-12-28) 57. Ethylene carbonate price trend and forecast.
procurementresource.com/resource-center/ethylene-carbonate-price-trends. (Accessed 2024-12-28) 58. Flores-Granobles, M. & Saeys, M. Minimizing CO2 emissions with renewable energy: a comparative study of emerging technologies in the steel industry. Energy Environ. Sci. 13, 1923–1932 (2020).
59. Ma, W. et al. Electrocatalytic reduction of CO2 to ethylene and ethanol through hydrogen-assisted C–C coupling over fluorine-modified copper. Nat. Catal. 3, 478–487 (2020).
60. Cao, J.-W. et al. One-step ethylene production from a four-component gas mixture by a single physisorbent. Nat. Commun. 12, 6507 (2021).
61.
Ren, T., Patel, M. & Blok, K. Olefins from conventional and heavy feedstocks: Energy use in steam cracking and alternative pro-cesses. Energy 31, 425–451 (2006).
62. Kim, D. et al. Single-step ethylene purification from ternary C2 hydrocarbon mixtures in a scalable metal–organic framework.
Chem. Eng. J. 470, 143858 (2023).
63. Jiang, Y. et al. Benchmark single-step ethylene purification from ternary mixtures by a customized fluorinated anion-embedded MOF. Nat. Commun. 14, 401 (2023).
64. Chuah, C. Y. Membranes for Gas Separation and Purification Pro-cesses. Membranes 12, 622 (2022).
65. Ma, S. One-step electrosynthesis of ethylene and ethanol from CO2 in an alkaline electrolyzer. J. Power Sources. 301, 219 (2016).
Acknowledgements We acknowledge National Natural Science Foundation of China (22478335 and 22108242) (Y.M.), the National Key R&D Program of China (2021YFA1502700) (Y.M.), China Postdoctoral Science Foundation (2022M722728) (M.C.), and Fundamental Research Funds for the Zhe-jiang Provincial Universities (226-2024-00113) (Y.M.) for providing sup-port for this work.
Author contributions Y.M. supervised the project. Y.M. and M.C. conceived the project and designed the experiments. M.C. conducted the experiments as well as performed the technoeconomic analysis. M.C. and Y.M. analyzed the data and prepared the manuscript. Y.M., S.D. and J.X. contributed to the manuscript editing. All the authors discussed the results and assisted during the manuscript preparation.
Competing interests The authors declare no competing interests.
Additional information Supplementary information The online version contains supplementary material available at
Correspondence and requests for materials should be addressed to Yiming Mo.
Peer review information Nature Communications thanks Hussain Almajed, Omar Guerra Fernández, and the other, anonymous, reviewers for their contribution to the peer review of this work. A peer review file is available.
Reprints and permissions information is available at Publisher’s note Springer Nature remains neutral with regard to jur-isdictional claims in published maps and institutional affiliations.
Article Nature Communications| (2025) 16:3285 10 Open Access This article is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, which permits any non-commercial use, sharing, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if you modified the licensed material. You do not have permission under this licence toshare adapted material derived from this article or parts of it. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http:// creativecommons.org/licenses/by-nc-nd/4.0/.
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190252 | https://math.stackexchange.com/questions/929005/derivation-of-derangement-with-inclusion-exclusion | Skip to main content
Derivation of derangement with inclusion-exclusion
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I read many articles on the derivation of the derangement formula but I can't understand them clearly. At first I read the Wikipedia article. I understand the recursive derivation of derangement. But I am interested in understanding the inclusion-exclusion based formula for derangements.
inclusion-exclusion
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edited Dec 28, 2019 at 20:06
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Suppose we want to count Dn, the number of derangements of {1,⋯,n}.
Let S be the set of all permutations of {1,⋯,n}, and
let Ti be the set of permutations which leave i in its natural position.
Then Dn=|Tc1∩⋯∩Tcn|
=|S|−∑i|Ti|+∑i<j|Ti∩Tj|−∑i<j<k|Ti∩Tj∩Tk|+⋯+(−1)n|T1∩⋯∩Tn|
=n!−(n1)(n−1)!+(n2)(n−2)!−(n3)(n−3)!+⋯+(−1)n(nn)(n−n)!
=n!−n!1!+n!2!−n!3!+⋯+(−1)nn!n!=n![1−11!+12! −13!+⋯+(−1)n1n!].
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answered Sep 12, 2014 at 21:01
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I couldn't understand it completely. I am familiarized with these notations but what is the meaning of the notations .
– Shahed al mamun
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Sep 13, 2014 at 5:08
Let me know which notation you would like me to explain further, and I'll try to do that.
– user84413
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Straightforward method! +1
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190253 | https://www.cuemath.com/ncert-solutions/which-term-of-the-ap-53-48-43-is-the-first-negative-term/ | Which term of the AP: 53, 48, 43,... is the first negative term
Solution:
It is given that
AP: 53, 48, 43,…
First term (a) = 53 and
Common difference (d) = 48 - 53 = -5
Consider the nth term of the AP as the first negative term.
i.e., Tn < 0
We know that the nth term of an AP,
Tn = a + (n - 1)d
Here
⇒ [a + (n - 1 )d] < 0
Substituting the values
⇒ 53 + (n - 1)(-5) < 0
⇒ 53 - 5n + 5 < 0
So we get
⇒ 58 - 5n < 0
⇒ 5n > 58
⇒ n > 11.6
⇒ n = 12
So the 12th term is the first negative term of the given AP
T12 = a + (12 - 1)d
= 53 + 11 (-5)
= 53 - 55
= - 2 < 0
Therefore, the first negative term is the 12th term.
✦ Try This: Which term of the AP: 50, 45, 40,... is the first negative term
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5
NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 17
Which term of the AP: 53, 48, 43,... is the first negative term
Summary:
An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same. In AP: 53, 48, 43,... the first negative term is 12th term.
☛ Related Questions: |
190254 | https://mathcenter.oxford.emory.edu/site/math111/inverseTrig/ | Inverse Trigonometric Functions
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Inverse Trigonometric Functions
We will often have need to ask the question: "For what angle does the sine (or cosine, tangent, ...) equal such-and-such?"
Such a question is normally answered by finding and applying an inverse function. However, due to their periodic nature, none of the six trigonometric functions pass the horizontal line test, and as such, none of these functions have a true inverse function.
That said, we can find something almost as useful...
Let us start by considering sinx. If we restrict the domain of the sine function to [−π/2,π/2], the resulting function will pass the horizontal line test, and consequently has an inverse. The graph of this inverse is shown below in red.
Restricting the domain to [−π/2,π/2] is useful on a couple of fronts. Not only do we produce an invertible function, but we also produce one that has the exact same range as the sine function with an unrestricted domain.
Additionally, while there are several intervals that do both of these things (e.g., [π/2,3 π/2], [−3 π/2,−5 π/2], etc.), the interval [−π/2,π/2] minimizes the magnitudes of the angles involved.
The function drawn above in red is called the arcsine function and is denoted either by arcsinx, or sin−1x. This function behaves in many ways (but not all) as an inverse to the unrestricted domain version of sinx.
For instance,
Like most inverse function pairs, the domain of sin−1x is identical to the range of the sinx, both being [−1,1]. However, the range of sin−1x is only [π/2,π/2], a small subset of the domain of the sinx.
For any x in the domain of sin−1, we have sin(sin−1x)=x. However, sin−1(sinx)=x is true only if x∈[−π/2,π/2] and false otherwise!
In a like manner, the remaining five trigonometric functions have "inverses":
The arccosine function, denoted by arccosx or cos−1x is the inverse to the cosine function with a restricted domain of [0,π], as shown below in red.
The arctangent function, denoted by arctanx or tan−1x is the inverse to the tangent function with a restricted domain of (−π/2,π/2), as shown below in red.
Similarly, we define the arccotangent function, denoted by arccot x or cot−1x to be the inverse to the cotangent function with a restricted domain of (0,π), as shown below in red.
Interestingly, our choice for which "piece" of the cotangent function to invert to produce the arccotangent function is not universally agreed upon. The advantage of our selection of (0,π) is that the resulting arccotangent function is continuous and defined everywhere, and behaves in a manner more consistent with that seen in the related tan−1 function.
We also define the arcsecant function, denoted by arcsec x or sec−1x to be the inverse to the secant function with a restricted domain of [0,π], as shown below in red.
Here again, there is not universal agreement that [0,π] should be the "piece" of the secant function that should be inverted to produce the arcsecant function. James Stewart's Single Variable Calculus, for instance, uses [0,π/2)∪[π,3 π/2) instead. Both choices result in a discontinuous graph, however ours minimizes the magnitudes of the angles involved.
Lastly, we define the arccosecant function, denoted by arccsc x or csc−1x to be the inverse of the cosecant function with a restricted domain of [−π/2,0)∪(0,π/2], as shown below in red.
Once again, some authors define the arccosecant to be the inverse of a different "piece" of the cosecant function, such as (0,π/2]∪(π,3 π/2]. As was the case with the arcsec, both choices result in a discontinuous graph. Our choice, however, again minimizes the magnitudes of the angles involved. |
190255 | https://internal.medicine.ufl.edu/files/2012/07/5.16.22.-Does-this-patient-have-A.I.-review.pdf | THE RATIONAL CLINICAL EXAMINATION Does This Patient Have Aortic Regurgitation?
Niteesh K. Choudhry, MD Edward E. Etchells, MD, MSc CLINICAL SCENARIO You are asked to see a 59-year-old woman with liver cirrhosis who will be undergoing sclerotherapy for esopha-geal varices. When she was examined by her primary care physician, she had a pulse pressure of 70 mm Hg. The pri-mary care physician is concerned about the possibility of aortic regurgitation (AR) and asks you whether endocardi-tis prophylaxis is necessary for sclero-therapy. You conduct a complete physi-cal examination and hear no early diastolic murmur in the third or fourth intercostal spaces at the left-sternal bor-der. You feel that the patient is unlikely to have AR and that endocarditis pro-phylaxis is not needed. You suspect that the wide pulse pressure is a peripheral hemodynamic consequence of cirrho-sis, not AR. The primary care physi-cian,however,wonderswhetherthepro-cedure should be delayed until an echocardiogram can be obtained.
WHY IS THE CLINICAL EXAMINATION IMPORTANT IN EVALUATING FOR AR?
Aortic regurgitation is a potentially se-rious cardiac abnormality that may be caused by important underlying disor-ders. Patients with AR require careful clinical monitoring to identify the op-timal time for surgical intervention.
Asymptomatic patients with severe AR may benefit from vasodilator therapy.1 Endocarditis prophylaxis may be indi-cated for patients with AR who are un-dergoing various procedures.2 The use of noninvasive cardiac test-ing, such as echocardiography, has in-creased in recent years. It is estimated that 2% of the general population un-dergoes noninvasive cardiac diagnos-tic evaluation annually.3 If a careful clinical examination can exclude the presence of AR, then there would be no need to proceed with further cardiac evaluation.
Anatomical and Physiological Origins of Diastolic Murmurs The cardinal manifestation of AR is a diastolic murmur. Diastolic murmurs are important indicators of structural cardiac abnormalities or pathological states of increased flow (TABLE 1). As discussed in a previous article in this series,4 heart murmurs are produced when turbulent blood flow causes pro-longed auditory vibrations of cardiac structures. The intensity of the mur-mur depends on many factors, includ-ing blood viscosity, blood flow veloc-ityandturbulence,thedistancebetween the vibrations and the stethoscope, the angle at which the vibrations meet the stethoscope, the transmission quali-ties of the tissue between the vibra-tion and the stethoscope, and the au-ditory capabilities of the examiner.5 How to Examine for AR A complete clinical history and physi-cal examination are essential in the Author Affiliations: Division of General Internal Medi-cine and Clinical Epidemiology, Department of Medi-cine, University of Toronto and the University Health Network, Toronto, Ontario.
Corresponding Author: Edward E. Etchells, MD, MSc, University Health Network, 200 Elizabeth St, eng-248, Toronto, Ontario, Canada M5G 2C4 (e-mail: eetchells@torhosp.toronto.on.ca).
The Rational Clinical Examination Section Editors: David L. Simel, MD, MHS, Durham Veterans Affairs Medical Center and Duke University Medical Center, Durham, NC; Drummond Rennie, MD, Deputy Edi-tor (West), JAMA.
Objective To review evidence as to the precision and accuracy of clinical examina-tion for aortic regurgitation (AR).
Methods We conducted a structured MEDLINE search of English-language articles (January 1966-July 1997), manually reviewed all reference lists of potentially relevant articles, and contacted authors of relevant studies for additional information. Each study (n = 16) was independently reviewed by both authors and graded for methodological quality.
Results Most studies assessed cardiologists as examiners. Cardiologists’ precision for detecting diastolic murmurs was moderate using audiotapes (k = 0.51) and was good in the clinical setting (simple agreement, 94%). The most useful finding for ruling in AR is the presence of an early diastolic murmur (positive likelihood ratio [LR], 8.8-32.0 [95% confidence interval {CI}, 2.8-32 to 16-63] for detecting mild or greater AR and 4.0-8.3 [95% CI, 2.5-6.9 to 6.2-11] for detecting moderate or greater AR) (2 grade A studies). The most useful finding for ruling out AR is the absence of early di-astolic murmur (negative LR, 0.2-0.3 [95% CI, 0.1-0.3 to 0.2-0.4) for mild or greater AR and 0.1 [95% CI, 0.0-0.3] for moderate or greater AR) (2 grade A studies). Except for a test evaluating the response to transient arterial occlusion (positive LR, 8.4 [95% CI, 1.3-81.0]; negative LR, 0.3 [95% CI, 0.1-0.8]), most signs display poor sensitivity and specificity for AR.
Conclusion Clinical examination by cardiologists is accurate for detecting AR, but not enough is known about the examinations of less-expert clinicians.
JAMA. 1999;281:2231-2238 www.jama.com ©1999 American Medical Association. All rights reserved.
JAMA, June 16, 1999—Vol 281, No. 23 2231 at University of Florida, on May 24, 2005 www.jama.com Downloaded from evaluation of patients with a diastolic murmur. A diastolic murmur in a pa-tient with renal failure and volume over-load will have different significance than a diastolic murmur in a patient with a history of rheumatic fever and atrial fi-brillation.
The examiner’s ability to detect a di-astolic murmur can be undermined by environmental factors such as noisy rooms, examiner factors such as fa-tigue or haste, and patient factors such as dyspnea or tachycardia.6 If examin-ing conditions are not optimal, the ex-amination should be repeated when conditions improve.
The precision and accuracy of many individual components of the examina-tion for AR, including all of the cardiac history and most of the physical exami-nation, have not been adequately evalu-ated. This article will focus on aspects of the cardiac physical examination that have been sufficiently assessed for pre-cision or accuracy.
Cardiac Auscultation During routine auscultation, the ex-aminer attempts to detect a diastolic murmur. Diastole is the period that be-gins with the closure of the aortic and pulmonic valves (second heart sound [S2]) and ends with the closure of the mitral and tricuspid valves (first heart sound [S1]). A common maneuver used to identify diastole is to palpate the ca-rotid artery pulse during auscultation; S1 is synchronous with the carotid ar-tery pulsation while S2 follows the pulse.
A diastolic murmur is a diastolic sound longer than a heart sound. Examiners should describe the grade, location of maximal intensity (FIGURE 1), timing (FIGURE 2), duration, pitch, and radia-tion of the murmur.
The Levine grading system,7 with slight modifications,8 was developed for systolic murmurs but may also be used to describe diastolic murmurs. A grade 1 murmur is not heard immediately on auscultation, but is heard after the ex-aminer focuses for a few seconds. Grade 2 murmurs are heard immediately on auscultation but are softer than the loud grade 3. Grade 4 murmurs are associ-ated with a palpable precordial vibra-tion called a thrill. (Grade 5 and 6 mur-murs are also associated with a thrill. A grade 5 murmur is audible when only 1 edge of the stethoscope is on the chest, and a grade 6 murmur is audible with the entire stethoscope lifted off the chest.) The typical murmur of AR is an early diastolic, decrescendo blowing sound (Figure 2), which may be accentuated with the patient sitting upright and lean-ing forward.9 In some cases, S2 can be obscured by the murmur. Most AR mur-mursarehighpitchedandarebestheard with the diaphragm of the stethoscope placedfirmlyonthechestwall.SomeAR murmurs are low pitched and are bet-ter heard with the bell of the stetho-scope placed lightly on the chest wall.
For example, the AR murmur associ-ated with endocarditis and a fenes-trated aortic valve can be low pitched.
Theexaminershouldapplythestetho-scope to the chest wall in the third or Table 1. Selected Causes of Diastolic Murmurs Abnormal cardiac structure Aortic regurgitation Mitral stenosis Pulmonic regurgitation Tricuspid stenosis Atrial myxoma Ventricular septal defect Atrial septal defect Mitral regurgitation Normal cardiac structure, increased flow Renal failure with volume overload Thyrotoxicosis Anemia Sepsis Diastolic murmurs are caused by abnormally increased diastolic flow across the mitral or tricuspid valves.
Figure 1. Typical Location of Abnormal Diastolic Murmurs 1 2 3 There are 3 important areas to auscultate for diastolic murmurs. Area 1 is the second and third intercostal spaces at the right-sternal border. Area 2 is the second and fourth intercostal spaces at the left-sternal border. Aortic regurgitation murmurs may be heard in both areas 1 and 2. If the murmur is loudest in area 1, then the un-derlying cause of aortic regurgitation may be an ascending aortic aneurysm or aortic dissection. Pulmonic re-gurgitation murmurs are loudest in the superior part of area 2, and may radiate downward. The murmur of mitral stenosis and the Flint murmur of aortic regurgitation are best heard at the apex (area 3).
Figure 2. Selected Features of Diastolic Murmors S2 S1 S2 OS S1 Diastolic murmurs are classified based on the time of onset of the murmur.15 An early diastolic murmur be-gins with the second heart sound (S2). Top, Early di-astolic murmurs typically decrease in intensity (decre-scendo) and disappear before the first heart sound (S1).
In some cases, an early diastolic murmur can con-tinue through diastole. Bottom, A mid-diastolic mur-mur begins clearly after S2 (in mitral stenosis, classi-cally after an opening snap [OS]). A late diastolic (or presystolic) murmur begins in the interval immedi-ately before S1. In mitral stenosis, the mid-diastolic mur-mur may merge with the late systolic murmur.
AORTIC REGURGITATION 2232 JAMA, June 16, 1999—Vol 281, No. 23 ©1999 American Medical Association. All rights reserved. at University of Florida, on May 24, 2005 www.jama.com Downloaded from fourth intercostal space at the left ster-nal border and listen while the patient stops his/her breath at end of expira-tion. The patient should not hold his/ her breath because he/she may inadvert-ently do a Valsalva maneuver. If the murmur is louder at the second to third right intercostal space, the underlying cause of AR may be an ascending aortic aneurysm or aortic dissection.10 Aortic regurgitation also may be as-sociated with a systolic murmur,11 which results from the flow of an ab-normally large volume of blood through a nonstenotic aortic valve or a bicus-pid aortic valve. The murmur is an early peaking, crescendo-decrescendo sys-tolic sound that is best heard with the diaphragm of the stethoscope applied to the second-right intercostal space.
The Flint murmur is a low-pitched late-diastolicapicalmurmur,whichisas-sociated with AR. The murmur is likely produced when the regurgitant jet of blood collides with the left ventricular endocardium.12 The murmur may have amid-diastoliccomponent,buttheorigi-nal description by Flint referred only to “presystolicblubbering.”13Itisbestheard with the patient in the left-lateral decu-bituspositionusingthebellofthestetho-scope. Differentiating the Flint mur-mur from the murmur of mitral stenosis can be difficult. The murmur of mitral stenosis is primarily mid-diastolic (pos-siblywithalatediastoliccomponent)and may be associated with an opening snap (OS) and a loud S1 (Figure 2).14 The typical murmur of pulmonic re-gurgitation (PR) is an early diastolic de-crescendomurmurheardbestinthesec-ond-left intercostal space at the sternal border. The murmur may radiate to the third- and fourth-left intercostal spaces, and may increase during quiet inspira-tion. If there is splitting of S2, the astute examiner may note that the murmur be-gins after the pulmonic valve compo-nent (P2) of S2 rather than the aortic component. The murmur of PR may be lower pitched than the murmur of AR, unless pulmonary hypertension is pres-ent. A right-sided Flint murmur can be heard, particularly in patients with pul-monary hypertension.
Mitral stenosis is associated with a mid-diastolic decrescendo low-frequency rumble, which, if the patient is in sinus rhythm, may be followed by late-diastolic(presystolic)crescendothat ends with the mitral component of S1 (Figure 2). It is best heard using the bell of the stethoscope placed at the apex soon after moving the patient into the left-lateral decubitus position. This will both bring the left ventricle closer to the chest wall and serve as a form of exer-cise, which will increase flow across the mitral valve and, therefore, increase the intensity of the murmur.10 The mur-mur of mitral stenosis may be inaudible in patients with low cardiac output.
The S1 may be increased in intensity in mitral stenosis.14 A normal S1 is best appreciatedneartheapexwhereitshould be louder than S2. The S1 is normally softerthanS2inthesecond-rightandsec-ond-leftintercostalspacesadjacenttothe sternum. If S1 is as loud as or louder than the S2 in these areas, then the S1 is in-creased in intensity.
An OS is a high-frequency early dias-tolicsoundthatisassociatedwiththeop-ening of a stenotic mitral valve. It oc-curs 50 to 100 milliseconds after the aortic valve component (A2) of S2 and is best heard in the area from the left ster-nal border to the apex. Much like the murmur of mitral stenosis, it may be ac-centuated by listening to the patient in theleft-lateraldecubituspositionshortly after the patient has performed exer-cise.TheA2-OSintervalshortenswithin-creasing severity of mitral stenosis. The OS may be absent in the case of a heavily calcified immobile mitral valve. It is of-ten difficult to differentiate an OS from the P2 of S2. The OS usually decreases in intensity with inspiration and S2-OS interval widens on standing. Con-versely, P2 becomes louder with inspi-ration and the A2-P2 interval remains the same or narrows with standing.14 Addi-tionally, P2 is not expected to be heard at the apex unless the patient has pul-monary hypertension.
Maneuvers Selective use of maneuvers can en-hance the detection and interpretation of diastolic murmurs. There is no point in doing maneuvers if a loud AR mur-mur has been detected during routine auscultation. However, if the clinician is unsure about the presence of a faint diastolic murmur, then a maneuver that increases murmur intensity may clarify thesituation.Iftheclinicianhasaheight-ened suspicion for AR (eg, after hear-ing an aortic ejection sound), or if ex-aminingconditionsarenotoptimal,then a maneuver to augment murmur inten-sity might bring out an otherwise inau-dible murmur. Finally, if the clinician is unsure whether a murmur repre-sents PR or AR, then these maneuvers may help distinguish between the 2. In this latter situation, the clinician should listen where the murmur is just barely audible, so that it is easy to detect a de-crease or increase in murmur intensity during the maneuver.
Quiet inspiration increases venous re-turn and is intended to augment right-sided heart murmurs such as PR. To de-termine the effect of inspiration on the intensity of the murmur, the exam-iner should listen during quiet inspi-ration rather than asking the patient to breathe deeply, as the murmur may be obscured by breath sounds.
Transient arterial occlusion primar-ily increases systemic arterial resistance that results in an intensification of left-sidedregurgitantlesionssuchasARand may help distinguish the murmur from that of PR. To perform this maneuver, sphygmomanometersareplacedaround bothofthepatient’sarmsandareinflated to 20 to 40 mm Hg above the previously recorded systolic blood pressure. Any changes in murmur intensity are noted 20 seconds after cuff inflation.16 Peripheral Hemodynamic Signs There are a variety of peripheral hemo-dynamic signs traditionally associated with AR. Some of these signs have been adequatelyevaluated,includingdeMus-set head bobbing sign,17 a wide pulse pressure,18 the brachial-popliteal pulse gradient (Hill sign19), Duroziez fem-oral murmur,17 the femoral pistol shot murmur,14 and Corrigan water ham-mer pulse.20 The de Musset head bob-AORTIC REGURGITATION ©1999 American Medical Association. All rights reserved.
JAMA, June 16, 1999—Vol 281, No. 23 2233 at University of Florida, on May 24, 2005 www.jama.com Downloaded from bing sign consists of a forward shaking oftheheadwitheveryheartbeat;itisbest observed in patients who are sitting.17 Pulse pressure refers to the differ-ence between systolic and diastolic blood pressures. A widened pulse pres-sure may be defined as greater than 50 mm Hg.21 Other definitions include a pulse pressure greater than 50% of the systolic pressure.18 Determination of the blood pressure has been described in another article in this series.22 The brachial-popliteal pulse gradient (Hill sign) can be defined as a systolic blood pressure in the lower extremities thatisatleast20mmHghigherthanthat in the arms.21 To determine a popliteal blood pressure, an appropriately sized blood pressure cuff should be placed on the patient’s thigh14 with the artery marker over the popliteal artery; the cuff should be inflated and the systolic pres-sure can then be determined in the pop-liteal fossa either by palpation, as judged by the point where the pulse reappears as the cuff is deflated, or by ausculta-tion,listeningforKorotkoffsoundstoap-pear. Both the brachial and popliteal blood pressures should be measured while the patient is supine. The average of repeated readings should be used, es-pecially in patients with irregular heart rates, such as atrial fibrillation.
Duroziez double intermittent fem-oral bruit is elicited by first gently com-pressing the femoral artery with the dia-phragm of the stethoscope. This will yield a systolic bruit in all patients. As increasing pressure is applied to the dia-phragm, an early diastolic bruit will be-come audible in patients with AR.
While listening to the diastolic bruit, the stethoscope should then be tilted such that the distal rim (closest to the patient’s feet) is compressing the fem-oral artery. If the bruit becomes louder with this maneuver then the diastolic bruit is due to the retrograde flow of blood toward the heart in AR. The stethoscope should then be tilted such that the proximal rim (closest to the pa-tient’s head) is compressing the fem-oral artery. If the diastolic bruit be-comes softer, this can be taken as supportive evidence of the presence of retrograde blood flow. If, however, the bruit becomes louder with proximal pressure (and softer with distal pres-sure), then this sign should not be used as evidence of AR but may indicate the presence of a high-flow state such as re-nal failure with volume overload.23 Femoral pistol shot sounds are elic-ited by auscultating with the dia-phragm of the stethoscope over the fem-oral arteries. A high-pitched pistol shot sound may be heard in AR. Corrigan water hammer pulse refers to an in-creased volume and rate of rise of the radial pulse when the wrist is elevated perpendicular to the body of a supine patient. The radial pulse should first be assessed while the patient is lying su-pine with his/her arms resting at the sides. Sufficient pressure should be ap-plied to obliterate the pulse. While maintaining this pressure, the pa-tient’s arm should be elevated such that it is perpendicular to the plane of the body. In AR, the pulse will become pal-pable despite applying an equivalent amount of pressure as when the arm was at the patient’s side.
Otherperipheralhemodynamicsigns, such as Mayne sign (a decrease in di-astolic blood pressure of 15 mm Hg when the arm is held above the head compared with when the arm is held at the level of the heart),24 Quinke cap-illary pulsation, Muller pulsatile uvula, and Rosenbach liver pulsation, have not been adequately evaluated for preci-sion or accuracy.
METHODS To identify articles pertaining to the pre-cision and accuracy of the physical ex-amination for AR, we used standard methods for conducting research over-views.25 Our data collection strategy in-volved3stepsandwasdeliberatelybroad to reduce the possibility of overlook-ingimportantarticles.First,wesearched MEDLINE for English-language ar-ticles from 1966 through July 1997 using a structured search strategy (avail-able on request from the authors). Sec-ond, we manually reviewed potentially relevant articles and their reference lists.
Third, we contacted the authors of rel-evant studies for additional informa-tion. Studies were excluded if they were review articles, involved patients younger than 18 years, were small (ie, ,20 participants), involved prosthetic heart valves, if no clinical examination was performed or reported, or if there was no acceptable reference standard (Doppler echocardiography or cardiac catheterization).
Studies were independently re-viewedformethodologicalqualitybythe 2 authors and disagreements were re-solvedbyconsensus.Qualitygradeswere assigned using published guidelines.26 Grade A studies involve the indepen-dent comparison of a sign or symptom with a reference standard of diagnosis amongalargenumberofconsecutivepa-tientssuspectedofhavingthetargetcon-dition. Grade B studies meet the criteria forgradeAstudiesbuthaveasmallnum-ber of patients. Grade C studies involve nonconsecutive patients, patients who are known to have the target condition and healthy individuals, nonindepen-dent comparisons between the sign or symptom and the reference standard, or nonindependent comparison with a ref-erence standard of uncertain validity.
GradeCstudiestendtooverestimatethe accuracy of the sign or symptom.
We recreated contingency tables for all studies and determined the likeli-hood ratios (LRs) for the cardiac dis-ease of interest.27,28 When a cell from the table included a value of zero, con-fidence intervals (CIs) for the LRs were estimated using an iterative approach.
We also sought information on the ex-amination for other causes of diastolic murmurs, such as mitral stenosis or PR.
Unfortunately, we found few studies of sufficient methodological quality for these conditions. This relative lack of information implies that methodologi-cally sound studies are needed, but does not imply that the clinical examina-tion for these conditions is imprecise, inaccurate, or unimportant.
Precision of the Examination Related to Diastolic Murmurs Precision refers to agreement regarding a particular clinical finding between dif-AORTIC REGURGITATION 2234 JAMA, June 16, 1999—Vol 281, No. 23 ©1999 American Medical Association. All rights reserved. at University of Florida, on May 24, 2005 www.jama.com Downloaded from ferent physicians (interobserver) or be-tween multiple assessments by the same physician (intraobserver). The preci-sion of the clinical examination for di-astolic murmurs has been evaluated in usual clinical situations by auscultating patients29,30 or in controlled nonclinical circumstances by listening to recorded audiotapes (TABLE 2).31 Therehavebeen4studiesthataddress theinterobserverprecisionofcardiacaus-cultation to detect diastolic murmurs (Table2).Whilesimpleagreementishigh in these studies, the 1 study for which it was possible to calculate agreement ad-justedforchance(k)showedonlymod-erate agreement. The experience of ob-servers likely affects precision. The 1 study29thatcomparedcardiologistswith noncardiologists found a higher simple agreement for cardiologists.
The interobserver agreement be-tween examiners on the intensity of heart sounds is excellent (92%).30 In this study, examiners progressively in-serted 0.5-mm-thick paper disks be-tween the patient’s chest and the stetho-scope. The total thickness of the disks was used as a measure of heart sound intensity. Murmur intensity was also as-sessed using this technique (Table 2).
The Bottom Line for Precision The interobserver precision of cardi-ologists examining for any diastolic murmur is moderate using audiotapes (k = 0.51) and good in the clinical set-ting (simple agreement, 94%). Noncar-diologists may be less precise than car-diologists. The precision of examining for the intensity of murmurs and heart sounds using a standardized series of paper disks to assess intensity is good (simple agreement, 92%-96%).
Accuracy of the Examination for AR We consider Doppler echocardiogra-phy and cardiac catheterization to be acceptable reference standards for AR (TABLE3).In1study,thereferencestan-dard was open-heart surgery. Some studies explicitly graded the severity of AR detected by the reference standard (for example, on the basis of the abso-lute height of the regurgitant jet or the ratiooftheheightofthejettotheheight of the left ventricular outflow tract on Doppler echocardiography), which allowed us to calculate LRs for detect-ing both mild (or worse) and moder-ate (or worse) AR.
Cardiologists conducted the clinical examinations in most studies. Too few studies,usingfewpatients,allowforrea-sonableestimatesoftheaccuracyofnon-cardiologists, although noncardiolo-gists are likely less adept at detecting the diastolic murmur of AR. Approxi-mately20%ofresidentsandmedicalstu-dents correctly identified the murmur of AR on high-fidelity digitized audio-tapes,32 while 46% of internal medicine residentscorrectlyidentifiedanARmur-mur using a patient simulator.33 The best-studied physical finding is the typical early diastolic murmur of AR.34-47 If an examiner does not hear a typical AR murmur then the likeli-hood that the patient has moderate or greater AR is significantly reduced (negative LR, 0.1 for grade A studies); the likelihood of mild or greater AR is also significantly reduced (negative LR, 0.2-0.3 for grade A studies).34,35 If an examiner hears the typical AR murmur, the likelihood that the pa-tient has moderate or greater AR is in-creased (positive LR, 4.0-8.3 for grade A studies); the likelihood of mild or greater AR is also significantly in-creased (positive LR, 8.8-32.0 for grade A studies).34,35 The intensity of the murmur corre-lates with the severity of echocardio-graphic AR. Desjardins et al48 studied 40 patients with echocardiographic AR, including 17 with severe AR. A grade 3 diastolic murmur had an LR of 4.5 (95% CI, 1.6-14.0) for distinguishing severe AR from less severe AR, while a grade 2 murmur had an LR of 1.1 (95% CI, 0.5-2.4), a grade 1 murmur had an LR of 0.0 (95% CI, 0.0-0.9), and ab-sence of a diastolic murmur had an LR of 0.0 (95% CI, 0.0-1.1).48 TwogradeCstudiesoftheFlintmur-mur and some peripheral hemody-namic findings are reported in Table 3.
Grade C studies tend to overestimate diagnostic test accuracy. Despite this tendency, 1 study found that absence of a Flint murmur did not rule out AR (negative LR, 0.5-0.8).49 Another study of patients with mild-to-severe AR only found that a wide pulse pressure or peripheral hemodynamic sign (Duro-ziezbruit,femoralpistolshots,andCor-rigan pulses) was not helpful for dis-tinguishing mild AR from moderate or severe AR.21 One small study (grade C) evaluated peripheral hemodynamic signs in pa-tientsexclusivelywithprovenARofvary-ing severity defined by aortography or surgery. These studies provide an esti-mate of the sensitivity of the peripheral hemodynamicsigns.ThedeMussethead bobbingsignwasseeninonly1of20pa-tients(sensitivity,5%)andDuroziezfem-oral bruit was observed in 8 of 12 pa-tients (sensitivity, 67%).17 THE BOTTOM LINE FOR AR When a cardiologist hears the typical murmur of AR, the likelihood of mild or greater AR is increased signifi-cantly (2 grade A studies). The ab-sence of a typical diastolic murmur sig-nificantly reduces the likelihood of AR (2 grade A studies). Noncardiologists may be less proficient than cardiolo-gists at detecting the murmur of AR.
Table 2. Interobserver Reliability (Precision) for Detecting Diastolic Murmurs Finding Type of Examiner No. of Examiners No. of Patients k Simple Agreement, % Murmur absent vs present Cardiologists (tapes)31 5 100 0.51 79 Cardiologists29 2 32 . . .
94 Noncardiologists29 3 32 . . .
78 Intensity of murmur Not stated30† 5 25 . . .
92 Ellipses indicate data not available.
†Examiners used paper disks, 0.5 mm in thickness, that were progressively inserted between the chest wall and the stethoscope until the murmur became inaudible. The total thickness of the disks used was used as the measure of intensity. For example, if a murmur was inaudible after inserting 3 disks, then this was a 1.5-mm murmur.
AORTIC REGURGITATION ©1999 American Medical Association. All rights reserved.
JAMA, June 16, 1999—Vol 281, No. 23 2235 at University of Florida, on May 24, 2005 www.jama.com Downloaded from Mitral Stenosis and PR In 1 grade A study of 529 unselected nursing home residents (31 with mi-tral stenosis), a cardiologist detected a mid-diastolic murmur in all cases of mi-tral stenosis, with no false-positive or -negative examinations (W. A. Aronow, MD, written communication, 1997).50 Only 1 patient had an audible OS.
Noncardiologists may be less profi-cient at detecting the physical find-ings of mitral stenosis. Less than 10% of residents and medical students cor-rectly identified a mid-diastolic mur-mur of mitral stenosis on a high-fidelity digitized audiotape,32 while 43% of medical residents identified a mid-diastolic murmur of mitral stenosis us-ing a patient simulator. In the latter study, only 21% identified the OS of mi-tral stenosis.33 The only evaluated element of the clinical examination for PR is the pres-ence of a typical diastolic decrescendo murmur best audible in the second in-tercostal space at the left-upper sternal border, which may increase in intensity with quiet inspiration. All studies used cardiologists as examiners and were of poor methodologic quality (grade C).
When a cardiologist hears the mur-mur of PR, the likelihood of PR in-creases (positive LR, 17 in both stud-ies), but absence of a PR murmur was not helpful for ruling out PR (LR, 0.9 in both studies).37,43 Table 3. Accuracy of the Physical Examination for Detecting Aortic Regurgitation Study Patient Population Reference Standard No. of Patients With AR Positive Likelihood Ratio (95% CI)† Negative Likelihood Ratio (95% CI)‡ Quality Grade Typical Murmur With Severity of AR Specified Aronow and Kronzon,34 1989 Elderly patients Echocardiography (n = 450) A Mild or greater AR 131 32 (16-63) 0.2 (0.1-0.3) Moderate or greater AR 74 8.3 (6.2-11) 0.1 (0.0-0.2) Grayburn et al,35 1986 Referred for catheterization Catheterization (n = 106) A Mild or greater AR 82 8.8 (2.8-32) 0.3 (0.2-0.4) Moderate or greater AR 57 4.0 (2.5-6.9) 0.1 (0.1-0.3) Roldan et al,36 1996 Asymptomatic connective Echocardiography (n = 143) C tissue disease and Mild or greater AR 10 80 (14-470) 0.4 (0.2-0.7) controls Moderate or greater AR 5 69 (18-270) 0.0 (0.0-0.6) Rahko,37 1989 Referred for echocardiogram Echocardiography (n = 403) C Mild or greater AR 134 27 (13-60) 0.4 (0.3-0.5) Moderate or greater AR 82 12 (8.1-19) 0.2 (0.1-0.3) Cohn et al,38 1967 Mitral valve repair Open-heart surgery (n = 156) C Mild or greater AR 50 5.2 (3.3-8.4) 0.3 (0.2-0.4) Moderate or greater AR 37 3.9 (2.6-5.7) 0.2 (0.1-0.4) Meyers et al,39 1982 Referred for aortography Catheterization (n = 75) C Mild or greater AR 66 3.3 (1.3-12) 0.4 (0.2-0.7) Moderate or greater AR 39 1.6 (1.2-2.4) 0.4 (0.2-0.7) Dittmann et al,40 1987 Valvular heart disease Catheterization (n = 55) C§ Mild or greater AR 42 16 (2.1-155) 0.4 (0.3-0.6) Severe AR 19 3.6 (2.1-6.6) 0.1 (0.0-0.4) Meyers et al,41 1985 Valvular heart disease Catheterization (n = 20) C Mild or greater AR 11 9.8 (1.3-96) 0.5 (0.2-0.9) Moderate or greater AR 3 5.7 (1.4-14) 0.0 (0.0-0.9) Linhart,42 1971 Mitral stenosis Catheterization (n = 28) C Mild or greater AR 11 6.2 (1.9-23) 0.3 (0.1-0.7) Moderate or greater AR 7 7.0 (2.5-20) 0.0 (0.0-1.3) Typical Murmur Without AR Severity Specified (May Include Trivial AR) Come et al,43 1986 Mitral valve prolapse, plus patients with systolic flow murmurs Echocardiography (n = 165) 7 90 (8-982) 0.7 (0.4-0.9) C Nienaber et al,44 1993 Clinically suspected aortic dissection Echocardiography (n = 110) 32 33 (9.4-120) 0.2 (0.1-0.3) C\ Ward et al,45 1977 Clinically suspected aortic dissection Catheterization (n = 65) 49 13 (2.9-75) 0.2 (0.1-0.3) C\ Esper,46 1982 AR and other heart disease Echocardiography (n = 43) 24 12 (2.4-67) 0.4 (0.3-0.7) C Saal et al,47 1985 Mitral stenosis Catheterization (n = 45) 35 8.0 (1.9-45) 0.2 (0.1-0.4) C AORTIC REGURGITATION 2236 JAMA, June 16, 1999—Vol 281, No. 23 ©1999 American Medical Association. All rights reserved. at University of Florida, on May 24, 2005 www.jama.com Downloaded from The Bottom Line for Mitral Stenosis and PR Presence of a mid-diastolic murmur sig-nificantly increases the likelihood of mi-tral stenosis, while absence of a mid-diastolic murmur significantly reduces the likelihood of mitral stenosis (1 grade A study). When a cardiologist hears a typical PR murmur, the likelihood of PR increases significantly. Absence of a typical murmur does not alter the like-lihood of PR (2 grade C studies). Non-cardiologists may be less proficient at detecting the mid-diastolic murmur of mitral stenosis.
Diastolic Murmurs in Patients With Renal Failure Diastolicmurmursduetoabnormalflow states, rather than abnormal cardiac structure, may be associated with a va-riety of conditions. Renal failure with volume overload is the only abnormal flow state associated with diastolic mur-murs that has been evaluated.
Upto9%ofpatientswithend-stagere-nal disease have diastolic murmurs, par-ticularly when these patients also have volume overload, anemia, and hyper-tension.51 These murmurs typically dis-appear after the treatment of volume overload,aswasdemonstratedin2small studies(gradeC).51,52Thesemurmursare probablyduetotransientpulmonaryhy-pertension and dilation of the pulmo-nary artery root, leading to PR.52 The Bottom Line for Diastolic Murmurs in Patients With Renal Failure Although there is an insufficient amount of data on which to make rig-orous recommendations, if an early di-astolic murmur is heard in a dialysis pa-tient with volume overload, the patient should be reexamined after treatment because the murmur may disappear.
When to Examine for AR There are no evaluative data on which to base a recommendation regarding whentoexamineforAR.UndetectedAR maybecommoninelderlypersons:13% (n = 552) of asymptomatic elderly Finn-ishpersonshadmoderateorsevereecho-cardiographic AR.53 Unfortunately, that study does not indicate how many of these patients had audible diastolicmur-murs.Audiblediastolicmurmursmaybe relativelyuncommonfindingsinasymp-tomatic persons. In 1 study only 1% (n = 103) of elderly asymptomatic nurs-ing home residents had an audible dias-tolic murmur.54 Despite the lack of evaluative data, we think that a prudent clinician will examine for AR in most clinical set-tings. Aortic regurgitation is a serious cardiac abnormality, which may be caused by underlying disorders andmay be asymptomatic. The clinician’s sus-picion for AR may be heightened by evi-dence of systemic disease, such as an-kylosing spondylitis, a peripheral hemodynamic finding (although these are by no means indicative of underly-ing AR), or an abnormality detected during routine auscultation (such as an aortic ejection sound). Other findings may suggest different cardiac abnor-malities associated with diastolic mur-murs, such as evidence of pulmonary hypertension (for PR), a wide-fixed split S2 (for atrial-septal defect), or a holo-Table 3. Accuracy of the Physical Examination for Detecting Aortic Regurgitation (cont) Study Patient Population Reference Standard No. of Patients With AR Positive Likelihood Ratio (95% CI)† Negative Likelihood Ratio (95% CI)‡ Quality Grade Maneuver With transient arterial occlusion murmur increases in intensity16 Patients with AR, mitral stenosis, and pulmonic regurgitation Catheterization or echocardiography (n = 16) 10 8.4 (1.3-81) 0.3 (0.1-0.8) C Associated Physical Findings Flint murmur 49 Isolated AR and controls Echocardiography (n = 36) C Mild or greater AR 28 4 (0.5-40) 0.8 (0.6-1.3) Moderate or greater AR 13 25 (2.8-243) 0.5 (0.2-0.7) Any systolic murmur 49 Isolated AR and controls Echocardiography (n = 36) C Mild or greater AR 28 1.3 (0.9-2.7) 0.5 (0.2-1.6) Moderate or greater AR 13 1.5 (1.0-2.1) 0.0 (0.0-1.0) Popliteal-brachial gradient Mild to severe AR Catheterization (n = 33) C .20 mm Hg21 Moderate or greater AR 28 8.2 (1.5-78) 0.2 (0.1-0.5) Peripheral hemodynamic signs21¶ Mild to severe AR Catheterization (n = 34) C Moderate or greater AR 28 2.1 (0.3-22) 0.8 (0.7-1.7) Pulse pressure .50 mm Hg21 Mild to severe AR Catheterization (n = 33) C Moderate or greater AR 28 1.0 (0.7-2.2) 0.9 (0.2-5.5) AR indicates aortic regurgitation; CI, confidence interval.
†The applicable likelihood ratio when the finding is present.
‡The applicable likelihood ratio when the finding is absent.
§Grade B study except catheterization results not interpreted independently of clinical findings.
\Grade B study except echocardiogram not interpreted independently of clinical findings.
¶Included Duroziez bruit, femoral pistol shots, and Corrigan pulses.
AORTIC REGURGITATION ©1999 American Medical Association. All rights reserved.
JAMA, June 16, 1999—Vol 281, No. 23 2237 at University of Florida, on May 24, 2005 www.jama.com Downloaded from systolic apical murmur (for mitral re-gurgitation).
Recommendations for Further Research Most studies used cardiologists to con-duct clinical examinations. There are somedatathatsuggestthatnoncardiolo-gistsmaybelessaccuratethencardiolo-gists, so studies evaluating techniques toimprovetheskillsofnoncardiologists areneeded.Therearealsonostudiesde-finingtheoptimalexaminationtechnique for detecting the AR murmur.
SCENARIO RESOLUTION Your patient, who will be undergoing sclerotherapy for esophageal varices, has a wide pulse pressure but no typi-cal early diastolic murmur. The likeli-hood of mild or moderate AR is signifi-cantly reduced by the absence of a typical early diastolic murmur (nega-tive LR, 0.1-0.3; 2 grade A studies). You perform transient arterial occlusion and no diastolic murmur appears, which en-hances your confidence (negative LR, 0.3). You are confident in your assess-ment because it was conducted in a quiet room with a comfortable and co-operative patient. You can advise the primary care physician that AR is un-likely and echocardiography is not nec-essary.
Funding/Support: This work was funded in part by grant 98-01R from Physicians Services Incorporated Foundation.
Disclaimer: The views expressed in this article are those of the authors and do not reflect those of any spon-soring or supporting agency.
Acknowledgment: We thank Eugene Oddone, MD, MHS, for his helpful comments on earlier drafts of this article, and W. S. Aronow, MD, and C. A. Nienaber, MD, for providing additional data and methodologi-cal information about their studies.
REFERENCES 1. Scognamiglio R, Rahimtoola SH, Fasoli G, Nistri S, Volta SD. Nifedipine in asymptomatic patients with severe aortic regurgitation and normal left ventricu-lar function. N Engl J Med. 1994;331:689-694.
2. Dajani AS, Taubert KA, Wilson W, et al. Prevention ofbacterialendocarditis.JAMA.1997;211:1794-1801.
3. Ballard DJ, Khandeira BK, Tajik AJ, Seward JB, Weber VP, Melton LJ. Population-based study of echocardi-ography. Int J Technol Assess Health Care. 1989;5: 249-261.
4. Etchells E, Bell C, Robb K. Does this patient have an abnormal systolic murmur? JAMA. 1997;277: 564-571.
5. Rushmer RF, Sparkman DR, Polley RFL, et al. Vari-ability in detection and interpretation of heart mur-murs. AJDC. 1952;83:740-754.
6. Feinstein AR, Di Massa R. The unheard diastolic murmur in acute rheumatic fever. N Engl J Med. 1959; 260:1331-1333.
7. Freeman AR, Levine SA. The clinical significance of the systolic murmur: a study of 1000 consecutive cases. Ann Intern Med. 1933;6:1371-1385.
8. Constant J, Lippschultz EJ. Diagramming and grad-ing heart sounds and murmurs. Am Heart J. 1965;70: 322-326.
9. Constant J. Bedside Cardiology. 3rd ed. Boston, Mass: Little Brown & Co Inc; 1985.
10. Harvey WP. Cardiac pearls. Dis Mon. 1994;40: 43-113.
11. Shaver JA. Cardiac auscultation. Curr Probl Car-diol. 1995;20:445-530.
12. Landzberg JS, Pflugfelder PW, Cassidy MM, Schiller NB, Higgins CB, Cheitlin MD. Etiology of the Austin Flint murmur. J Am Coll Cardiol. 1992;20:408-413.
13. Flint A. On cardiac murmurs. Am J Med Sci. 1862; 44:29-54.
14. Sapira JD. The Art and Science of Bedside Diag-nosis. Baltimore, Md: Urban & Schwarzenberg; 1990.
15. Soffer A, Feinstein A, Rosener S, et al. Glossary of cardiologic terms related to physical diagnosis and history. Am J Cardiol. 1967;20:285-286.
16. Lembo NJ, Dell’Italia LJ, Crawford MH, O’Rourke RA. Diagnosis of left-sided regurgitant murmurs by tran-sient arterial occlusion: a new maneuver using blood pressure cuffs. Ann Intern Med. 1986;105:368-370.
17. Sapira JD. Quincke, de Musset, Duroziez, and Hill.
South Med J. 1981;74:459-467.
18. Hegglin R, Scheu H, Rothlin M. Aortic insuffi-ciency. Circulation. 1968;37(suppl V):V77-V92.
19. Hill L. The measurement of systolic blood pres-sure in man. Heart. 1909;1:73-82.
20. Warnes CA, Harris PC, Fritts HW. Effect of el-evating the wrist on the radial pulse in aortic regur-gitation. Am J Cardiol. 1983;51:1551-1553.
21. Frank MJ, Casanegra P, Migliori AJ, Levinson GE.
The clinical evaluation of aortic regurgitation. Arch In-tern Med. 1965;116:357-365.
22. Reeves RA. Does this patient have hyperten-sion? JAMA. 1995;273:1211-1218.
23. Blumgart HL, Ernstene AC. Two mechanisms in the production of Duroziez’s sign. JAMA. 1933;100: 173-177.
24. Abbas F, Sapira JD. Mayne’s sign is not patho-gnomonic of aortic insufficiency. South Med J. 1987; 80:1051-1052.
25. L’Abbe KA, Detsky AS, O’Rourke K. Meta-analysis in clinical research. Ann Intern Med. 1987; 107:224-233.
26. Holleman DR, Simel DL. Does the clinical exami-nation predict airflow limitation? JAMA. 1995;273: 313-319.
27. Koopman PAR. Confidence intervals for the ra-tio of two binomial proportions. Biometrics. 1984;40: 513-517.
28. Centor RM, Keightley J. The Two by Two Ana-lyzer. v1.0. 1992.
29. Raftery EB, Holland WW. Examination of the heart.
Am J Epidemiol. 1967;85:438-444.
30. Sharma RS. Measurement of intensity of heart sounds and murmurs. Lancet. 1981;2:612.
31. Dubrow RJ, Calatayud JB, Abraham S, Caceres CA.
A study of physician variation in heart-sound inter-pretation. Med Ann Dist Columbia. 1964;33:305-308, 355.
32. Mangione S, Nieman LZ. Cardiac auscultatory skills of internal medicine and family practice trainees. JAMA.
1997;278:717-722.
33. St Clair EW, Oddone EZ, Waugh RA, Corey GR, Feussner JR. Assessing housestaff diagnostic skills us-ing a cardiology patient simulator. Ann Intern Med.
1992;117:751-756.
34. Aronow WS, Kronzon I. Correlation of preva-lence and severity of aortic regurgitation detected by pulsed doppler echocardiography with the murmur of aortic regurgitation in elderly patients in a long-term health care facility. Am J Cardiol. 1989;63:128-129.
35. Grayburn PA, Smith MD, Handshoe R, Fried-man BJ, DeMarie AN. Detection of aortic insuffi-ciency by standard echocardiography, pulsed dop-pler echocardiography, and auscultation. Ann Intern Med. 1986;104:599-605.
36. Roldan CA, Shively BK, Crawford MH. Value of the cardiovascular physical examination for detect-ing valvular heart disease in asymptomatic subjects.
Am J Cardiol. 1996;77:1327-1331.
37. Rahko PS. Prevalence of regurgitant murmurs in patients with valvular regurgitation detected by dop-pler echocardiography. Ann Intern Med. 1989;111: 466-472.
38. Cohn LH, Mason DT, Ross J, Morrow AG, Braun-wald E. Preoperative assessment of aortic regurgita-tion in patients with mitral valve disease. Am J Car-diol. 1967;19:177-182.
39. Meyers DG, Sagar KB, Ingram RF, Paulsen WJH, RomhiltDW.Diagnosisofaorticinsufficiency.SouthMed J. 1982;75:1192-1194.
40. Dittmann H, Karsch KR, Seipel L. Diagnosis and quantification of aortic regurgitation by pulsed dop-pler echocardiography in patients with mitral valve dis-ease. Eur Heart J. 1987;8(suppl C):53-57.
41. Meyers DG, Olson TS, Hansen DA. Ausculta-tion, M-mode echocardiography and pulsed doppler echocardiography compared with angiography for di-agnosis of chronic aortic regurgitation. Am J Cardiol.
1985;56:811-812.
42. Linhart JW. Aortic regurgitation. Ann Thorac Surg.
1971;11:27-37.
43. Come PC, Riley MF, Carl LV, Nakao S. Pulsed dop-pler echocardiographic evaluation of valvular regur-gitation in patients with mitral valve prolpase. J Am Coll Cardiol. 1986;8:1355-1364.
44. Nienaber CA, von Kodolitsch Y, Nicolas V, et al. The diagnosis of thoracic aortic dissection by noninvasive im-aging procedures. N Engl J Med. 1993;328:1-9.
45. Ward JM, Baker CW, Rubenstein SA, Johnston SL. Detection of aortic insufficiency by pulse doppler echocardiography. J Clin Ultrasound. 1977;5:5-10.
46. Esper RJ. Detection of mild aortic regurgitation by range-gated pulsed doppler echocardiography. Am J Cardiol. 1982;50:1037-1043.
47. Saal AK, Gross BW, Franklin DW, Pearlman AS.
Noninvasive detection of aortic insufficiency in pa-tients with mitral stenosis by pulsed doppler echocar-diography. J Am Coll Cardiol. 1985;5:176-181.
48. Desjardins VA, Enriquez-Sarano M, Tajiik AJ, Bai-ley KR, Seward JB. Intensity of murmurs correlates with severity of valvular regurgitation. Am J Med. 1996; 100:149-156.
49. Lee D, Chen CH, Hsu TL, Chiang CE, Wang SP, Chang MS. Reappraisal of cardiac murmurs related to aortic regurgitation. Chin Med J (Engl). 1995;56:152.
50. Aronow WS, Schwartz KS, Koenigsberg M. Cor-relation of murmurs of mitral stenosis and mitral regur-gitation with presence or absence of mitral annular cal-cium in person older than 62 years in a long-term health care facility. Am J Cardiol. 1987;59:181-182.
51. Alexander WD, Polak A. Early diastolic murmurs in end-stage renal failure. Br Heart J. 1977;39:900.
52. Perez JE, Smith CA, Meltzer VN. Pulmonic valve insufficiency. Ann Intern Med. 1985;103:497-502.
53. Lindroos M, Kupari M, Heikkila J, Tilvis R. Preva-lence of aortic valve abnormalities in the elderly. J Am Coll Cardiol. 1993;21:1220-1225.
54. Bethel CS. Heart sounds in the aged. Am J Car-diol. 1963;11:763-767.
AORTIC REGURGITATION 2238 JAMA, June 16, 1999—Vol 281, No. 23 ©1999 American Medical Association. All rights reserved. at University of Florida, on May 24, 2005 www.jama.com Downloaded from |
190256 | https://arxiv.org/pdf/1405.7654 | Prepared for submission to JCAP
Detecting non-relativistic cosmic neutrinos by capture on tritium: phenomenology and physics potential
Andrew J. Long, Cecilia Lunardini, Eray Sabancilar
Physics Department, Arizona State University, Tempe, Arizona 85287, USA. E-mail: andrewjlong@asu.edu, Cecilia.Lunardini@asu.edu, Eray.Sabancilar@asu.edu
Abstract. We study the physics potential of the detection of the Cosmic Neutrino Background via neutrino capture on tritium, taking the proposed PTOLEMY experiment as a case study. With the projected energy resolution of ∆ ∼ 0.15 eV, the experiment will be sensitive to neutrino masses with degenerate spectrum, m1 ' m2 ' m3 = mν & 0.1 eV. These neutrinos are non-relativistic today; detecting them would be a unique opportunity to probe this unexplored kinematical regime. The signature of neutrino capture is a peak in the electron spectrum that is displaced by 2 mν above the beta decay endpoint. The signal would exceed the background from beta decay if the energy resolution is ∆ . 0.7 mν . Interestingly, the total capture rate depends on the origin of the neutrino mass, being Γ D ' 4 and Γ M ' 8 events per year (for a 100 g tritium target) for unclustered Dirac and Majorana neutrinos, respectively. An enhancement of the rate of up to O(1) is expected due to gravitational clustering, with the unique potential to probe the local overdensity of neutrinos. Turning to more exotic neutrino physics, PTOLEMY could be sensitive to a lepton asymmetry, and reveal the eV-scale sterile neutrino that is favored by short baseline oscillation searches. The experiment would also be sensitive to a neutrino lifetime on the order of the age of the universe and break the degeneracy between neutrino mass and lifetime which affects existing bounds.
arXiv:1405.7654v2 [hep-ph] 12 Nov 2014
Contents
1 Introduction 12 Cosmic background neutrinos and their capture on tritum 3
2.1 Thermal history of the C νB 32.2 Helicity composition of the C νB 52.3 Detection of the C νB 7
3 Detection prospects at a PTOLEMY-like experiment 10 4 Detection prospects for varying neutrino properties 13
4.1 Majorana vs. Dirac neutrinos 13 4.2 Clustering and annual modulation 14 4.3 The hierarchical mass spectrum 15
5 Probing sterile neutrinos 16
5.1 eV-scale sterile neutrinos 16 5.2 keV-scale warm dark matter sterile neutrinos 17
6 Sensitivity to other non-standard neutrino physics 18
6.1 Lepton asymmetry 18 6.2 Neutrino decay 19 6.3 Non-standard thermal history 21
7 Discussion 22 A Amplitude and cross section for polarized neutrinos 23
A.1 Kinematics 23 A.2 The polarized neutrino capture amplitude 25
1 Introduction
The Cosmic Neutrino Background (C νB) is a cardinal feature of early universe cosmology, and holds the key to understanding many of its most interesting and well-studied phenomena: from the primor-dial synthesis of elements, to the anisotropies of the cosmic microwave background (CMB), and even to the formation of dark matter halos (for a review see, e.g. , [1–4]). The body of information from cosmological probes, on the composition and distribution of matter and energy in the early universe, constitutes a very strong indirect evidence that the C νB exists and confirms the Standard Model’s prediction of its energy density. Specifically, measurements of the CMB anisotropies and the large scale distribution of galaxies have already supplied two key pieces of data: a measurement of the effective number of neutrino species, Neff , and a strikingly strong upper bound on the sum of the neutrino masses, ∑ mν . The most recent values from the Planck satellite read as follows :
Neff = 3 .30 ± 0.27 and ∑ mν < 0.23 eV at 95% CL . (1.1) With the next generation of CMB telescopes, the sensitivity to ∑ mν will be reduced to the 0 .05 eV level, which could allow for a measurement . At this time, however, we still lack the truly golden signature of the C νB that only a laboratory-controlled, direct detection experiment could provide. Such a detection would not only complement other cosmological probes, and thereby help to resolve degeneracies among the neutrino model pa-rameters, but it would access a whole array of phenomena that are beyond the reach of cosmological – 1 – Electron Kinetic Energy H Ke L
Electron Spectrum H dG ê dE e L
+m4
+mΝ-mΝ
Kend 0 ª 18.6 keV
Β-decay endpoint H Kend L
CΝB
Sterile ΝFigure 1 . A cartoon illustrating the expected signal from the three active C νB neutrinos of mass m1 '
m2 ' m3 = mν (solid line), and from a hypothetical, mostly sterile, neutrino mass state, ν4, of mass m4
(dashed line). The C νB signal is displaced from the beta decay endpoint by 2 mν , and the ν4 signal would be displaced by mν + m4. The signal and background are not represented to scale. Here Ke = Ee − me is the electron kinetic energy, and K0end denoted by the vertical dashed line refers to the beta decay end point kinetic energy in the mν = 0 limit. For a details, see Sec. 2.3 and Sec. 3.
measurements. In the first place, a direct detection would confirm that the relic neutrinos are still present in the universe today – a reasonable assumption if the neutrinos are stable, but one which has no empirical confirmation from cosmological observations alone. To put this less dramatically, a direct detection of the C νB would probe late time effects, those occurring after recombination, such as neutrino clustering (and therefore the neutrino coupling to gravity), changes in the C νBflavor composition or number density due to neutrino decay, or decay of heavy relics into neutri-nos, and so on. Perhaps even more importantly, a direct detection of the C νB would constitute the first probe of non-relativistic neutrinos (since current detectors are only sensitive to relatively large neutrino masses), and thereby open the window onto an entirely new kinematical regime. Studying non-relativistic neutrinos could allow for tests of certain neutrino properties that are difficult to access at high momentum such as the Dirac or Majorana character of neutrinos. Given the importance of a direct detection of the C νB, it is not surprising that research in this field has been active and uninterrupted. In 1962 Weinberg was the first to advocate for C νB detection via neutrino capture on beta-decaying nuclei (NCB) since this process requires no threshold energy . The NCB technique is primarily limited by availability of the target material and by the need for extremely high precision in measuring the electron energy 1. Other detection methods have their own challenges. The Stodolsky effect, for instance, could allow C νB neutrinos to be detected by their coherent scattering on a torsion balance [8, 9], but the expected accelerations are well below the sensitivity of current detectors [10, 11], and vanishes if the C νB is lepton-symmetric. In the last few years, attention has focused again on Weinberg’s NCB technique, and a number of detailed studies have assessed the prospects for detection with a tritium target [12–16]. In this type of an experiment, the smoking gun signature of C νB capture, ν + H3 → He 3 + e−, is a peak in the electron spectrum at an energy of 2 mν above the beta decay endpoint; see Fig. 1. Detecting this peak requires an energy resolution below the level of mν = O(0 .1 eV). Compared to other beta-decaying nuclei, tritium makes a particularly attractive candidate target because of its availability, high neutrino capture cross section, long lifetime (12 years), and low Q-value . For a 100 gram target, the expected
1In his paper, Weinberg reports of an experimental attempt being carried out by R. W. P. Drever at the University of Glasgow at the time of his writing, resulting in a preliminary bound on the C νB Fermi energy EF<500 eV. We have been unable to retrieve any other information on this early experiment.
– 2 – capture rate is approximately 10 events per year . So far, however, difficulties in achieving the necessary sub-eV energy resolution, and in controlling broadening of the electron energy distribution have precluded any serious experimental effort. In 2012/2013 the Princeton Tritium Observatory for Light, Early-Universe, Massive-Neutrino Yield (PTOLEMY), located at the Princeton Plasma Physics Laboratory, began developing a tech-nology that could help to solve the energy resolution challenges . The tritium nuclei will be deposited onto a source disk, such as a graphene substrate. This geometry helps to reduce electron backscatter, and thereby achieve an energy resolution of ∆ ∼ 0.15 eV, of the order of the neutrino mass scale. With this resolution and a 100 gram sample of tritium, PTOLEMY could transform C νBdetection from fantasy into reality. These recent advances, and especially the prospect of an having an experimental search in the near future, motivate studying the phenomenology of NCB in more detail. This is the spirit of our paper. In particular, the main novelties of our study are the sensitivity to the Dirac or Majorana nature of the neutrino, a more detailed analysis of the background rate, and the potential of the NCB to study a number of effects ranging from expected standard phenomenology, such as gravitational clustering and mass hierarchy, to more exotic ideas like lepton asymmetry, sterile neutrinos, neutrino decay and non-standard thermal history. The plan of the paper is as follows. In Sec. 2, we discuss the creation and evolution of the CνB neutrinos, and calculate the polarized neutrino capture cross section and the capture rate for tritium nucleus to clarify the difference between the Dirac and Majorana neutrinos. A detailed calculation of the neutrino capture kinematics and the polarized neutrino scattering amplitude is given in Appendix A. In Sec. 3, we focus on a PTOLEMY-like experiment, and treat the tritium beta decay as the main background for the tritium neutrino capture signal. In particular, we study the signal to noise ratio by taking into account the finite energy resolution of the detector, and find the required energy resolution for various neutrino masses. In Sec. 4, we discuss the difference between the Dirac and the Majorana neutrinos, the effect of the mass hierarchy, and gravitational clustering of neutrinos. In Sec. 5, we discuss the sensitivity to an eV (and sub-eV) scale sterile neutrino and a keV-scale warm dark matter sterile neutrino. In Sec. 6, we discuss various effects of new physics that can lead to an enhancement or suppression of the C νB number density, such as lepton asymmetry in the neutrino sector, neutrino decay, and late time entropy injection. A summary and discussion follow in Sec. 7.
2 Cosmic background neutrinos and their capture on tritum
In this section we will trace the history of a C νB neutrino, considering its production, propagation and detection. In reviewing the physics of these, we emphasize two critical points: the distinction between Dirac and Majorana neutrinos and the distinction between helicity and chirality. These are important to derive one of the main conclusions, namely that the C νB capture rate for Dirac and Majorana neutrinos differ by a factor of 2.
2.1 Thermal history of the C νB
Let us first discuss the production of neutrinos in the early universe, i.e. , their properties up to the point when they start free streaming. In the hot, dense conditions of the early universe, the neutrinos maintained thermal equilibrium with the plasma (electrons, positrons, and photons) through scattering processes such as
νe ←→ νe and e+e− ←→ ν ¯ν . (2.1) These processes are mediated by the weak interaction, therefore the neutrinos are produced as flavor eigenstates, νe, ν μ, ν τ , ¯νe, ¯νμ, ¯ντ . The scattering rate of the processes in Eq. (2.1) depends strongly on the temperature T , as Γ ≈ G2
F
T 5, where GF ≈ 1.2 × 10 −5 GeV −2 is the Fermi constant. At this time the spectrum of the neutrinos is thermal, given by the Fermi-Dirac distribution, fFD (p, T ) = – 3 – (1 + eE/T )−1, where E = √p2 + m2
ν
and T is the temperature of the plasma. Integrating over the phase space gives the number density of neutrinos per degree of freedom (flavor and spin):
nν (T ) = 3ζ(3) 4π2 T 3 . (2.2) (We will neglect the possibility of a lepton asymmetry for now, and return to this point in Sec. 6.1.) At a temperature of Tfo ∼ MeV, the scattering rate dropped below the Hubble expansion rate,
H ≈ T 2/M P (where MP ≈ 2.4 × 10 18 GeV), and as a consequence the neutrinos fell out of thermal equilibrium (“freeze out”). Effectively, the time of freeze out can be considered as the instant of production of the C νB neutrinos that we hope to detect today, since after this time the neutrinos simply free stream. In any case, it is easy to recognize that our conclusions do not depend on the exact instant of production of each neutrino. Between freeze out and the present epoch, neutrinos undergo a number of interesting effects, that we summarize below.
(i) redshift.
In the sudden freeze out approximation, the phase space distribution function after decoupling is given by an appropriate redshifting of the distribution function that was realized at decoupling. This leads to a modified Fermi-Dirac distribution 23
fν [p(z) , T ν (z)] = 1
ep(z)/T ν (z) + 1 , dn ν = d3p(z)(2 π)3 fν [p(z) , T ν (z)] , (2.3) where
p(z) = 1 + z
1 + zfo
pfo , Tν (z) = 1 + z
1 + zfo
Tfo (2.4) are the neutrino momentum and the effective neutrino temperature, respectively. Here they are expressed in terms of the momentum variable pfo , the neutrino temperature and redshift at freeze out, Tfo and zfo ' 6 × 10 10 .After neutrino freeze out, the C νB relic abundance is given by Eq. (2.2), where Eq. (2.4) gives the effective neutrino temperature. As the universe expands, z decreases and so too does Tν . Meanwhile the photons redshift like
Tγ (z) = 1 + z
1 + zfo
g∗(zfo )1/3
g∗(z)1/3 Tfo , (2.5) where g∗(z) = 45 s(z)/[2 π2T (z)3] and s(z) is the entropy density at epoch z. After electron-positron annihilation freezes out at T ≈ 100 keV, this entropy is transferred to the photons, which causes them to cool less quickly. This leaves the C νB at a relatively lower temperature,
Tν ≈ (4 /11) 1/3Tγ . (2.6) We can extrapolate until today when the temperature of the CMB is measured to be Tγ = 0 .235 meV . Then, the relationship above predicts the current temperature of the C νB to be Tν = 0 .168 meV. Using Eq. (2.2) this corresponds to a number density of
nν (z) = n0(1 + z)3, (2.7) where
n0 ≈ 56 cm −3 (2.8)
2This approximation agrees with exact solutions of the Boltzmann equation to within O(0 .2%) .
3Note that Eq. (2.3) is valid for any value of pand of the neutrino mass. In it, the mass term is suppressed by a factor of (1 + z)/(1 + zfo ) 1, which we neglect.
– 4 – per degree of freedom or 6 n0 ≈ 336 cm −3 for the entire C νB. Using Eq. (2.3), the root mean square momentum of neutrinos in the present epoch can be found to be
p0 ≈ 0.603 meV . (2.9) Since we are only interested in mν & 0.1 eV for the direct detection purposes, and p0 mν ∼ 0.1 eV, we assume that the C νB neutrinos are extremely non-relativistic today.
(ii) quantum decoherence.
As previously mentioned, neutrinos are produced as flavor eigenstates, να, which are a coherent superposition of mass eigenstates, νi: να = ∑
i
Uαi νi, with U being the Pontecorvo-Maki-Nakagawa-Sakata (PMNS) matrix [19–21] probed by oscillation experiments. Over time, the neutrino wavepacket decoheres as the different mass eigenstates νi propagate at different velocities . The timescale for this decoherence, ∆ t, can be estimated by solving (v1 − v2)∆ t ≈ λ where vi ≈ p/ √p2 + m2
i
≈ 1 − m2
i
/2p2 are the velocities of two mass eigenstates and
λ ≈ p−1 is the Compton wavelength of the wavepacket. The solution for ∆ t, in units of Hubble time (H−1 ≈ MP /T 2), is: ∆tH−1 ≈ 2pm22 − m21
T 2
MP
≈ 10 −7 , (2.10) where we used m2 ≈ 2m1 ≈ 0.1 eV and p ≈ Tfo ≈ 1 MeV. It is found that the flavor eigenstate C νBneutrinos quickly decohere into their mass eigenstates on a time scale much less than one Hubble time . Since we do not expect the decoherence to affect the relative abundances, we then conclude that neutrinos with the mass values of interest here, are present in the universe today as mass eigenstates, equally populated with an abundance given by Eq. (2.2).
2.2 Helicity composition of the C νB
Next, let us turn to the question of the neutrino spin state at production. Recall that a field’s chirality determines its transformation property under the Lorentz group, and that the weak interaction is chiral in nature, e.g. , the left-chiral component of the electron interacts with the weak bosons, but the right-chiral component does not. Therefore neutrinos (anti-neutrinos) are only produced in the left-chiral (right-chiral) state. Chirality should not be confused with a particle’s helicity, which is given by the projection of its momentum vector onto its spin vector. Since the C νB neutrinos are ultra-relativistic at freeze out ( Tfo mν ), we do not (yet) need to explicitly distinguish helicity and chirality, which exactly coincide for massless particles. For simplicity, here we will use the terminology “left-handed” to refer to a relativistic state that is left-helical and left-chiral, and we do similarly with the right-handed states. At this point is it convenient to enumerate all possible spin states. If the neutrinos are Dirac particles then we have four degrees of freedom per generation, which we will label as
νL left-handed active neutrino ¯νR right-handed active anti-neutrino
νR right-handed sterile neutrino ¯νL left-handed sterile anti-neutrino
. (2.11) Neutrinos and anti-neutrinos are distinguished by their lepton number, which is a conserved quantity. The states νL and ¯ νR are active in the sense that they interact via the weak interaction, while in contrast νR and ¯ νL are labeled as sterile because they interact only via the Higgs boson ( i.e. , the mass term). This interaction is suppressed by a very small Yukawa coupling yν ≈ mν /v ≈ 10 −12 , where
v ≈ 246 GeV is the vacuum expectation value of the Higgs field. The production mechanisms we have discussed above clearly apply only to the active states, which therefore acquire the abundance, nν (z), given by Eq. (2.7). Meanwhile, the sterile neutrinos can not come into thermal equilibrium with the SM, so it is reasonable to assume that their relic – 5 – abundance is negligible compared to that of the active states 4. Then, for the Dirac case, we expect the spin state abundances to be
n(νL) = nν (z)
n(¯ νR) = nν (z)
n(νR) ≈ 0
n(¯ νL) ≈ 0(2.12) where nν (z) is given by Eq. (2.7). The total C νB abundance is given by 6 nν (z) after summing over spin and flavor states. If the neutrinos are Majorana particles then lepton number is not a good quantum number, and we should avoid using the language “neutrino” and “anti-neutrino” 5. Instead, we will label the degrees of freedom as
νL left-handed active neutrino
νR right-handed active neutrino
NR right-handed sterile neutrino
NL left-handed sterile neutrino
. (2.13) As in the Dirac case, the active neutrinos interact weakly, and both the left- and right-handed states are populated at freeze out. The sterile neutrinos interact only through the Higgs boson, like in the Dirac case, but now they are typically much heavier than even the electroweak scale (see, e.g. ,[24–26]). As such, they will decay into a Higgs boson and a lepton, and their relic abundance today is zero. To summarize the Majorana case, we have
n(νL) = nν (z)
n(νR) = nν (z)
n(NR) = 0
n(NL) = 0 (2.14) where once again the total C νB abundance is 6 nν (t). Let us discuss how the neutrino quantum states evolve starting from the composition at freezeout, Eqs. (2.12) and (2.14). To describe the cooling of neutrinos down to the present time, we need to abandon the ultrarelativistic approximation, and therefore study the regime where helicity and chirality do not coincide. To do so, a key point to consider is that the helicity operator commutes with the free particle Hamiltonian, and its conservation is tied to the conservation of angular momentum. Instead, the chirality operator does not commute because of the mass term. Consequently, while the neutrinos are freely streaming, it is their helicity and not their chirality that is conserved . Thus, we can determine the abundances today from Eqs. (2.12) and (2.14) upon recognizing that “handedness” at freeze out translates into “helicity” today. Let us denote n(νhL ) as the number density of left-helical neutrinos, n(νhR ) as the number density of right-helical neutrinos, and so on. Then the abundances today are, for Dirac neutrinos:
n(νhL ) = n0
n(¯ νhR ) = n0
n(νhR ) ≈ 0
n(¯ νhL ) ≈ 0(2.15)
4One cannot exclude the possibility that there was a primordial abundance of sterile neutrinos, and to answer this question unambiguously one would have to specify the physics of the reheating phase that followed inflation. Nevertheless, it seems unlikely that this abundance was as large as nν(z) at the time of neutrino freeze out. As each of the SM fermion species froze out during the thermal history, they transferred their entropy to the remaining thermal species. Each of these entropy injections would have diluted the decoupled sterile neutrinos. (The physics is identical to the suppression of the C νB abundance relative to the CMB abundance after e+e−annihilation.)
5Our language here differs from conventions in the literature. When discussing Majorana neutrinos, it is customary to equate lepton number with chirality, such that the left-chiral particle is called a neutrino and the right-chiral particle is called an anti-neutrino. This language is very useful for discussing relativistic neutrinos, but impractical for non-relativistic neutrinos, for which we must distinguish helicity and chirality.
– 6 – and, for Majorana neutrinos:
n(νhL ) = n0
n(νhR ) = n0
n(NhR ) = 0
n(NhL ) = 0 (2.16) where n0 is given by Eq. (2.8). Note that the total abundance is the same, 6 n0, in both cases. However, the C νB contains both left- and right-helical active neutrinos in the Majorana case, but only left-helical active neutrinos in the Dirac case. Finally, we note that, if the neutrinos are not exactly free streaming, but instead they are allowed to interact, then the helicity can be flipped. This leads to a redistribution of the abundances in the Dirac case, n(νhL ) = n(νhR ) = n(¯ νhR ) = n(¯ νhL ) = n0/2, but no change in the Majorana case since the heavy neutrinos are decoupled. We will return to this point in Sec. 4.2 when we discuss gravitational clustering.
2.3 Detection of the C νB
In this section the rate of C νB capture on tritium is worked out. To best illustrate the role of helicity eigenstates, we start by discussing the case of the more elementary process of neutrino scattering on a neutron, and then generalize to the case of tritium.
(i) neutrino absorption on a free neutron.
Let us consider the process
νj + n → p + e− , (2.17) where the incident neutrino is taken to be in a mass eigenstate νj , following the discussion in the previous section. For this process, the kinematics can be easily worked out in the rest frame of the neutron. As per the discussion of Sec. 2.2, the neutrino is very non-relativistic, so we can take
Eν ≈ mν . After properly including the recoil of the proton, we find that the electron is ejected with a kinetic energy Ke = Ee − me, given by (see Appendix A.1)
Kcνb
e
≈ Kend + 2 mν , (2.18) where
Kend = (mn − me)2 − (mν + mp)2
2mn
= Q − meQmn
− Q2
2mp
(2.19) is the beta decay endpoint energy 6 and Q ≡ mn − mp − me − mν .We calculate the scattering amplitude for the processes in Eq. (2.17). Due to the low ener-gies involved, we can safely work in the four-fermion interaction approximation, and obtain (see Appendix A.2 for details):
iMj = −i GF
√2 Vud U ∗
ej
[ueγα(1 − γ5)uνj
] [ upγβ (f (0) − g(0) γ5) un
] ηαβ , (2.20) where ux is the Dirac spinor for species x, and Vud ≈ 0.97425 is an element of the Cabibbo-Kobayashi-Maskawa (CKM) matrix . The element Uej of the PMNS matrix appears because only the electron component of each mass eigenstate can participate in the process (2.17). The functions f (q) and g(q)are nuclear form factors, and in the limit of small momentum transfer they approach f ≡ f (0) ≈ 1and g ≡ g(0) ≈ 1.2695 .
6Neglecting nucleon recoil is equivalent to neglecting the last two terms in in Eq. (2.19), and gives the more familiar result Kcνb
e≈Q+ 2 mν. This approximation is not really legitimate, however, since the size of the neglected terms exceeds the neutrino mass: e.g. , for mν= 0 we get Q0≈0.7823 MeV, K0end ≈0.7816 MeV, and therefore
K0end −Q0≈ − 0.7 keV.
– 7 – We proceed to calculate the cross section by squaring the amplitude and performing the ap-propriate spin sums. In the neutrino capture experiment under consideration, the spins of the final state electron and nucleus are not measured, and therefore we must sum over the possible final states. Similarly, the initial nucleus is not prepared with a definite spin, and therefore we must sum over its two possible spins. However, as we discussed in Sec. 2.2, Dirac neutrinos are prepared in a definite spin state, they are left-helical, whereas both helicities are present if the neutrinos are Majorana. We will keep the calculation general for now. We denote the neutrino helicity by sν where sν = +1 /2corresponds to right-handed helicity and −1/2 to left-handed. Having performed the spin sums as discussed above, one finds the squared matrix element to be (see Appendix A.2 for details)
|M| 2
j
(sν ) = 8 G2
F
|Vud |2|Uej |2mnmpEeEν
[
A(sν )( f 2 + 3 g2) + B(sν )( f 2 − g2)ve cos θ
]
, (2.21) where θ is the angle between the neutrino and electron momenta, cos θ = pe · pνj /(|pe| ∣∣pνj
∣∣), and vi
is the velocity of the species i: vi ≡ | pi| /E i. The spin-dependent factors are
A(sν ) ≡ 1 − 2sν vνj =
{
1 − vνj , sν = +1 /2 right helical 1 + vνj , sν = −1/2 left helical ,B(sν ) ≡ vνj − 2sν =
{
vνj − 1 , sν = +1 /2 right helical
vνj + 1 , sν = −1/2 left helical . (2.22) If the neutrinos were relativistic, vνj ' 1, then we would find A = B = 0 for right-helical neutrinos, which implies that these particles cannot be captured, and A = B = 2 for left-helical neutrinos. This reproduces the familiar finding that in the relativistic limit helicity and chirality coincide, and only the left-chiral neutrinos interact with the weak force. In the non-relativistic limit, which is relevant here, we have A(±1/2) = ∓B(±1/2) = 1, indicating that both left- and right-helical neutrinos can be captured. We calculate the differential cross section from the squared amplitude, Eq. (2.21), in the standard way (see Appendix A.2), and get:
dσ j (sν )
d cos θ = G2
F
4π |Vud |2|Uej |2F (Z, E e) mpEepe
mnvνj
[
A(sν )( f 2 + 3 g2) + B(sν )( f 2 − g2)ve cos θ
]
(2.23) where F (Z, E e) is the Fermi function describing the enhancement of the cross section due to the Coulombic attraction between the outgoing electron and proton. It can be modeled as
F (Z, E e) = 2πη
1 − e−2πη , (2.24) with η = ZαE e/p e, and Z being the atomic number of the daughter nucleus ( Z = 1 here); α ≈
1/137 .036 is the fine structure constant. Since the incoming neutrino is practically at rest, pν pe, the kinematics allow for isotropic emission of the electron. Then the integral over θ is trivial, and one obtains the total capture cross section multiplied by the neutrino velocity, which is the quantity relevant for the capture rate:
σj (sν )vνj = G2
F
2π |Vud |2|Uej |2F (Z, E e) mp
mn
Ee pe A(sν )( f 2 + 3 g2) . (2.25) Since A(±1/2) = 1 in the approximation vνj 1, the cross section is identical for the two spin states. Therefore any differences in the capture rate of different spin states must arise from their abundance today, as will be seen below.
(ii) neutrino absorption on tritium.
Finally, let us generalize our results to the process
νj + H3 → He 3 + e− . (2.26) – 8 – The calculation of the cross section runs parallel to the derivation of Eq. (2.25), upon replacing n → H3
and p → He 3 . The neutron and proton masses are replaced with the nuclear masses of the species involved: mn → m H3 ≈ 2808 .92 MeV and mp → m He 3 ≈ 2808 .39 MeV. The same replacement must be done in Eqs. (2.18) and (2.19) to find the Q-value and the beta spectrum endpoint. Neglecting the neutrino mass, these evaluate to 7:
Q0 ≈ 18 .6 keV and K0end − Q0 ≈ − 3.4 eV . (2.27) Instead of the form factors, f (q) and g(q), one now encounters nuclear matrix elements that quantify the probability of finding a neutron in the H3 , on which the neutrino can scatter, and a proton in the He 3 . This requires the replacement f 2 → 〈 fF 〉2 ≈ 0.9987 and 3 g2 → (gA/g V )2〈gGT 〉2 where
〈gGT 〉2 ≈ 2.788, gA ≈ 1.2695, and gV ≈ 1 . After making the replacements described above, we obtain the velocity-multiplied capture cross section for mass eigenstate j:
σj (sν )vνj = A(sν ) |Uej |2 ¯σ , (2.28) where ¯σ ≡ G2
F
2π |Vud |2F (Z, E e) m He 3
m H3
Ee pe
(
〈fF 〉2 + ( gA/g V )2 〈gGT 〉2)
' 3.834 × 10 −45 cm 2 . (2.29) In the numerical estimate we use Ee = me + KCνB e and Eq. (A.21). Considering that for non-relativistic neutrinos, A(+1 /2) = A(−1/2) = 1, we obtain again that the capture cross section is the same for the left- and right-helical states, and is given by:
∑
j=1 ,2,3
σj (sν = ±1/2) vνj
∣∣∣vνj 1
= ¯ σ , (2.30) after summing over the mass eigenstates and using the unitarity of the PMNS matrix, ∑
j
|Uej |2 = 1. To clarify possible confusions, it is worth noting how this result is related to other commonly encountered cross sections, namely:
(i) the spin-averaged and mass summed cross section
This cross section is velocity-independent, because A(+1 /2) + A(−1/2) = 2 independent of vνj , and is: 12
∑
sν=±1/2
∑
j=1 ,2,3
σj (sν )vνj = ¯ σ , (2.31)
(ii) the cross section to capture relativistic neutrinos
This cross section vanishes for the right-helical state and for the left-helical state it is equal to twice our result:
∑
j=1 ,2,3
σj (sν = −1/2) vνj
∣∣∣vνj =1
= 2¯ σ = 7 .6 × 10 −45 cm 2 . (2.32) A cross section of this value has been used before in the context of C νB capture on tritium in both Refs. and , and the followup works in Refs. [14–16]. We emphasize that this is leads to an overestimate of the capture rate, and therefore it should be avoided. Moving on, finally we can calculate the total capture rate expected in a sample of tritium with mass MT. In Eq. (2.28) we have the capture cross section for a given neutrino mass and helicity
7We would like to stress that one expects to find the C νB signal at an energy that is displaced by 2 mν=O(0 .1 eV) above the beta decay endpoint, Ke=Kend , and that the endpoint itself is displaced by 3 .4 eV below the Q-value of the decay. Since 3 .4 eV mνone should take care not to confuse the endpoint and the Q-value.
– 9 – eigenstate. This requires summing over the cross section for each of the six initial states ( j = 1 , 2, 3and sν = ±1/2) weighted by the appropriate flux: Γcνb =
3
∑
j=1
[σj (+1 /2) vνj nj (νhR ) + σj (−1/2) vνj nj (νhL )] NT , (2.33) where NT = MT/m H3 is the approximate number of nuclei in the sample. Using Eq. (2.28) the capture rate can be written as Γcνb =
3
∑
j=1
|Uej |2 ¯σ [nj (νhR ) + nj (νhL )] NT = ¯ σ[n(νhR ) + n(νhL )]NT , (2.34) where ¯ σ was given by Eq. (2.29), and we used the fact that different neutrino mass eigenstates are equally populated to perform the sum over j. Here n(νhL ) and n(νhR ) are the number densities of left- and right-helical neutrinos per degree of freedom. We have also used A(−1/2) ≈ A(+1 /2) ≈ 1in the non-relativistic limit. Eq. (2.34) is the central result of this section. Let us see how it applies to the cases of Dirac and Majorana neutrinos, using the results of Sec. 2.2. If the neutrinos are Dirac particles, we saw that
n(νhL ) = n0 and n(νhR ) = 0, and the capture rate becomes ΓD
cνb
= ¯ σn 0NT . (2.35) Alternatively, for the Majorana case we found n(νhL ) = n(νhR ) = n0, and the capture rate becomes ΓM
cνb
= 2¯ σn 0NT . (2.36) That is, the capture rate in the Majorana case is twice that in the Dirac case: ΓM
cνb
= 2 Γ D
cνb
. (2.37) The relative factor of 2 is a central result of our paper. It can be understood as follows. In the Dirac case, we found that the C νB consists of only left-helical neutrinos and right-helical anti-neutrinos. If these neutrinos were in the relativistic limit, where helicity and chirality coincide, only the left-helical states could interact weakly. The right-helical states would be sterile, and only half of the background neutrinos would be available for capture. Since the C νB is non-relativistic, both the left- and right-helical states contain some left-chiral component, and therefore they both interact. The right-helical anti-neutrinos cannot be captured because the process ¯ ν + p → n + e+ is kinematically forbidden: it requires Eν > (mn + me − mp) ≈ 2 MeV in the proton rest frame, but the C νB neutrinos only carry
Eν ≈ mν . eV (similarly for the tritium). Thus in the Dirac case, only half of the C νB abundance is available for capture. On the other hand, for the Majorana case one does not distinguish neutrinos and anti-neutrinos; instead we find that the C νB consists of left-helical neutrinos and right-helical neutrinos, which both interact weakly and therefore are available for capture.
3 Detection prospects at a PTOLEMY-like experiment
Let us now turn to the phenomenology of a tritium-based experiment. Considering a target mass of 100 g, as is proposed for PTOLEMY , Eqs. (2.35) and (2.36) evaluate to ΓM
cνb
≈ 8.12 yr −1 and ΓD
cνb
≈ 4.06 yr −1 (3.1) for the Dirac and Majorana neutrino cases, respectively. These rates are limited only by the sample size, since they are independent of the neutrino mass (as long as the neutrinos are non-relativistic), and the C νB neutrino flux is fixed (in absence of exotica). One of the main challenges for a neutrino capture experiment is the energy resolution. The resolution of a detector quantifies the smallest separation at which two spectral features ( e.g. , two – 10 – peaks) can be distinguished. For instance, two Gaussian curves centered at E1 and E2, having equal amplitude, and having equal standard deviation σ can be distinguished provided that |E1 − E2| & ∆where ∆ = √8 ln 2 σ ≈ 2.35 σ (3.2) is the full width at half maximum (FWHM) of the Gaussian . The FWHM is conventionally taken to be the detector resolution. Applied to our case, this argument means that the spectral excess due to the C νB can be resolved if its separation from the beta endpoint exceeds the resolution: ∆Ke = 2 mν >∼ ∆. PTOLEMY is expected to achieve an energy resolution of ∆ = 0 .15 eV , just enough to probe the upper end of the neutrino mass spectrum, where the three masses mj are degenerate or quasi degenerate: |mi − mj | mj , i.e. , m1 ' m2 ' m3 = mν . In this situation, the mass splittings can not be resolved by the detector, and the signature of the C νB reduces to a single excess corresponding to the effective mass mν . Most of the discussion from here on will refer to this case. A brief discussion on possibly resolving the individual masses is given in Sec. 4.3. Tritium beta decay is the best known and likely the main source of background 8 for the C νBneutrino capture events. The effect of the finite energy resolution is that the most energetic electrons from beta decay might have measured energy that extends beyond the endpoint Kend , into the region where the signal is expected. To estimate the rate of such events, consider first the beta decay spectrum :
dΓβ
dE e
=
3
∑
j=1
|Uej |2 ¯σπ2 H(Ee, m νj )NT , (3.3) where
H(Ee, m νj ) ≡ 1 − m2
e
/(Eem H3 )(1 − 2Ee/m H3 + m2
e
/m 2H3 )2
√
y
(
y + 2mνj m He 3
m H3
) [
y + mνj
m H3
(m He 3 + mνj
)]
, (3.4) and y = me + Kend − Ee and the other variables are as in Sec. 2.3. After integrating over energy, the total tritium beta decay rate is found to be Γβ =
∫ me+Kend
me
dE e
dΓβ
dE e
≈ 10 24
( MT
100 g
)
yr −1 . (3.5) Comparing with the signal rate in Eq. (3.1), it appears immediately that even an extremely small contamination of beta decay events in the signal region can represent a serious challenge for C νBdetection. To calculate the number of background events, we model the observed spectrum by convolving the beta decay and C νB event “true” spectra with a Gaussian envelope of FWHM ∆ [Eq. (3.2)]:
d˜ΓM
cνb
dE e
= 1
√2π σ
∫ ∞−∞
dE ′
e
ΓM
cνb
(E′
e
) δ[E′
e
− (Eend + 2 mν )] exp
[
− (E′
e
− Ee)2
2σ2
]
(3.6)
d˜Γβ
dE e
= 1
√2π σ
∫ ∞−∞
dE ′
e
dΓβ
dE e
(E′
e
) exp
[
− (E′
e
− Ee)2
2σ2
]
. (3.7) In Fig. 2 we show the smoothed spectra and their sum for various different combinations of detector resolution and neutrino mass. For ∆ ≈ mν , the smoothed beta decay spectrum extends well beyond the endpoint energy at Ke − K0end ≈ − mν and contaminates the neutrino capture signal region at
Ke − K0end ≈ +mν .To estimate the potential to distinguish the signal from the background, we calculate the signal-to-noise ratio. Following , the calculation is done for an (observed) energy bin of width ∆ that is
8See Ref. for a discussion of additional backgrounds.
– 11 – -0.4 -0.2 0.0 0.2 0.4 0.6
0.1
100
10 5
10 8
10 11
Ke-Kend 0@eV D
dG ê dE e @ yr -1 eV -1 D
0.0 0.1 0.2 0.3 0.4 0.5 0.6
0
10
20
30
40
50
D = 0.2 eV
mΝ=0.2 eV
mΝ=0.3 eV
mΝ=0.4 eV -0.4 -0.2 0.0 0.2 0.4
0.1
100
10 5
10 8
10 11
Ke-Kend 0@eV D
dG ê dE e @ yr -1 eV -1 D
-0.1 0.0 0.1 0.2 0.3 0.4 0.5
0
20
40
60
80
D = 0.1 eV
mΝ=0.1 eV
mΝ=0.2 eV
mΝ=0.3 eV
Figure 2 . Solid lines: the expected spectrum of electrons in terms of observed energy, obtained from Eqs. (3.6) and (3.7), for detector resolution (FWHM) ∆ and neutrino mass mν . The dashed lines give the two contributions (signal and background) separately. The dotted lines show the spectrum of beta decay electrons for the ideal case of perfect energy resolution, ∆ ' 0. The zero of the horizontal axis coincides with beta decay endpoint (for perfect resolution) for massless neutrinos.
centered on the neutrino capture signal peak. In this bin, the signal and background event rates are: ˜ΓM
cνb
(∆) =
∫ Ecνb
e+∆ /2
Ecνb
e−∆/2
dE e
d˜Γcνb
dE e
(Ee) , (3.8) ˜Γβ (∆) =
∫ Ecνb
e+∆ /2
Ecνb
e−∆/2
dE e
d˜Γβ
dE e
(Ee) , (3.9) respectively, where Ecνb
e
≡ Kcνb
e
me + 2 mν , and their ratio is:
rsn = ˜ΓM
cνb
(∆) ˜Γβ (∆) . (3.10) In Fig. 3, contour plot of rsn = 1 is shown for a range of detector resolutions and neutrino masses. Successful detection of the C νB signal is impossible if rsn 1, and it is very likely if rsn 1. For a given ∆, the signal-to-noise ratio is a rapidly rising function of the neutrino mass (and therefore of the width of the gap in energy between the C νB signal and the beta spectrum endpoint), because the endpoint electrons are exponentially suppressed in the tail of the Gaussian. As a rule of thumb, for Majorana neutrinos we find that
rsn & 1 for ∆ . 0.7mν . (3.11) This condition is only slightly different for Dirac neutrinos, although the signal rate itself is lower by a factor of 2 [Eq. (3.1)]. This conclusion on the signal-to-noise ratio differs slightly from that in the similar analysis of Ref. . The difference is due to two aspects: (i) here rsn is obtained by numerically evaluating Eqs. (3.8) and (3.9). Instead, in Ref. the convolution integral is approximated by a factorized form for the beta decay background, which tends to underestimate rsn , and the C νB signal was not convolved with a Gaussian, which tends to overestimate the signal. (ii) here ∆, is identified with the Gaussian FWHM (under the advice of the PTOLEMY collaboration, ), and not with the Gaussian standard deviation σ as in Ref. . In terms of σ, our condition reads mν & 1.4( √8 ln 2 σ) ≈ 3.3σ,which is compatible with Ref. . In Table 1 we consider various values for the detector resolution and neutrino masses, and we show the expected signal event rates and signal-to-noise ratios for the Dirac and Majorana cases. We also show the effect of neutrino clustering; see Sec. 4.2 below. If rsn is large then the systematic error – 12 – 0.10 0.15 0.20 0.25 0.30 0.35 0.40
0.05
0.10
0.15
0.20
0.25
Neutrino Mass: mΝ @eV D
Detector Resolution: D @eV D
rSN << 1
rSN >> 1
CΝB signal hidden in Β-decay bkg.
CΝB signal stands out over Β-decay bkg.
Figure 3 . Contour plot of signal to noise ratio, rsn = 1 [Eq. (3.10)], for a range of detector resolutions, ∆, and neutrino masses, mν , for Majorana neutrinos with a degenerate neutrino mass spectrum. In the region below the rsn = 1 line, the C νB signal stands out over the beta decay background, and in the region above this line, the background events dominate. For Majorana neutrinos, rsn & 1 corresponds to ∆ . 0.7mν .
arising from the beta decay endpoint is negligible, and (in the absence of other systematic errors) the limiting factor is statistics. If N events are detected, then the counting error is expected to go like √N , and the statistical significance can be estimated as N/ √N = √N . A 3 σ detection requires
N ≈ 9 events in the signal region, and a 5 σ detection requires N ≈ 25 events. With an event rate of Γ M
cνb
≈ 8 yr −1 these significances would require approximately 1 and 3 years of data taking, respectively. ∆ (eV) mν (eV) ΓD
cνb
(yr −1) rD
sn
ΓM
cνb
(yr −1) rM
sn
f NFW c f MW c
0.10 0.15 4.1 (3.1) 37 8.1 (6.2) 74 1.4 1.6 0.20 0.30 4.6 9.2 3.1 4.4 0.30 0.45 1.4 2.8 6.4 10 0.40 0.60 0.6 1.2 12 20
Table 1 . The signal to noise ratio rsn [Eq. (3.10)] the neutrino capture rate, and the enhancement factor due to the gravitational clustering [Eq. (4.1)] for various values of the detector resolution, ∆, and of the neutrino mass mν . M and D stand for Majorana and Dirac neutrinos, respectively. The rates in parentheses refer to the neutrino capture event rates in the bin of width ∆ centered at Kcνb
e
= Kend + 2 mν [see Eq. (3.8)].
4 Detection prospects for varying neutrino properties
So far, we have discussed the simplest, “base” case of capture of neutrinos with a single mass and known density given by the cosmological prediction, n0 = 56 cm −3 per species. Here we elaborate further, and give a more detailed discussion of the phenomenology that is expected depending on the neutrino properties. Specifically, we discuss the distinction between Majorana and Dirac neutrinos, the correction to the rate due to neutrino clustering, and the effect of the neutrino mass hierarchy.
4.1 Majorana vs. Dirac neutrinos
When neutrinos are non-relativistic, the distinction between the Dirac and Majorana character be-comes pronounced. It is critical to recognize that the C νB represents the only known source of – 13 – non-relativistic neutrinos in the universe. As we saw in Sec. 2.3, the Dirac or Majorana character of the C νB neutrinos has a significant effect on C νB neutrino capture: the capture rate for Majorana neutrinos is double that of Dirac neutrinos [see Eq. (2.37)]. The factor of two difference can be under-stood as follows. For the Dirac case, only the left-helical neutrinos are available for capture since the right-helical neutrino population is absent from the C νB, and the anti-neutrinos cannot be captured. For the Majorana case, the C νB contains both left- and right-helical neutrinos, and the capture rate is doubled. In Table 1 we compare the signal rates for Dirac and Majorana neutrinos, as well as the corresponding signal to noise ratios, rsn .
4.2 Clustering and annual modulation
Like all massive particles, neutrinos should cluster in the gravitational potential wells of galaxies and clusters of galaxies. Due to clustering, the local number density, ncν , is larger than the unclustered case, n0, and the capture rate should therefore be enhanced by a factor
fc = nc
ν
n0
. (4.1) The calculation of fc requires solving the Boltzmann equation for the cosmic evolution of a system consisting of both cold dark matter and neutrinos, where they are treated as warm dark matter. A variety of approaches, based on different approximations and numerical techniques, have been presented [33, 34]. We show the results of Ref. in the last two columns of Table 1. There, fc is given for two different models of the dark matter halo of our galaxy, the so called Milky Way model and the Navarro-Frenk-White profile . For masses of the order of mν ∼ 0.1 eV, the effect of clustering should be at the level of few tens of per cent, comparable to the 1 σ statistical error expected at PTOLEMY in a few years or running (see Sec. 3). Therefore, the experiment may not be able to measure the local value of fc, but at least it will place a first stringent constraint on it. If the effect of clustering is indeed modest, it may be subdominant to the factor of 2 difference expected between Dirac and Majorana neutrinos, which could still be distinguished. An additional consequence of clustering is the mixing of neutrino helicities . As a gravi-tationally bound – but otherwise non-interacting – neutrino orbits around the halo, its momentum changes direction and magnitude, but its spin remains fixed. This causes helicity to change, so that a population of neutrinos initially prepared in a given helicity state ( e.g. , 100% initially right-helical) will in time grow a component of the opposite helicity, and ultimately reach an equilibrium where the right-helical and left-helical states are equally populated. We saw in Sec. 2.2 that the cosmolog-ical population of Dirac neutrinos (anti-neutrinos) consists of 100% left-helical (right-helical) states [Eq. (2.15)]. Assuming complete clustering ( i.e. , all the neutrinos available for capture are bound grav-itationally to the halo), the populations will equilibrate: n(νhL ) = n(¯ νhR ) = n(νhR ) = n(¯ νhL ) = n0/2. Majorana, neutrinos on the other hand are already equilibrated initially [Eq. (2.16)] and clustering will simply conserve the equilibrium: n(νhL ) = n(νhR ) = n0. After repeating the argument in Sec. 2.3, one finds that even with complete clustering the Majorana capture rate is still double that of the Dirac neutrinos. This is because for clustered Dirac neutrinos, the new population of right-helical states,
n(νhR ), compensates for the loss of the left-helical ones in Eq. (2.34). Finally, let us consider the possibility that the C νB signal rate could exhibit an annual modula-tion, similar to the one predicted for dark matter direct detection. This modulation could be due to the fact that if neutrinos are substantially clustered, then their velocity distribution relative to Earth is not isotropic and static, as it is usually assumed. The modulation should then follow the relative velocity of the Earth’s motion with respect to the galactic disk 9.In fact, the answer to the question of modulation is negative . As we saw in Eq. (2.34), the capture rate depends on the product of number density, cross section and neutrino velocity, vν . Since
9Clustering also produces a modified momentum distribution compared to unclustered neutrinos, specifically, for strong clustering the average momentum will be higher than that of the Femi-Dirac prediction . Additionally, the momentum distribution in the rest frame of the Earth will depend on the Earth’s motion relative to the galactic plane. As long as the neutrinos are non-relativistic, however, changes in the neutrino momentum distribution do not affect the capture rate.
– 14 – neutrino capture is an exothermic process, i.e. , some of the nuclear binding energy is liberated, the cross section scales as σ ∝ 1/v ν [13, 38]. Since the velocity cancels in Eq. (2.34), the rate is insensitive to the neutrino velocity, and thus, there should be no annual modulation of the signal. This is different from DM direct detection, which is an elastic scattering process, with Γ ∝ v. In contrast with DM, then, for C νB detection the astrophysical uncertainties on the velocity profile are not an issue. In this sense, C νB detection is cleaner than DM detection. If an annual modulation does appear at a C νBdetector, its origin would have to be traced elsewhere. For instance, an O(0 .1 − 1%) modulation may arise from the gravitational focusing from the Sun , even if the neutrinos are not clustered on the scale of the Milky Way.
4.3 The hierarchical mass spectrum
Let us now consider the mass differences between the different neutrino states. From the observation of oscillations, the degeneracy splitting is measured to be : ∆m221 ≈ (8 .66 meV) 2 and ∣∣∆m232
∣∣ ≈ ∣∣∆m231
∣∣ ≈ (48 meV) 2 . (4.2) The sign of ∆ m231 is yet unknown, allowing for two possible mass hierarchies (or “orderings”): normal hierarchy (NH): ∆m231 > 0 m1 < m 2 < m 3 (4.3) inverted hierarchy (IH): ∆m231 < 0 m3 < m 1 < m 2 . (4.4) In the coming years, long baseline experiments hope to distinguish these two scenarios . If the masses mj are comparable with the largest splitting, mj ∼ √|∆m231 | ≈ 0.05 eV, the degenerate, single-mass, approximation used so far becomes inadequate. This is likely to be the case: indeed, if the stringent cosmological bound on the masses, Eq. (1.1), is saturated then the spectrum can only be marginally degenerate, mνj ≈ 0.07 eV. In the hierarchical regime, C νB detection will not be possible without a significant improvement in the detector resolution. Nevertheless, we feel that it is illustrative to discuss how the signal qualitatively changes in this case. A detailed discussion is also given in Refs. [14, 15]. For a detector with an arbitrarily good energy resolution, ∆ mν , each mass eigenstate νj would make a distinguishable contribution to the C νB capture and to the beta decay spectrum as well. The beta decay spectrum would be the sum of three spectra, and its endpoint would be determined by the lightest neutrino mass, mmin = min[ mj ] ( mmin = m1 for NH, mmin = m3 for IH): Kend = K0end −mmin .For the C νB capture signal, each state νj would produce a distinct line at an electron kinetic energy of Ke j = K0end + mνj , or, equivalently:
Kcνb
e j
= Kend + mmin + mνj , (4.5) which recovers Eq. (2.18) in the degenerate regime. The total signal rate is still given by Eq. (2.34), but the three terms of the sum will appear as three separate excesses in the energy spectrum, each with weight |Uej |2, where :
|Ue1|2 ' 0.68 , |Ue2|2 ' 0.30 , and |Ue3|2 ' 0.02 . (4.6) Therefore, the signal is the strongest for ν1, weaker for ν2, and the weakest for ν3, as shown in Fig. 4. From the figure one can clearly see that, if we consider the effect of finite detector resolution, the CνB detection is easier in the IH case than for NH. Indeed, the IH case, ν1 and ν2 have the largest separation from the beta decay endpoint, and they have the strongest signal, making them easier to distinguish from the background. In the NH case, ν3 has the largest separation, but it has the weakest signal. Note that the intensity of the beta decay background also differs between the IH and NH cases. For the IH case, the endpoint is determined by ν3, which however contributes only proportionally to
|Ue3|2, hence the lower background rate. Instead, for the NH case the suppression of the beta spectrum near the endpoint is only |Ue1|2 [see Eq. (3.3)], corresponding to a higher background. In Fig. 4, two values of ∆ are considered. For ∆ = 0 .01 eV, in the NH case the signal is lost behind the background, but in the IH case the signal is clearly seen. The ν2 and ν1 eigenstates appear – 15 – -0.02 0.00 0.02 0.04 0.06
0
200
400
600
800
Ke-Kend 0@eV D
dG ê dE e @ yr -1 eV -1 D
D = 0.01 eV
mmin =0.001 eV
NH Hsolid L
IH Hdashed L-0.02 0.00 0.02 0.04 0.06
0
1000
2000
3000
4000
5000
6000
7000
Ke-Kend 0@eV D
dG ê dE e @ yr -1 eV -1 D
D = 0.001 eV
mmin =0.001 eV
NH Hsolid L
IH Hdashed L
Figure 4 . The C νB signal at an ultra-high-resolution detector. Each panel shows both hierarchies (NH and IH), with lightest neutrino being almost massless, mmin ≈ 1 meV. The Gaussian peaks are the C νB signal and the sloped lines are the beta decay background. The detector resolution is ∆ = 0 .01 eV in the left panel and ∆ = 0 .001 eV in the right panel.
as a single peak, because the resolution is insufficient to resolve the small mass gap between them: √∆m221 ≈ 8.66 meV < 0.015 eV. For an even more ambitious resolution, ∆ = 0 .001 eV, and NH, we can see the signal, and resolve both the ν2 and ν3 eigenstates. For IH, the signal is still visible, but the ν2 and ν1 eigenstates are still not resolved.
5 Probing sterile neutrinos
5.1 eV-scale sterile neutrinos
In addition to the three known flavor eigenstates of active neutrinos, there might exist other states that are inert, or “sterile” with respect to the Standard Model gauge interactions. Here we discuss sterile states that mix with the active states, and share their same helicity, so that they can be produced via active-sterile oscillations. Within this scenario, the most interesting case is that of a sterile neutrino, νs, and its corresponding mass eigenstate, ν4, with mass at the eV scale, m4 ∼ 1 eV. This additional sterile neutrino state is the favored interpretation of the anomalous excess of νe and ¯ νe
observed in νμ and ¯ νμ beams at LSND [40, 41] and MiniBooNE . It is also a possible explanation of the flux deficits observed in reactor neutrinos [43–45] and at solar neutrino calibration tests using gallium [46–48]. In presence of a fourth state, flavor mixing is described by a 4 × 4 matrix, with the elements Uα4
(α = e, μ, τ, s ) describing the flavor composition of ν4. The LSND / MiniBooNE experiments favor sin 2 2θ = 4 |Ue4|2|Uμ4|2 ∼ (1 − 10) · 10 −3 and ∆m241 = m24 − m21 ∼ (0 .1 − 10) eV 2 , (5.1) while global fits of all the anomalies favor the “democratic” value
|Uμ4|2 ∼ | Ue4|2 ' 3 × 10 −2 . (5.2) Here the electron-sterile mixing, Ue4, is of interest. With the values of mixings and masses given above, and in absence of other exotica, νs should be produced (via νμ→ νs and νe→ νs oscillations) before BBN with abundance at or close to thermal, so that its contribution to the radiation energy density is comparable to that of the active neutrinos. Interestingly, this is compatible with, or even favored by, recent cosmological data. Roughly, the situation is as follows: – 16 – (i) recent cosmological observations of an excess of radiation, Neff > 3, from both the BBN [51, 52] and CMB data [53–55], which therefore further support the indication of the existence of νs. (ii) The measurement of the Hubble constant by Planck is at tension with the local H0 data . (iii) The measurement of tensor perturbations by BICEP2 , is at tension with bounds on tensors from Planck’s CMB temperature data . It has been argued very recently that including a sterile neutrino yielding ∑
j
mνj ∼ 0.5 eV and ∆Neff ∼ 0.96 can resolve both the tensions at (ii) [58, 59] and (iii) [60–65]. It has to be noted, however, that data lends themselves to multiple interpretations and the situation is still evolving at this time (see e.g. , Ref. for a different view). The signature of ν4 at a tritium neutrino capture experiment is a line displaced by ∆Ke = m4 + mν (5.3) above the endpoint of the beta decay spectrum [see Eq. (4.5)] (see also Ref. ). The detection rate is proportional to the local number density of sterile neutrinos, n(νs), and to the appropriate mixing factor, |Ue4|2. Let us consider a basic scenario in which νs is produced via oscillations, in absence of other exotica, and accounts for the entire excess of radiation, ∆ Neff = Neff − 3.046. It can be shown (see, e.g. , [67, 68]) that its momentum distribution is the same as the one of the active neutrinos, up to a constant scaling factor, and therefore the local number density of ν4 is
n(νs) ' fc n0 ∆Neff , (5.4) where fc . 50 is the enhancement factor due to gravitational clustering (see Sec. 4.2 and Table 1). Thus, the ratio of the ν4 capture rate to the C νB active (Majorana) neutrino capture rate is Γν4
ΓM
cνb
≈ 0.6 ∆ Neff
( |Ue4|2
3 × 10 −2
) ( fc
20
)
, (5.5) or Γ ν4 ≈ 4.9 yr −1. The result in Eq. (5.5) refers to rather optimistic parameters, and therefore should be considered as the best case scenario. Although the rate is smaller than for the active species, its significance in the detector might be boosted by its larger separation from the the endpoint of the beta decay spectrum. The reason is twofold: first, the excess due to ν4 would be more easily resolved, even with a worse resolution than PTOLEMY; second, the region near the ν4 peak would be nearly background-free, since the beta decay spectrum falls exponentially with energy. These aspects are illustrated in Fig. 1.
5.2 keV-scale warm dark matter sterile neutrinos
The above discussion carries over for a sterile neutrino in the keV mass range (see Ref. ), which is a candidate for warm dark matter, and has number of interesting manifestations depending on its mixing with the active species. The strongest constraints on Ue4 in this mass range are
|Ue4|2 . O(10 −9) ; (5.6) they come from bounds on the abundance of νs in the early universe, and specifically from data on the spectrum of Large Scale Structures, on observations of the Lyman-α forest, and from X-ray observations constraining ν4 radiative decay (see e.g. , [70, 71] and references therein). Besides bounds, there are positive claims hinting at the existence of a keV-scale νs. Recently a 3 .5 keV X-ray line has been identified in various galaxy clusters [72, 73]. Interpreting this line with a decaying sterile neutrino state yields the parameters m4 ' 7 keV and mixing sin 2 2θ = 4 |Uα4|2 ' (2 − 20) × 10 −11 [72, 73]. Such small mixing values will lead to a corresponding suppression of the neutrino detection rate at PTOLEMY. However, this suppression is partially offset by an enhancement: with its larger mass,
ν4 can cluster much more efficiently, and therefore its local abundance could be much larger than the unclustered C νB abundance. Specifically, if we assume that ν4 account for 100% of the dark matter local density, ρDM ' 0.3 GeV cm −3 , the clustering enhancement factor is:
fc ≈ ρDM /m 4
n0
' 7.6 × 10 2
( 7 keV
m4
)
, (5.7) – 17 – Taking both the mixing suppression and the clustering enhancement into account, the expected rate at PTOLEMY is given by Γν4
ΓM
cνb
' | Ue4|2fc ' 7.6 × 10 −9
( |Ue4|2
10 −11
) ( 7 keV
m4
)
. (5.8) Thus, we conclude that the interesting region of the parameter space is out of reach of this type of experiment, although interesting, complementary bounds on νs could be obtained .
6 Sensitivity to other non-standard neutrino physics
We now turn to other possible effects that might enhance or suppress the C νB capture signal, such as the lepton asymmetry in the neutrino sector, neutrino decay, and the entropy injection after the neutrino decoupling.
6.1 Lepton asymmetry
It is established that the universe possesses a cosmic baryon asymmetry, defined as the difference between the number density of baryons and that of anti-baryons: nB = nb − n¯b. Normalized to the photon density, the asymmetry is nB /n γ ≈ 10 −10 . A neutrino asymmetry, nL = nν − n¯ν , is also expected in many models of baryogenesis. In most models it is expected to be comparable to nB ,however there are cases ( e.g. , [75–77]), where O(10 −3)−O(1) lepton asymmetry in the neutrino sector can be created, and the current constraints are at the level of nL/n γ . 0.1 − 0.5 . In Eqs. (2.11) and (2.13) we enumerated the degrees of freedom for Dirac and Majorana neutrinos. An asymmetry may arise between states which are CP conjugates to one another. If the neutrinos are Dirac particles, then this asymmetry is manifest as n(νhL ) 6 = n(¯ νhR ), and is conserved in the absence of lepton-number violating interactions. In the Majorana case, the asymmetry means n(νhL ) 6 =
n(νhR ), and is approximately conserved as long as the helicity-flipping rate is smaller than the Hubble expansion rate . As discussed in Sec. 2.2, this is the case for free-streaming neutrinos. Let us start by considering the Dirac case, and generalize the neutrino distribution function, Eq. (2.3), to include an asymmetry. We will assume that each of the three mass eigenstates carries the same asymmetry, because equilibration of flavor is generally expected due to oscillations (see e.g. ,[80, 81]). Let μν be the chemical potential and ξν = μν /T ν . Then the number density and energy density of neutrinos are:
n(νhL ) = Nf
∫ d3p
(2 π)3
1
e(p−μν )/T ν + 1 ≈ 3Nf ζ(3) 4π2 T 3
ν
Nf
12 ξν T 3
ν
O(μ2
ν
Tν ) , (6.1)
ρ(νhL ) = Nf
∫ d3p
(2 π)3
pe(p−μν )/T ν + 1 ≈ 7Nf π2
240 T 4
ν
9Nf ζ(3) 4π2 ξν T 4
ν
Nf
8 ξ2
ν
T 4
ν
O(μ3
ν
Tν ) ,
where Nf = 3 reflects the sum over flavors, p ≡ | p|, and we have assumed ξν 1 in the expansions on the left side. The corresponding quantities for anti-neutrinos are given by a change of the sign in
ξν : n(¯ νhR ) = n(νhL )|ξν →− ξν , etc. As we saw in Sec. 2.3, only n(νhL ) is relevant for C νB detection. We immediately see that, compared to the symmetric case ( ξν = 0) n(νhL ) is enhanced (suppressed) if ξν > 0 ( ξν < 0). Therefore, the C νB capture rate will have a corresponding enhancement (suppression) factor:
f D
ξ
= n(νhL )
n(νhL )|ξν =0
' 1 + π2
9ζ(3) ξν ≈ 1 + 0 .91 ξν . (6.2) For Majorana neutrinos, the calculation proceeds from Eq. (6.1) in a similar way, however here the quantity relevant to C νB detection is the sum n(νhL ) + n(νhR ) [see Eq. (2.34)]. Upon summing, the term linear in ξν cancels out, and the enhancement factor in this case is instead
f M
ξ
= n(νhL ) + n(νhR )[n(νhL ) + n(νhR )] ξν =0
' 1 + 2 ln 2 3ζ(3) ξ2
ν
≈ 1 + 0 .38 ξ2
ν
, (6.3) – 18 – ξν f D
ξ
f M
ξ
∆Neff
0.30 1.31 1.03 0.12 0.45 1.50 1.08 0.27 0.60 1.71 1.14 0.48 0.90 2.21 1.32 1.10 -0.30 0.76 1.03 0.12 -0.45 0.66 1.08 0.27 -0.60 0.57 1.14 0.48 -0.90 0.43 1.32 1.10
Table 2 .The Dirac and Majorana capture enhancement factors, fD
ξand fM
ξ, as well as ∆ Neff , for given values of the lepton asymmetry parameter ξν.Results are obtained by exact calculation [phase space integrations on the left hand side in Eq. (6.1)].
therefore, for Majorana neutrinos capture is always enhanced by asymmetry. The lepton asymmetry also translates into an additional energy density,
ρtot
ν
= ρ(νhL ) + ρ(¯ νhR ) ≈ 7Nf π2
120 T 4
ν
Nf
4 ξ2
ν
T 4
ν
, (6.4) that increases regardless of the sign of ξν . In cosmology the proxy for ρtot
ν
is the commonly quoted effective number of neutrinos, Neff [Eq. (1.1)]:
Neff = Nf
ρtot
ν
ρtot
ν
|ξν =0
≈ Nf + 30 Nf
7π2 ξ2
ν
, (6.5) where we can immediately read the excess due to the asymmetry: ∆Neff = Neff − Nf ' 30 Nf
7π2 ξ2
ν
' 1.3 ξ2
ν
. (6.6) The bound on Neff , Eq. (1.1), thus imply a bound on ξν . Additionally, a strong bound on ξν arises from constrains on the neutron to proton ratio at BBN. Table 2 shows f D
ξ
, f M
ξ
and ∆ Neff for a set of values of ξν . From this table, and from Eq. (6.6), we can infer the maximum capture enhancement allowed by cosmology. The Planck satellite constraint on ∆Neff , Eq. (1.1), translates into |ξν | . 0.5. A more careful analysis of CMB data (WMAP9, SPT, and ACT) finds that an anti-neutrino excess is preferred, roughly −0.4 . ξν . 0.2 , where the exact range depends on the combination of data sets used. This interval corresponds to 0 .6 . f D
ξ
. 1.2 and 1.0 . f M
ξ
. 1.1. When the He 4 abundance is folded in, the bound tightens to −0.091 . ξν . 0.051 , corresponding to a negligible effect on the neutrino capture rate.
6.2 Neutrino decay
Being massive and lepton flavor-violating, neutrinos could be unstable. Given the neutrino mass eigenstate νi, with proper lifetime τi, observational constraints on its decay are usually expressed in terms of lifetime-to-mass ratio, τi/m i (see, e.g. , Ref. for a collection of the current limits). The best model-independent constraint derives from the measured supernova neutrino flux in SN1987A :
τm > 10 5 s · eV −1 (6.7) for the mass eigenstates ν1 and ν2. In order to discuss model-dependent constraints, it is convenient to classify the decay channels as:
• Radiative, “visible”, decay. One of the decay products is a photon. – 19 – • “Weak” decay. One of the decay products is a (lighter) neutrino, and the other products are invisible. For example, it could be that all the neutrinos ultimately decay down to the lightest neutrino species.
• Invisible decay. The decay products are exotic, non-interacting particles such as sterile neutrinos. Very strong limits are placed on the radiative decay channel from solar ν and γ fluxes
τm & 7 × 10 9 s · eV −1 (6.8) for the ν1 ≈ νe mass eigenstate. Because visible decay channels are already strongly constrained, we will focus on the weak and invisible decay channels.
(i) invisible decay.
If a neutrino completely decays into invisible particles, then the expected C νB capture rate will be suppressed or vanish completely depending on the lifetime. For neutrinos with proper lifetime τ 0
ν
, the suppression factor due to the decay of into invisible particles is (see, e.g., )
f inv d = e−λν , (6.9) where
λν =
∫ dt τν
=
∫ zfo
0
dz
(1 + z)H(z)γ(z)τ 0
ν
. (6.10) Here τν (z) = τ 0
ν
γ(z) is the Lorentz-dilated lifetime at epoch z, zfo ' 6×10 9 is the neutrino decoupling epoch, and the Hubble parameter and the Lorentz factor of a neutrino are respectively given by
H(z) = H0
√Ωr(1 + z)4 + Ω m(1 + z)3 + Ω Λ , γ(z) = Eν
mν
=
√
p20
m2
ν
(1 + z)2 + 1 , (6.11) where H0 = 67 .04 km s −1 Mpc −1, Ω r = 9 .35 × 10 −5, Ω m = 0 .3183, Ω Λ = 0 .6817 and p0 is the neutrino momentum in the present epoch [Eq. (2.9)]. The calculation of λν is greatly simplified by considering that the integral in Eq. (6.10) is dom-inated by the recent epoch, z 1, and that for masses of interest here, the neutrinos were already non-relativistic at that time: mν ∼ 0.1 eV p0 [Eq. (2.9)]. Thus, one expects (and the full calculation confirms this) the non-relativistic result λν ∼ t0/τ 0
ν
, and hence,
f inv d ∼ e−t0/τ 0
ν
, (6.12) where the age of the universe is t0 = 4 .36 × 10 17 s. From Eq. (6.12) it follows that a detection of the C νB at PTOLEMY, at a rate consistent with the standard value, would place constraints on the invisible decay rate at the order τ 0
ν
∼ t0. Instead, a significant suppression, resulting in a negative search, could be evidence for neutrino decay implying a upper bound on the lifetime, τ 0
ν
. t0.Interestingly, the sensitivity to the lifetime is not of the usual form τ /m ν : we can really con-strain the lifetime regardless of the mass, provided that the mass is in the range of sensitivity of the experiment. This is because this decay test is done with non-relativistic neutrinos, a unique aspect of this setup. For comparison with currently available limits, however, we can express the sensitivity as
τ /m ν ∼ t0/m ν ≈ 4.36 × 10 18 s · eV −1
( 0.1 eV
mν
)
, (6.13) which is enormously better than the current model-independent limit, Eq. (6.7), and competitive with the cosmological limit for radiative decay, Eq. (6.8) 10 . In this way, a C νB direct detection experiment would serve as a complementary probe to other astrophysical searches for neutrino decay.
10 Strong indirect limits are available, see for instance [27, 85].
– 20 – (ii) weak decay.
Let us consider the case of complete decay of all the C νB neutrinos down into the lightest mass eigenstate, which is ν1 for NH or ν3 for IH, see Sec. 4.3. As a consequence, the neutrino population today is entirely made of this state, which is therefore three times more abundant than for stable neutrinos. This means that Eq. (2.34) should be modified by replacing ∑
j
|Uej |2 = 1 with 3 |Uei |2,where i = 1 for NH and 3 for IH. The result is that the capture rate is enhanced or suppressed by a factor
f wd = 3|Uei |2
∑
j
|Uej |2 =
{
3|Ue1|2 ≈ 2.03 NH 3|Ue3|2 ≈ 0.068 IH . (6.14) For the IH case, neutrino weak decay would lead to a null result. On the other hand, detection would be enhanced in the NH case, provided that the detector resolution is good enough to resolve m1. The observation of an anomalous rate compatible with Eq. (6.14) would result in a lower or upper bound on the neutrino proper lifetime, along the same argument as in the case of invisible decay. In case of an incomplete decay, the value f wd is intermediate between 1 and the results in Eq. (6.14).
6.3 Non-standard thermal history
The predicted C νB detection rate depends sensitively on the temperature of the relic neutrinos, via the relationship between the temperature and the number density [Eq. (2.2)]. For example, in the Majorana neutrino case our calculated rate is ΓM
cνb
' 8 yr −1
( Tν
1.9◦ K
)3
. (6.15) Supposing that new physics were to affect the C νB temperature (while maintaining the thermal distribution), it is immediately clear from Eq. (6.15) that the C νB detection rate could be altered dramatically with even a small temperature change: for Tν ' 4◦ K we would have Γ cνb ' 64 yr −1.Conversely, a colder C νB leads to a smaller capture rate. Needless to say, the C νB temperature has never been directly measured. Its value is predicted to be Tν = T std
ν
' 1.9◦ K using the observed temperature of the CMB, Tγ ' 2.7◦ K, and the relationship between Tν and Tγ [see Eq. (2.5)]:
Tν
Tγ
= g1/3
∗
(0)
g1/3
∗
(zfo ) =
( 411
)1/3
, (6.16) Here g∗(z) is the effective number of relativistic species. After neutrino freeze out, the plasma consisted of electrons, positrons, and photons giving g∗(zfo ) = 2 + (7 /8)4 = 11 /2. After e+e− annihilation all the the entropy is transferred to the photons for which g∗(0) = 2. It is possible that the C νB temperature could be substantially different than Tν if the thermal history of the universe were modified. Specifically, we will suppose that physics beyond the Standard Model is responsible for an entropy injection. For example, in analogy with the e+e− annihilation scenario, we can consider a new species of particle that is initially coupled to the plasma but decouples and transfers its entropy to the remaining thermalized species. Alternatively, the entropy injection could arise from an out-of-equilibrium decay or a first order phase transition. If the injection occurs before neutrino decoupling, then both the photons and the neutrinos are heated. This delays neutrino decoupling, but once the neutrinos have frozen out, the ratio Tν /T γ is unaffected; it is still controlled by e+e− annihilation. Next suppose that entropy is injected into the photons after neutrino decoupling but before recombination. This heats the photons, which must cool for a longer time to reach the measured value of 2 .7◦ K, and causes the neutrinos to be relatively colder. The C νB temperature is calculated using Eq. (6.16) where g∗(0) = 2 and g∗(zfo ) = 11 /2 + ∆ g where ∆ g counts the additional degrees of freedom that were in equilibrium prior to the entropy injection. For instance, if the entropy arises – 21 – from the freeze out of a single Dirac species then ∆ g = (7 /8)4 and Tν /T γ = (2 /9) 1/3. This implies a colder C νB, Tν ' 1.6◦ K, and a lower C νB capture rate, Γ cνb ' 5 yr −1.It seems unlikely that an entropy injection could result in a heating of the C νB neutrinos. Even if the species that freezes out decays into neutrinos (see, e.g. , ), this will not increase the C νBtemperature, but instead it will lead to a non-thermal spectrum, since the neutrinos are already free streaming. A constraint on the C νB temperature, and therefore on entropy injection, arises from the mea-surement of Neff ' 3 from the CMB. Recall that Neff gives the energy density of relativistic species at the surface of last scattering normalized to the expected C νB temperature. In the standard thermal history, the C νB temperature is equal to T std
ν
at the surface of last scattering, and the neutrinos con-tribute Neff ' 3. If the neutrinos had a non-standard temperature Tν < T std
ν
then their contribution is suppressed as Neff ' 3( Tν /T std
ν
)4. The Planck measurement of Neff , Eq. (1.1), translates into the interval 1 .95 ◦ K < T ν < 2.03 ◦ K. To allow a larger deviation of Tν from the standard value, one would have to introduce new relativistic degrees of freedom with just the right energy density to compensate for the energy lost by considering the colder C νB.
7 Discussion
The detection of the C νB via capture on tritium is conceptually interesting, and, for the first time, possibly realistic. The existence of a specific experimental proposal, PTOLEMY, motivates the present study on the phenomenology of this technique. The planned active mass of PTOLEMY is 100 g of tritium, for which the predicted rate is Γ ' (4 − 8) yr −1.Some of the major challenges for a C νB capture experiment are the energy resolution and the background control. The signal (if any) due to the C νB will partially overlap with the background from beta decay, and it is reasonable to expect that the signal and the background might be comparable. The estimated energy resolution at PTOLEMY will be ∆ ∼ 0.15 eV; if the neutrino masses are on the order of 0 .07 eV close to the upper limit allowed by cosmology [Eq. (1.1)], then this resolution is nearly enough to distinguish the signal from the background [Eq. (3.11)], but it is not sufficient if the neutrinos are substantially lighter, in the hierarchical spectrum regime. Since PTOLEMY will probe only a portion of the parameter space, it is not guaranteed to succeed. Still, it will represent an important first step towards the development of more sophisticated technologies for C νB capture. The spirit of our study is to address the question of what fundamental physics can be learned from a C νB capture experiment, with emphasis on PTOLEMY, but an open mind towards even more ambitious possibilities. Below, the main results of our study are summarized. 1. For 100 grams of tritium, the C νB capture rate is found to be Γ D
cνb
' 4 yr −1 for Dirac neutrinos and Γ M
cνb
' 8 yr −1 for Majorana neutrinos [Eq. (3.1)]. This confirms previous calculations [12, 13] where the rate was also found to be 8 yr −1, although without distinguishing the nature of the neutrinos or working with the polarized capture cross section [see below Eq. (2.32)], as we have done here. This relative factor of 2 between the Dirac and Majorana cases has to be taken into account when planning an experimental setup, as it could spell the difference between an indication of the C νB and its discovery. 2. A C νB capture experiment will probe non-relativistic neutrinos. This kinematical regime is completely unexplored at this time, and may reveal interesting properties that are not accessible in the ordinary relativistic regime, such as the distinction between Dirac and Majorana nature of neutrinos, as we discussed above. This is in striking contrast with the smallness of corrections at the relativistic regime [87, 88]. In principle, the PTOLEMY concept combines two very attractive features that are traditionally separated: the kinematic measurement of the neutrino mass from nuclear decays (which is relatively well-understood but insensitive to the origin of neutrino mass) and the ability to distinguish between Dirac and Majorana neutrinos. The latter so far has been an exclusive feature of neutrinoless double beta decay [89–91]. 3. The m4 ∼ 1 eV sterile neutrino favored by MiniBooNE and other oscillation searches could appear at PTOLEMY with a remarkably clean and unambiguous signature: an excess of up to – 22 – 5 events per year [Eq. (5.5)] with an electron kinetic energy that should easily be distinguished from that caused by the active neutrinos and the beta decay endpoint [Eq. (5.3)]. In absence of other exotica, this neutrino should be produced copiously (at or close to thermal abundance) in the early universe. This detection could completely resolve the confused situation that we have inherited from oscillation experiments, where different searches lead to conflicting results and open questions exist on systematic uncertainties and parameter degeneracies. Additionally, it could also help to resolve the tension between cosmological probes of neutrinos, specifically measurements of Neff , as well as resolving the tension between B-mode polarization data from BICEP2 and the Planck bound. It should be noted, though, that the absence of an excess at the eV-scale would not exclude the LSND / MiniBooNE sterile neutrino, but instead restrict the allowed region of |Ue4|. The keV-scale sterile neutrinos require much smaller mixing angles if they are to be the dark matter, and this implies a correspondingly small capture rate [Eq. (5.8)]. 4. A direct detection of the C νB would be a unique probe of what happened to neutrinos since the CMB decoupling time. In principle, it can vastly improve constraints on neutrino decay, and a detection of the C νB would imply a neutrino lifetime longer than the age of the universe [Eq. (6.13)]. Interestingly, this bound would be on the neutrino lifetime itself, and not on the ratio of lifetime and mass that is probed with relativistic neutrinos. A direct detection would also provide the unique opportunity to probe the coupling of neutrinos to gravity through the local neutrino overdensity, and thereby explore late-time phenomena such as neutrino clustering. Since the C νB capture rate goes like the third power of the C νB temperature, direct detection may be used to test non-standard thermal histories in which the neutrinos are heated or cooled by a late-time entropy injection. 5. We have found that many of the variants on standard neutrino physics lead to enhancements or suppressions of the C νB capture rate. These include gravitational clustering [Eq. (4.1) and Ta-ble 1], weak decay of neutrinos [Eq. (6.14)], the presence of a lepton asymmetry [Eqs. (6.2) and (6.3) and Table 2], and a non-standard thermal history [Sec. 6.3]. Certainly, one has to be mindful of uncertainties and degeneracies. Since a direct detection of the C νB will only provide two pieces of data, the ν mass scale and the detection rate, it would be impossible to distinguish between different causes of enhancement or suppression of the C νB, unless the neutrino capture data are combined with the indirect information from cosmological measurements. By the time that the PTOLEMY experiment becomes operational, some of the neutrino param-eters will hopefully have been measured by other experiments, e.g. , the mass hierarchy by accelerator experiments, the mass scale via cosmology and beta decay, and the Dirac or Majorana character via neutrinoless double beta decay. This information will be a great advantage to PTOLEMY by helping to break the degeneracies in neutrino parameters (discussed above) and thereby allow PTOLEMY to draw more solid conclusions about the physics of the C νB. We want to emphasize that the detection of the C νB will not only be a boon to the field of neutrino physics and cosmology, but also could lead to interesting and unexpected new physics that could manifest itself in the regime where neutrinos are non-relativistic. Therefore, the agenda for PTOLEMY and similar proposals might become richer than previously considered.
Acknowledgments
We are grateful to D. Chung, R. Gran, G. Mangano, M. Messina, C. Quigg and C. Tully for very useful discussions. C.L. and E.S. acknowledge the National Science Foundation grant number PHY-1205745. A.J.L. is supported by the DOE under Grant No. DE-SC0008016.
A Amplitude and cross section for polarized neutrinos
A.1 Kinematics
In this section, we present the kinematic relations that arise in the calculations of tritium beta decay and neutrino capture on tritium. The calculation treats only the nuclear process (masses m H3 and – 23 – m He 3 are nuclear masses), and we comment on the role of the atomic electron at the end. We work in the rest frame of the tritium nucleus. Since tritium beta decay is a three-body process, H3 → He 3 + e− + νj , the electron can be emitted with a range of momenta. The maximum possible momentum is obtained when the electron is emitted anti-parallel to both the helium-3 nucleus and the neutrino. This momentum demarcates the beta decay endpoint,
pend = 12m H3
√
m2H3 − (m He 3 + mν + me)2
√
m2H3 − (m He 3 + mν − me)2 . (A.1) The corresponding electron kinetic energy is given by Kend = √p2end + m2
e
− me or
Kend = (m H3 − me)2 − (m He 3 + mν )2
2m H3
. (A.2) It is convenient to introduce the Q-value, defined by 11
Q ≡ m H3 − m He 3 − me − mν , (A.3) which corresponds to the total kinetic energy carried away by all three decay products. Note that no single particle can have a kinetic energy equal to Q because this would require a violation of momentum conservation, that it, it would neglect the recoil of the other decay products. In terms of
Q, the endpoint energy can be written
Kend = Q − Krecoil (A.4) where
Krecoil = me
m H3
Q + Q2
2m H3
(A.5) is the amount of kinetic energy unavailable to the electron, because it goes into the recoil of the helium-3 nucleus and the neutrino. It is also convenient to identify the energy
K0end = (m H3 − me)2 − m2He 3
2m H3
, (A.6) which is where the endpoint would be located if the neutrino were massless. Next let us consider the kinematics of the neutrino capture process, νj + H3 → He 3 + e−. The calculation simplifies greatly if we neglect the momentum of the incident neutrino. For typical C νBneutrinos, which have a momentum p0 ≈ 6 × 10 −4 eV and a mass mν ≈ 0.1 eV, this is a very good approximation. A simple calculation gives the kinetic energy of the emitted electron to be
KCνB
e
= (m H3 − me + mν )2 − m2He 3
2( m H3 + mν ) . (A.7) This is the energy at which the C νB signal will be located. Its displacement above the beta decay endpoint is given by ∆K = KCνB
e
− Kend = (m H3 + m He 3 + mν )2 − m2
e
2m H3 (m H3 + mν ) mν . (A.8) If we now make the well-justified approximations m H3 ≈ m He 3 me mν , we arrive at the simple result ∆ K ≈ 2mν or equivalently,
KCνB
e
≈ Kend + 2 mν . (A.9)
11 Alternatively, the Q-value may be defined as ¯Q=MH3−MHe 3where these are the atomic masses of tritium and helium-3. The difference between Qand ¯Qis the O(10 eV) atomic binding energies.
– 24 – Since mν p0, we were justified in dropping the neutrino momentum at the start. We have focused here on the kinematics of the nuclear processes, but the system we are really interested in is a neutral tritium atom converting into a helium ion. The energy of the emitted electron, however, should be insensitive to the presence of an atomic cloud, both in the beta decay and neutrino capture process. The nuclear process occurs on a short time scale, and on a much longer time scale the bound electron finds itself in an excited state of the helium atom. The helium ion relaxes to its ground state by emitting a photon. For this reason, one should not calculate the kinematics using the atomic states; the photon energy must be included as well, and this approach makes the calculation unnecessarily complicated. We will conclude this appendix by numerically evaluating the kinematical variables using the measured masses. Although it is not necessary to perform this exercise, since Eq. (A.9) depends only on the neutrino mass, we feel that it is illustrative to the reader. The nuclear masses of tritium and helium-3 are not provided directly in the AME2003 tables . Instead they must be derived from the atomic masses, which are
M H3 (atomic) = 3 016 049 . 2777(25) μu (A.10)
M He 3 (atomic) = 3 016 029 . 3191(26) μu (A.11) where u = 931 .4940090(71) MeV. The nuclear masses are then calculated as
m H3 (nuclear) = M H3 (atomic) − me + 13 .59811 eV (A.12)
m He 3 (nuclear) = M He 3 (atomic) − 2me + 24 .58678 eV (A.13) where the last term on each line is the atomic binding energy . The parenthetical values show the 1 σ errors, and the binding energies have negligible error. Taking also the measured electron mass from Ref. we have
me ≈ 510 . 998 910(13) keV (A.14)
m H3 ≈ 2 808 920 . 8205(23) keV (A.15)
m He 3 ≈ 2 808 391 . 2193(24) keV (A.16) and
Q ≈ 18 .6023(34) keV − mν (A.17)
Krecoil ≈ 0.003445729(86) keV + O(m2
ν
) (A.18)
K0end ≈ 18 .5988(34) keV (A.19)
Kend ≈ 18 .5988(34) keV − mν (A.20)
KCνB
e
≈ 18 .5988(34) keV + mν (A.21) ∆K ≈ 2mν . (A.22) The error is dominated by the uncertainty in the atomic masses. Although the error bars on Kend and
KCνB
e
are on the order of 3 .4 eV, and therefore much larger than the neutrino mass, the displacement ∆K is insensitive to these uncertainties.
A.2 The polarized neutrino capture amplitude
Here we provide some of the details behind the cross section calculation in Sec. 2.3. To our knowledge the literature does not contain an explicit calculation of the polarized neutrino capture cross section for this process. Starting with the matrix element in Eq. (2.20), we first calculate the modulus
|M| 2 = G2
F
2 |Vud |2|U ∗
ej
|2T αγ
1
T βδ
2
ηαβ ηγδ , (A.23) – 25 – where
T αγ
1
≡ Tr [γα(1 − γ5)uν uν γγ (1 − γ5)ueue
] (A.24)
T βδ
2
≡ Tr [γβ (f − γ5g) ununγδ (f − γ5g) upup
] . (A.25) To reduce notion clutter, we have dropped the index j that indicates the neutrino mass eigenstate. As described in the text, we will sum the spins of the final state electron and proton, and we will average the spin of the initial state neutron. Doing so gives
|M| 2 = 12
∑
sn,s e,s p=±1/2
|M| 2 = G2
F
4 |Vud |2|U ∗
ej
|2T αγ
1
T βδ
2
ηαβ ηγδ , (A.26) where
T αγ
1
= ∑
se=±1/2
Tr [γα(1 − γ5)uν uν γγ (1 − γ5)ueue
] , (A.27)
T βδ
2
= ∑
sn,s p=±1/2
Tr [γβ (f − γ5g) ununγδ (f − γ5g) upup
] . (A.28) We now require the completeness relations,
∑
si=±1/2
uiui = (/pi + Mi
) for i = n, p, e , uν uν = 12
(/pν + Mν
)( 1 + 2 sν γ5 /Sν
) , (A.29) where (Sν )α =
( |pν |
mν
, Eν
mν
ˆpν
)
(A.30) is the neutrino spin vector. Inserting Eq. (A.29) into Eq. (A.27) yields
T αγ
1
= 12 Tr
[
γα(1 − γ5)(/pν + mνj
)( 1 + 2 sν γ5 /Sν
)γγ (1 − γ5)(/pe + me
)]
(A.31)
T βδ
2
= Tr
[
γβ (f − γ5)( /pn + mn
)γδ (f − γ5g)( /pp + mp
)]
. (A.32) The traces are evaluated using the Mathematica package “Tracer” , and we find
T αγ
1
T βδ
2
ηαβ ηγδ = 32
{
(g + f )2[
(pe · pp)( pν · pn)
]
( g − f )2[
(pe · pn)( pν · pp)
]
( g2 − f 2)
[
mnmp(pe · pν )
]}
− 2sν mνj
{
(g + f )2[
(pe · pp)( Sν · pn)
]
( g − f )2[
(pe · pn)( Sν · pp)
]
( g2 − f 2)
[
mnmp(pe · Sν )
]}
. (A.33) The spin-independent terms ( sν = 0) match with Ref. , and the spin-dependent terms are new. We now specify to the rest frame of the neutron (parent nucleus) where (pn)μ = {mn; 0} , (pν )μ = {Eν ; pν } , (pp)μ = {Ep; pp} , (pe)μ = {Ee; pe} . (A.34) – 26 – Neglecting the proton (daughter nucleus) recoil, pp mp, we obtain
T αγ
1
T βδ
2
ηαβ ηγδ = 32 mnmpEeEν
{
2( g2 + f 2) + ( g2 − f 2)
[
1 − pe
Ee
· pν
Eν
]}
− 2sν mnmpEe |pν |
{
2( g2 + f 2) + ( g2 − f 2)
[
1 − Eν
|pν |
pe
Ee
· pν
|pν |
]}
= 32 mnmpEeEν
[
(f 2 + 3 g2)(1 − 2sν vν ) + ( f 2 − g2)( vν − 2sν )ve cos θ
]
, (A.35) where cos θ = pe · pνj /(|pe| ∣∣pνj
∣∣) and vi ≡ | pi| /E i. Inserting Eq. (A.35) into Eq. (A.26) gives Eq. (2.21). In the center of momentum frame the differential cross section is
dσ dt = 164 π
1
s
1
|pν cm |2 |M| 2 , (A.36) where s = ( pn + pν )2 and t = ( pe − pν )2. The Mandelstam variables can be evaluated in the lab frame using Eq. (A.34) to find
s = ( mn + Eν )2 − | pν |2 = m2
n
2 mnEν + m2
ν
' m2
n
, (A.37)
t = ( Ee − Eν )2 − | pe − pν |2 ' (me − mν )2 + 2 |pe|| pν | cos θ , (A.38) and dt/d cos θ = 2 |pe|| pν |. After also making the replacement pν cm = pν lab (mn/√s) ' pν , we obtain
dσ d cos θ = 132 π
1
m2
n
|pe||pν | |M| 2 . (A.39)
References
A. Dolgov, Neutrinos in cosmology , Phys.Rept. 370 (2002) 333–535, [ hep-ph/0202122 ]. J. Lesgourgues and S. Pastor, Massive neutrinos and cosmology , Phys.Rept. 429 (2006) 307–379, [astro-ph/0603494 ]. C. Quigg, Cosmic Neutrinos , arXiv:0802.0013 . A. Dolgov, Cosmology and Neutrino Properties , Phys.Atom.Nucl. 71 (2008) 2152–2164, [arXiv:0803.3887 ]. Planck Collaboration Collaboration, P. Ade et. al. , Planck 2013 results. XVI. Cosmological parameters , arXiv:1303.5076 . EPIC Collaboration Collaboration, J. Bock et. al. , Study of the Experimental Probe of Inflationary Cosmology (EPIC)-Intemediate Mission for NASA’s Einstein Inflation Probe , arXiv:0906.1188 . S. Weinberg, Universal Neutrino Degeneracy , Phys.Rev. 128 (1962) 1457–1473. L. Stodolsky, Speculations on Detection of the Neutrino Sea , Phys.Rev.Lett. 34 (1975) 110. B. Shvartsman, V. Braginsky, S. Gershtein, Y. Zeldovich, M. Y. Khlopov, et. al. , Possibility of detecting relic massive neutrinos , JETP Lett. 36 (1982) 277–279. G. Duda, G. Gelmini, and S. Nussinov, Expected signals in relic neutrino detectors , Phys.Rev. D64
(2001) 122001, [ hep-ph/0107027 ]. M. Hedman, Efficiently extracting energy from cosmological neutrinos , JCAP 1309 (2013) 029, [arXiv:1307.0652 ]. A. G. Cocco, G. Mangano, and M. Messina, Probing low energy neutrino backgrounds with neutrino capture on beta decaying nuclei , JCAP 0706 (2007) 015, [ hep-ph/0703075 ].
– 27 – R. Lazauskas, P. Vogel, and C. Volpe, Charged current cross section for massive cosmological neutrinos impinging on radioactive nuclei , J.Phys. G35 (2008) 025001, [ arXiv:0710.5312 ]. M. Blennow, Prospects for cosmic neutrino detection in tritium experiments in the case of hierarchical neutrino masses , Phys.Rev. D77 (2008) 113014, [ arXiv:0803.3762 ]. Y. Li, Z.-z. Xing, and S. Luo, Direct Detection of the Cosmic Neutrino Background Including Light Sterile Neutrinos , Phys.Lett. B692 (2010) 261–267, [ arXiv:1007.0914 ]. R. Hodak, S. Kovalenko, F. Simkovic, and A. Faessler, Beta decaying nuclei as a probe of cosmic neutrino background , arXiv:1102.1799 . S. Betts, W. Blanchard, R. Carnevale, C. Chang, C. Chen, et. al. , Development of a Relic Neutrino Detection Experiment at PTOLEMY: Princeton Tritium Observatory for Light, Early-Universe, Massive-Neutrino Yield , arXiv:1307.4738 . N. Y. Gnedin and O. Y. Gnedin, Cosmological neutrino background revisited , Astrophys.J. 509 (1998) 11–15, [ astro-ph/9712199 ]. B. Pontecorvo, Mesonium and anti-mesonium , Sov.Phys.JETP 6 (1957) 429. B. Pontecorvo, Neutrino Experiments and the Problem of Conservation of Leptonic Charge ,
Sov.Phys.JETP 26 (1968) 984–988. Z. Maki, M. Nakagawa, and S. Sakata, Remarks on the unified model of elementary particles ,
Prog.Theor.Phys. 28 (1962) 870–880. E. K. Akhmedov and A. Y. Smirnov, Paradoxes of neutrino oscillations , Phys.Atom.Nucl. 72 (2009) 1363–1381, [ arXiv:0905.1903 ]. B. Eberle, A. Ringwald, L. Song, and T. J. Weiler, Relic neutrino absorption spectroscopy , Phys.Rev.
D70 (2004) 023007, [ hep-ph/0401203 ]. M. Gell-Mann, P. Ramond, and R. Slansky, Complex Spinors and Unified Theories , Conf.Proc.
C790927 (1979) 315–321, [ arXiv:1306.4669 ]. R. N. Mohapatra and G. Senjanovic, Neutrino Mass and Spontaneous Parity Violation , Phys.Rev.Lett.
44 (1980) 912. T. Yanagida, Horizontal Symmetry and Masses of Neutrinos , Prog.Theor.Phys. 64 (1980) 1103. Particle Data Group Collaboration, J. Beringer et. al. , Review of Particle Physics (RPP) ,
Phys.Rev. D86 (2012) 010001. M. Fukugita and T. Yanagida, Physics of Neutrinos and Applications to Astrophysics .Springer-Verlag, Berlin, 2003. R. Schiavilla, V. Stoks, W. Gloeckle, H. Kamada, A. Nogga, et. al. , Weak capture of protons by protons , Phys.Rev. C58 (1998) 1263, [ nucl-th/9808010 ]. R. L. William, Techniques for Nuclear and Particle Physics Experiments: A How-To Approach .Springer, Berlin, 1994. S. Masood, S. Nasri, J. Schechter, M. Tortola, J. Valle, et. al. , Exact relativistic beta decay endpoint spectrum , Phys.Rev. C76 (2007) 045501, [ arXiv:0706.0897 ]. C. Tully, Private communication , . S. Singh and C.-P. Ma, Neutrino clustering in cold dark matter halos : Implications for ultrahigh-energy cosmic rays , Phys.Rev. D67 (2003) 023506, [ astro-ph/0208419 ]. A. Ringwald and Y. Y. Wong, Gravitational clustering of relic neutrinos and implications for their detection , JCAP 0412 (2004) 005, [ hep-ph/0408241 ]. A. Klypin, H. Zhao, and R. S. Somerville, Lambda CDM-based models for the Milky Way and M31 I: Dynamical models , Astrophys.J. 573 (2002) 597–613, [ astro-ph/0110390 ]. J. F. Navarro, C. S. Frenk, and S. D. M. White, A Universal Density Profile from Hierarchical Clustering , Astrophys. J. 490 (1997) 493–508, [ astro-ph/9611107 ].
– 28 – B. R. Safdi, M. Lisanti, J. Spitz, and J. A. Formaggio, Annual Modulation of Cosmic Relic Neutrinos ,
arXiv:1404.0680 . L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-relativistic Theory . Pergamon Press, Oxford, Berlin, 1977. S.-F. Ge, K. Hagiwara, N. Okamura, and Y. Takaesu, Determination of mass hierarchy with medium baseline reactor neutrino experiments , JHEP 1305 (2013) 131, [ arXiv:1210.8141 ]. LSND Collaboration Collaboration, A. Aguilar-Arevalo et. al. , Evidence for neutrino oscillations from the observation of anti-neutrino(electron) appearance in a anti-neutrino(muon) beam , Phys.Rev.
D64 (2001) 112007, [ hep-ex/0104049 ]. LSND Collaboration Collaboration, C. Athanassopoulos et. al. , Candidate events in a search for anti-muon-neutrino into anti-electron-neutrino oscillations , Phys.Rev.Lett. 75 (1995) 2650–2653, [nucl-ex/9504002 ]. MiniBooNE Collaboration Collaboration, A. Aguilar-Arevalo et. al. , Unexplained Excess of Electron-Like Events From a 1-GeV Neutrino Beam , Phys.Rev.Lett. 102 (2009) 101802, [arXiv:0812.2243 ]. G. Mention, M. Fechner, T. Lasserre, T. Mueller, D. Lhuillier, et. al. , The Reactor Antineutrino Anomaly , Phys.Rev. D83 (2011) 073006, [ arXiv:1101.2755 ]. T. Mueller, D. Lhuillier, M. Fallot, A. Letourneau, S. Cormon, et. al. , Improved Predictions of Reactor Antineutrino Spectra , Phys.Rev. C83 (2011) 054615, [ arXiv:1101.2663 ]. P. Huber, On the determination of anti-neutrino spectra from nuclear reactors , Phys.Rev. C84 (2011) 024617, [ arXiv:1106.0687 ]. GALLEX Collaboration. Collaboration, P. Anselmann et. al. , First results from the Cr-51 neutrino source experiment with the GALLEX detector , Phys.Lett. B342 (1995) 440–450. GALLEX Collaboration Collaboration, W. Hampel et. al. , Final results of the Cr-51 neutrino source experiments in GALLEX , Phys.Lett. B420 (1998) 114–126. SAGE Collaboration Collaboration, J. Abdurashitov et. al. , Measurement of the response of the Russian-American gallium experiment to neutrinos from a Cr-51 source , Phys.Rev. C59 (1999) 2246–2263, [ hep-ph/9803418 ]. MiniBooNE Collaboration Collaboration, A. Aguilar-Arevalo et. al. , Improved Search for ¯νμ → ¯νe
Oscillations in the MiniBooNE Experiment , Phys.Rev.Lett. 110 (2013) 161801, [ arXiv:1207.4809 ]. C. Giunti, M. Laveder, Y. Li, and H. Long, Pragmatic View of Short-Baseline Neutrino Oscillations ,
Phys.Rev. D88 (2013) 073008, [ arXiv:1308.5288 ]. R. Cooke, M. Pettini, R. A. Jorgenson, M. T. Murphy, and C. C. Steidel, Precision measures of the primordial abundance of deuterium , arXiv:1308.3240 . Y. Izotov, G. Stasinska, and N. Guseva, Primordial 4He abundance: a determination based on the largest sample of HII regions with a methodology tested on model HII regions , arXiv:1308.2100 . Z. Hou, C. Reichardt, K. Story, B. Follin, R. Keisler, et. al. , Constraints on Cosmology from the Cosmic Microwave Background Power Spectrum of the 2500-square degree SPT-SZ Survey ,
Astrophys.J. 782 (2014) 74, [ arXiv:1212.6267 ]. WMAP Collaboration, G. Hinshaw et. al. , Nine-Year Wilkinson Microwave Anisotropy Probe (WMAP) Observations: Cosmological Parameter Results , Astrophys.J.Suppl. 208 (2013) 19, [arXiv:1212.5226 ]. Atacama Cosmology Telescope Collaboration, J. L. Sievers et. al. , The Atacama Cosmology Telescope: Cosmological parameters from three seasons of data , JCAP 1310 (2013) 060, [arXiv:1301.0824 ]. A. G. Riess, L. Macri, S. Casertano, H. Lampeitl, H. C. Ferguson, et. al. , A 3 % Solution: Determination of the Hubble Constant with the Hubble Space Telescope and Wide Field Camera 3 ,
Astrophys.J. 730 (2011) 119, [ arXiv:1103.2976 ].
– 29 – BICEP2 Collaboration Collaboration, P. Ade et. al. , BICEP2 I: Detection Of B-mode Polarization at Degree Angular Scales , arXiv:1403.3985 . R. A. Battye and A. Moss, Evidence for massive neutrinos from CMB and lensing observations ,
Phys.Rev.Lett. 112 (2014) 051303, [ arXiv:1308.5870 ]. J.-W. Hu, R.-G. Cai, Z.-K. Guo, and B. Hu, Cosmological parameter estimation from CMB and X-ray clusters after Planck , arXiv:1401.0717 . J.-F. Zhang, Y.-H. Li, and X. Zhang, Sterile neutrinos help reconcile the observational results of primordial gravitational waves from Planck and BICEP2 , arXiv:1403.7028 . C. Dvorkin, M. Wyman, D. H. Rudd, and W. Hu, Neutrinos help reconcile Planck measurements with both Early and Local Universe , arXiv:1403.8049 . J.-F. Zhang, Y.-H. Li, and X. Zhang, Cosmological constraints on neutrinos after BICEP2 ,
arXiv:1404.3598 . H. Li, J.-Q. Xia, and X. Zhang, Global fitting analysis on cosmological models after BICEP2 ,
arXiv:1404.0238 . M. Archidiacono, N. Fornengo, S. Gariazzo, C. Giunti, S. Hannestad, et. al. , Light sterile neutrinos after BICEP-2 , arXiv:1404.1794 . E. Giusarma, E. Di Valentino, M. Lattanzi, A. Melchiorri, and O. Mena, Relic Neutrinos, thermal axions and cosmology in early 2014 , arXiv:1403.4852 . B. Leistedt, H. V. Peiris, and L. Verde, No new cosmological concordance with massive sterile neutrinos , arXiv:1404.5950 . S. Dodelson and L. M. Widrow, Sterile-neutrinos as dark matter , Phys.Rev.Lett. 72 (1994) 17–20, [hep-ph/9303287 ]. T. D. Jacques, L. M. Krauss, and C. Lunardini, Additional Light Sterile Neutrinos and Cosmology ,
Phys.Rev. D87 (2013), no. 8 083515, [ arXiv:1301.3119 ]. Y. Li and Z.-z. Xing, Possible Capture of keV Sterile Neutrino Dark Matter on Radioactive β-decaying Nuclei , Phys.Lett. B695 (2011) 205–210, [ arXiv:1009.5870 ]. A. Y. Smirnov and R. Zukanovich Funchal, Sterile neutrinos: Direct mixing effects versus induced mass matrix of active neutrinos , Phys.Rev. D74 (2006) 013001, [ hep-ph/0603009 ]. A. Kusenko, Sterile neutrinos: The Dark side of the light fermions , Phys.Rept. 481 (2009) 1–28, [arXiv:0906.2968 ]. E. Bulbul, M. Markevitch, A. Foster, R. K. Smith, M. Loewenstein, et. al. , Detection of An Unidentified Emission Line in the Stacked X-ray spectrum of Galaxy Clusters , arXiv:1402.2301 . A. Boyarsky, O. Ruchayskiy, D. Iakubovskyi, and J. Franse, An unidentified line in X-ray spectra of the Andromeda galaxy and Perseus galaxy cluster , arXiv:1402.4119 . J. Bovy and S. Tremaine, On the local dark matter density , Astrophys.J. 756 (2012) 89, [arXiv:1205.4033 ]. X.-D. Shi and G. M. Fuller, A New dark matter candidate: Nonthermal sterile neutrinos ,
Phys.Rev.Lett. 82 (1999) 2832–2835, [ astro-ph/9810076 ]. M. Laine and M. Shaposhnikov, Sterile neutrino dark matter as a consequence of nuMSM-induced lepton asymmetry , JCAP 0806 (2008) 031, [ arXiv:0804.4543 ]. A. Boyarsky, O. Ruchayskiy, and M. Shaposhnikov, The Role of sterile neutrinos in cosmology and astrophysics , Ann.Rev.Nucl.Part.Sci. 59 (2009) 191–214, [ arXiv:0901.0011 ]. D. J. Schwarz and M. Stuke, Does the CMB prefer a leptonic Universe? , New J. Phys. 15 (2013) 033021, [ arXiv:1211.6721 ]. P. Langacker, G. Segre, and S. Soni, Majorana Neutrinos, Nucleosynthesis, and the Lepton Asymmetry of the Universe , Phys.Rev. D26 (1982) 3425. C. Lunardini and A. Y. Smirnov, High-energy neutrino conversion and the lepton asymmetry in the universe , Phys.Rev. D64 (2001) 073006, [ hep-ph/0012056 ].
– 30 – A. Dolgov, S. Hansen, S. Pastor, S. Petcov, G. Raffelt, et. al. , Cosmological bounds on neutrino degeneracy improved by flavor oscillations , Nucl.Phys. B632 (2002) 363–382, [ hep-ph/0201287 ]. K. Hirata et al. [KAMIOKANDE-II Collaboration], “ Observation of a Neutrino Burst from the Supernova SN 1987a ”, Phys. Rev. Lett. 58 (1987) 1490. G. Raffelt, Neutrino radiative lifetime limits from the absence of solar gamma rays , Phys.Rev. D31
(1985) 3002–3004. P. Baerwald, M. Bustamante, and W. Winter, Neutrino Decays over Cosmological Distances and the Implications for Neutrino Telescopes , JCAP 1210 (2012) 020, [ arXiv:1208.4600 ]. S. Pakvasa, “ Do neutrinos decay? ”, AIP Conf. Proc. 542 (2000) 99 [hep-ph/0004077]. T. Kanzaki, M. Kawasaki, K. Kohri, and T. Moroi, Cosmological Constraints on Neutrino Injection ,
Phys.Rev. D76 (2007) 105017, [ arXiv:0705.1200 ]. M. Zralek, On the possibilities of distinguishing Dirac from Majorana neutrinos , Acta Phys.Polon.
B28 (1997) 2225–2257, [ hep-ph/9711506 ]. M. Czakon, J. Gluza, and M. Zralek, Nature of neutrinos in the light of present and future experiments , Phys.Lett. B465 (1999) 211–218, [ hep-ph/9906381 ]. CUORICINO Collaboration Collaboration, C. Arnaboldi et. al. , Results from a search for the 0 neutrino beta beta-decay of Te-130 , Phys.Rev. C78 (2008) 035502, [ arXiv:0802.3439 ]. F. Alessandria, E. Andreotti, R. Ardito, C. Arnaboldi, I. Avignone, F.T., et. al. , Sensitivity of CUORE to Neutrinoless Double-Beta Decay , arXiv:1109.0494 . GERDA Collaboration Collaboration, M. Agostini et. al. , Results on Neutrinoless Double-β Decay of 76 Ge from Phase I of the GERDA Experiment , Phys.Rev.Lett. 111 (2013), no. 12 122503, [arXiv:1307.4720 ]. G. Audi, A. Wapstra, and C. Thibault, The Ame2003 atomic mass evaluation (II). Tables, graphs and references , Nucl.Phys. A729 (2002) 337–676. J. A. BEARDEN and A. F. BURR, Reevaluation of x-ray atomic energy levels , Rev. Mod. Phys. 39
(Jan, 1967) 125–142. M. Jamin and M. E. Lautenbacher, TRACER: Version 1.1: A Mathematica package for gamma algebra in arbitrary dimensions , Comput.Phys.Commun. 74 (1993) 265–288.
– 31 – |
190257 | https://www.sciencedirect.com/science/article/abs/pii/S1078143906000536 | Intravesical bacillus Calmette-Guérin therapy: Experience with a reduced dwell-time in patients with pronounced side-effects: Andius P, Fehrling M, Holmäng S, Department of Urology, Sahlgrenska University Hospital, Göteborg, Sweden - ScienceDirect
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Urologic Oncology: Seminars and Original Investigations
Volume 24, Issue 3, May–June 2006, Page 276
Urologic oncology survey
Clinical research
Intravesical bacillus Calmette-Guérin therapy: Experience with a reduced dwell-time in patients with pronounced side-effects: Andius P, Fehrling M, Holmäng S, Department of Urology, Sahlgrenska University Hospital, Göteborg, Sweden
Author links open overlay panel William A.See M.D.
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Objective
To report the side-effects after a reduction in the dwell-time in patients who had pronounced symptoms after intravesical bacillus Calmette-Guérin (BCG) treatment, as side-effects such as fever, haematuria, and frequency are common and sometimes severe after BCG treatment in patients with bladder cancer.
Patients and Methods
The dwell-time was reduced to ≤30 min in 51 patients who had pronounced side-effects after the preceding BCG instillation. All patients self-reported side-effects after each instillation in a questionnaire.
Results
After reducing the BCG dwell-time, fever, chills, dysuria and the overall time-to-recovery were significantly reduced but frequency and haematuria were not influenced. Patients with carcinoma in situ had significantly less dysuria than patients with papillary tumours. There was no difference in the treatment results between patients who had a normal dwell-time and a reduced dwell-time, determined at the first and second follow-up cystoscopy.
Conclusion
Reducing the BCG dwell-time to ≤30 min could be an alternative to a dose reduction in patients who experience pronounced side-effects after BCG instillations. The long-term outcome after reducing dwell-time and after dose reduction has not been studied and warrants further investigation.
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Commentary
The number of unanswered questions regarding the use of bacillus Calmette-Guérin (BCG) as treatment for urothelial malignancy continues to outweigh resolved concerns. By evaluating the effect of reduced dwell time on BCG treatment efficacy and toxicity, this study by Andius et al. has improved upon this imbalance. Their results suggest that patients with significant initial treatment-associated toxicity see a reduction in their symptoms, while maintaining treatment efficacy, by a reduction in
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190258 | https://www.youtube.com/watch?v=DX_zkaK5PaI | Every Kind of Bridge Explained in 15 Minutes
Practical Engineering
4520000 subscribers
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Posted: 21 May 2024
See some cool bridges, learn some new words!
Errata: At 9:25, Edmonton is in Alberta, not Saskatchewan.
Without listing every bridge, there’s no true way to list every type of bridge. There’s too much nuance, creativity, and mixing and matching designs. But that’s part of the joy of paying attention to bridges. Once you understand the basics, you can start to puzzle out the more interesting details.
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Transcript:
The Earth is pretty cool and all, but many of
its most magnificent features make it tough for us to get around. When the topography is too
wet, steep, treacherous, or prone to disaster, sometimes the only way forward is up: our
roadways and walkways and railways break free from the surface using bridges. A lot of
the infrastructure we rely on day to day isn’t necessarily picturesque. It’s not that we can’t
build exquisite electrical transmission lines or stunning sanitary sewers. It’s just that we
rarely want to bear the cost. But bridges are different. To an enthusiast of constructed works,
many are downright breathtaking. There are so many ways to cross a gap, all kindred in function
but contrary in form. And the typical way that engineers classify and name them is in how each
design manages the incredible forces involved. Like everything in engineering, terminology
and categories vary. As Alfred Korzybski said, “The map is not the territory.” But,
trying to list them all is at least a chance to learn some new words and see some cool
bridges. And honestly, I can hardly think of anything more worthwhile than that. I’m
Grady, and this is Practical Engineering. One of the simplest structural crossings is the
beam bridge: just a horizontal member across two supports. That member can take a variety of forms,
including a rolled steel beam (sometimes called a stringer) or a larger steel member fabricated
from plates (often called a plate girder). Most modern bridges built as overpasses for grade
separation between traffic are beam bridges that use concrete girders. And instead of a group
of individual beams, many bridges use box girders, which are essentially closed structural tubes
that use material more efficiently (but can be more complicated to construct). Beam bridges
usually can’t span great distances because the girders required would be too large. At a
certain distance, the beams become so heavy, they can hardly support their own weight,
let alone the roadway and traffic on top. One way around the challenge of the structural
members’ self-weight is to use a truss instead of a girder. A truss is an assembly of smaller
elements that creates a rigid and lightweight structure. Unlike a beam, the members of a truss
don’t typically experience bending forces. The connections usually aren’t actual hinges that
permit free rotation, but they are close enough. So, all the load is axial (along their length)
in compression or tension. That simplifies the design process because it’s easier to predict
the forces within each structural member. The weight reduction allows trusses to span greater
distances than solid beams, and there are a wide variety of arrangements, many with their own
specific names. In general, a through truss puts the deck on the bottom level, and a deck truss
puts it on top, hiding the structural members below the road. A particularly photogenic
type of truss is a lenticular truss bridge, named because they resemble lenses, which
themselves are named because they resemble lentils! A Bailey bridge is a kind of temporary
truss bridge that is designed to be portable and easy to assemble. They were designed during
World War II, but Bailey bridges are still used today as temporary crossings when a bridge fails
or gets closed for construction. Most covered bridges are timber truss bridges. Since wood is
more susceptible to damage from exposure to the elements, the roof and siding are placed to keep
the structural elements truss-worthy. A trestle bridge is superficially similar to a truss: a
framework of smaller members. Trestle bridges don’t have long spans, but rather a continuous
series of short spans with frequent supports which are individually called trestles, but sometimes
the whole bridge is just called a trestle, so like so many other instances of structural
terminology, it can be a little confusing. This next bridge type uses a structural feature
that’s been a favorite of builders for millennia: the arch. Instead of beams loaded perpendicularly
or trusses that experience both compressive and tensile forces, arch bridges use a curved element
to transfer the bridge’s weight to supports using compression forces alone. Many of the oldest
bridges used arches because it was the only way to span a gap with materials available
at the time (stone and mortar). Even now, with the convenience of modern steel and
concrete, arches are a popular choice for bridges. They make efficient use of materials
but can be challenging to construct because the arch can’t provide its support until it is
complete. Temporary supports are often required during construction until the arch is connected
at its apex from both sides. In stone arches, the topmost stone is key to keeping the whole
thing standing, and, of course, it’s called the keystone. When the arch is below the roadway,
we call it a deck arch bridge. Vertical supports transfer the load of the deck onto the arch. The
area between the deck and arch has a great name: the spandrel. Open-spandrel bridges use columns
to transfer loads, and closed-spandrel bridges use continuous walls. If part of the arch extends
above the roadway with the deck suspended below, it’s called a through arch bridge. A moon bridge
is kind of an exaggerated arch bridge, usually reserved for pedestrians over narrow canals where
there’s not enough room for long approaches. They’re steep, so sometimes you have to use steps
or ladders to get up to the top and back down. One result of compressing an arch is that it
creates horizontal forces called thrusts. Arch bridges usually need strong abutments at either
side to push against that can withstand the extra horizontal loads. Alternatively, a tied arch
bridge uses a chord to connect both sides of the arch like a bowstring, so it can resist
the thrust forces. That means a tied arch is structurally more of a truss than an arch,
and that provides a lot of opportunities for creativity. For just one example, a network
arch bridge uses the tied arch design, plus criss-crossed suspension cables, to support
the deck. To tell an arch from a tied arch by eye, it’s usually enough to look at the supports. If
the end of each arch sits atop a spindly pier or some other structure that seems
insubstantial against horizontal forces, you can probably bet that they are tied together
and it’s not a true arch bridge. Similarly, a rigid-frame bridge integrates the superstructure
and substructure (in other words, the deck, supports, and everything else) into a single
unit. They don’t have to be arched, but many are. Another way to increase the span of a beam bridge
is to move the supports so that sections of the deck balance on their center instead of being
supported at each end. A cantilever bridge uses beams or trusses that project horizontally,
balancing most of the structure’s weight above the supports rather than in the center of the
span. This is such an effective technique that the Forth Bridge crossing the Firth of Forth in
Scotland took the title of longest span in the world away from the Brooklyn Bridge in 1890
and held the record for decades. This famous photograph demonstrates the principle of that
bridge perfectly: The two central piers bear the compression loads from the bridge. And, the
outer-most supports are anchors to provide the balancing force for each arm. This way,
you can suspend a load in the middle. The longest bridges take advantage of steel’s
ability to withstand incredible tension forces using cable supports. Cable-stayed bridges
support the deck from above through cables attached to tall towers or spars. The cables
(also called stays) form a fan pattern, giving this type of bridge its unique
appearance. Depending on the span, cable-stayed bridges can have one central
tower or more. Their simplicity allows for a wide variety of configurations, giving rise
to some dramatic (and often asymmetric) shapes. For shorter spans, you can combine the benefits
of a cable-stayed structure with girders to get an extradosed bridge. Imagine a concrete girder
bridge that uses internal tendons to keep the concrete in compression, then just pull those
tendons out of the girder and attach them to a short tower. Rather than holding the deck up
vertically like a cable-stayed bridge, they’re acting more horizontally to hold the girders in
compression, giving them the stiffness needed to support the deck. It’s a relatively new idea
compared to most of the other designs I’ve listed, but there are quite a few cool examples
of extradosed bridges across the globe. Where a cable-stayed bridge attaches the
deck directly to each tower, a suspension bridge uses cables or chains to dangle the
deck below. In a simple suspension bridge, the cables follow the curve of the deck.
This is your classic rope bridge. They’re not very stiff or strong, so simple suspension
bridges are usually only for pedestrians. A stressed ribbon bridge takes the concept
a step further by integrating the cables into the deck. The cables pull the deck
into compression, providing stiffness and stability so it doesn’t sway and bounce.
This design is also primarily used for smaller pedestrian bridges because it can’t span
long distances and the deck sags in the middle. Then you have the suspended deck bridge, the
design we most associate with the category with the longest spans in the world. Massive main
cables or chains dangle the road deck below with vertical hangers. Suspension bridges are iconic
structures because of their enormous spans and slender, graceful appearance. Towers on either
side prop up the main cables like broomsticks in a blanket fort. Most of the bridge’s weight
is transferred into the foundation through these towers. The rest is transferred into the
bridge’s abutments through immense anchorages keeping the cables from pulling out of the ground.
Alternatively, self-anchored suspension bridges connect the main cables to the deck on either
side, compressing it to resist the tension forces. Because they are so slender and lightweight,
most suspension bridges require stiffening with girders or trusses along the deck to reduce
movement from wind and traffic loads. These bridges are expensive to build and maintain, so
they’re really only used when no other structure will suffice. But you can hardly look at a
suspended deck bridge without being impressed. Bridges have to support the vehicles and people
that cross over the deck, but they often have to accommodate boats and ships passing underneath as
well. If it’s not feasible to build the bridge and its approaches high enough, another option is
just to have it get out of the way when a ship needs to pass. Moveable bridges come in all shapes
and sizes. A lot of people call them drawbridges after their medieval brethren over castle moats.
A bascule bridge is hinged so the deck can rotate upward. A swing bridge rotates horizontally
so a ship can pass on either side. A vertical lift bridge raises the entire deck upward, keeping
it horizontal like a table. A transporter bridge just has a small length of deck that is shuttled
back and forth across a river. That’s just a few, and in fact, every moveable bridge is unique
and customized for a specific location, so there are some truly interesting
structures if you keep an eye out. On the other hand, sometimes there’s no need for
ship passage or a lot of space below, and in that case, you can just float the bridge right on the
water. Floating bridges use buoyant supports, eliminating the need for a foundation.
These are used in military applications, but there are permanent examples too. Many use
hollow concrete structures as pontoons, with pumps inside to make sure they don’t fill up with
water and sink. And actually, a lot of bridges take advantage of buoyancy in their design,
even if it’s not the main source of support. A design like this presents a lot of interesting
engineering challenges, so there aren’t too many of them. Similarly, the pedestrian bridge at
Fort de Roovere in the Netherlands (probably pronounced that wrong) has its deck below the
water, giving it the nickname of the Moses Bridge. If space or funding is really tight,
one option to span a small stream is a low-water crossing. Unlike bridges
built above the typical flood level, low-water crossings are designed to be submerged
when water levels rise. They are most common in areas prone to flash floods, where runoff
in streams rises and falls quickly. Ideally, a crossing would be inaccessible only a few
times per year during heavy rainstorms. However, low-water crossings have some disadvantages. For
one, they can block the passage of fish just like a dam. And then there’s safety. A significant
proportion of flood-related fatalities occur when someone tries to drive a car or truck through
water overtopping a roadway. Water is heavy. It takes only a small but swift flow to push a
vehicle down into a river or creek, which means at least some of the resources saved by avoiding the
cost of a higher bridge are often spent to erect barricades during storms, install automatic
flood warning systems, and run advertisement campaigns encouraging motorists never to
drive through water overtopping a roadway. You may have heard the term viaduct before.
It’s not so much a specific type of bridge, but really about the length. Bridges that span a
wide valley need multiple intermediate supports. So, a viaduct is really just a long bridge
with multiple spans that are mostly above land. There’s really not a lot of agreement on
what is one and what isn’t. Some are singular and impressive structures. But many modern cities
have viaducts that are, although equally amazing from an engineering standpoint, a little less
beautiful. So, you’re more likely to hear them called elevated expressways. And that gets to the
heart of a topic like this: without listing every bridge, there’s no true way to list every type
of bridge. There’s too much nuance, creativity, and mixing and matching designs. The Phyllis J.
Tilly bridge in Fort Worth, Texas combines an arch and stressed ribbons. The Third Millennium Bridge
in Spain uses a concrete tied arch with suspension cables holding up the deck which is stiffened
with box girders. The Yavuz Sultan Selim Bridge in Turkey combines a cable-stayed and suspension
design. In some parts of India and Indonesia, living tree roots are used as simple suspension
bridges over rivers. There are bridges for pipelines, bridges for water, bridges for animals,
and I could go on. But that’s part of the joy of paying attention to bridges. Once you understand
the basics, you can start to puzzle out the more interesting details. Eventually, you’ll see
the Akashi Kaikyo Bridge on a calendar in your accountant’s office, and let him know
it’s a twin-hinged, three-span continuous, stiffened truss girder suspension bridge with
a double-tower system. Or maybe that’s just me. We care a lot about bridges. My previous video
covered the engineering that goes into vessel collision design for bridges, focusing on the
recent collapse of the Francis Scott Key Bridge in Baltimore that was a huge story in the news
covered by nearly every major outlet across the globe. Over 400 sources reported the even from
every side of the political spectrum. Since it was so widely reported, there’s a pretty even mix
between left-leaning, center, and right-leaning outlets, but if you look at the headlines, you’ll
see all kinds of ways the story was painted with political and ideological biases from both sides
of the aisle. By focusing on different details of the story - the victims' nationalities, the DEI
policies of the ship operator, the response by prominent politicians - the framing can subtly, or
not-so-subtly, change how you interpret the facts. Seeing all this in one place is possible thanks
to my sponsor, Ground News. They aggregate major news stories and add context to make reading the
news easier and more effective. Every story comes with a quick visual breakdowns and tags for
political bias, factuality, and ownership of the sources backed by ratings from independent
news monitoring organizations. For this story, you can see that nearly half of the reporting
outlets are media conglomerates and just over half of those outlets have been rated “High
Factuality.” They also have a feature called the Blind Spot that shows you stories mainly
covered by one side of the political spectrum: stuff you might totally miss if you only
follow a few main sources for your news. I don’t necessarily agree with how every
story on the Key Bridge is painted, but it’s important to me to get a broad
perspective on issues like this. It’s not just because I was trying to find
the right way to tell the story myself, but because stories like this are how we shape
our view of the world around us. In that way, journalism has a lot of power over us, and
Ground News hands some of that power back to you. If you’d like a more transparent media
landscape, they’re offering a huge discount right now at the link in the description: 40
percent off the Vantage subscription, which includes unlimited access to all their features.
That’s ground dot news slash practicalengineering or just click the link in the description. Thank
you for watching, and let me know what you think! |
190259 | https://www.youtube.com/watch?v=2YjlbaQupZA | Locus of the Midpoint of Chords Subtending Equal Central Angle in a Given Circle | JEE | CBSE | ISC
FineMath
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Posted: 8 Sep 2024
In this video, I have explained in details how to determine the locus of the midpoint of chords (in a given circle) that subtend equal central angle. Also, I have solved an example step by step in this video.
mathematics #jeemains #jeeadvanced #iitjee #olympiad #coordinategeometry #cbsemaths #iscmathematics #coordinategeometry #circle
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Transcript:
hi everyone in this video we're going to see how to find the locus of the midpoint of a cord that subtains a central angle of 2 Theta for a given Circle let's suppose you have a circle like this having the center at Point C and there is a cord AB whose midpoint is p and this cord subtains a central angle of 2 Theta at the center of course within the circle we can actually change the position of this cord maintaining the same central angle and that way we can easily imagine that the position of the midpoint will also be changing now what are the other possible positions of this cord let me show you a few examples the position of the cord could have been like this also where the central angle is still 2 Theta similarly we could have another position of the cord where the central angle is still 2 Theta and here is another possible position of the chord where the central angle is still 2 Theta also I have done another possible position of the cord where the central angle is 2 Theta and in fact there could be infinite number of positions like this here in the diagram I'm I'm showing just a few examples but if you can imagine we can slowly and slowly move this cord inside the circle and there could be infinite number of positions of this cord that means there will be infinite number of positions of the midpoint of this cord so in that case what would be the locus of the midpoint how would the locus look like is it going to look like a straight line is it going to look like a curved line is it going to look like a circle or ellipse or Parabola how is it going to look like well we're going to prove that theoretically and here I'm I'm going to show you a sample where you will realize that actually the locus of the midpoint of this cord is going to look like a circle and that circle is the yellow dotted Circle that I have drawn that circle is actually going to be the locus of the midpoint of this Cod no matter where you position the cord as long as the cord subtains a central angle of 2 Theta its Locus is going to be a circle like this whose Center also coincides with the center of the original Circle so let's see how we can prove it theoretically if the of the given Circle the original circle is actually X2 + y^2 + 2 GX + 2 FY + C = 0 then the coordinates of the center would be G comma F and let's assume that the coordinates of the midpoint of the cord AB the point is actually P that's the midpoint and let's assume its coordinates are x sub 1 comma y sub 1 next I'm going to join the points C and P with a straight line and the line CP is actually joining the midpoint of a cord from the center that means this line will actually BCT the central angle we have seen that in a previous theorem if we join the midpoint of a cord with the center then that line segment will actually BCT the central angle in this case the central angle was taken as 2 Theta so here because it bisects each of those angles angle ACP and angle BCP they will be equal to Theta now if we assume the coordinates of the center to be H comma k then obviously H is equal to Z and K is is equal to F and also the radius of the given Circle would be like this it will bequ < TK of G2 + F2 - c and since p is the midpoint of a cord the line segment CP will actually be perpendicular to AB that also we have proved in a previous theorem so in that case if you think about the little triangle which is triangle cap a that is a right triangle where the angle CPA will be the right angle now what will be the cosine of theta well we can say cosine of theta will be equal to the adjacent side over the hypotenuse in this case the adjacent side is CP and the hypotenuse is actually CA and from here we can say then CP will be equal to ca cosine of theta and if you think about CA that is nothing but the radius of the original Circle so we can say that CP will be equal to the radius of the original Circle let's assume it is the lower case r time cosine of theta up to this point if you think about it we know the value of R because the equation of the original Circle has been provided so we definitely know the value of r and also the central angle has been provided as 2 Theta so we know the value of theta so then the value of CP is actually known that means R cosine Theta it's a non value right now let's think about the length of the line segment CP using the distance formula using the distance formula we can say CP will be equal to S otk of x sub 1 - h 2 + y sub 1 - K squ now from those two equations we can say then CP will be equal to R cosine of theta that means Square < TK of x sub 1 - h 2 + y sub 1 - k 2 is = to R cosine of theta and now if we Square both sides of this equation we are going to get x sub 1 - h 2 + y sub 1 - K2 is equal to R cosine of theta s and this is actually the equation of the locus of the midpoint of this cord and if we try to write a generalized form of this equation in place of x sub 1 we can write X in place of Y sub one we can write y that will give us the general equation of the locus of the midpoint p and that equation is going to look like this x - h² + y - k 2 is = to R cosine of theta 2 so you can clearly see that this is the equation of a circle that has the center at the same point c as the original Circle the only difference is in the radius for the original Circle the radius is R but for this new Circle which is actually thec F of the midpoint the radius of the new circle is actually R cosine of theta that's the only difference otherwise the equation of this new Circle looks pretty much identical to the equation of the original Circle so from here can we say the locus of the midpoint is a concentric circle with radius equals R cosine of theta again what is r r is the radius of the original Circle and Theta is half of the central angle subtended by this cord so theoretically also we are able to prove that the locus of the midpoint of a cord like this is nothing but a concentric Circle having radius R cosine of theta next we're going to take an example let's suppose we have an example like this that says find the locus of the midpoint of a cord that subtains a central angle of 2 piun / 3 in the circle x^2 + y^ 2 - 8 x - 10 y + 5 = 0 so the equation of the original Circle has been provided the value of 2 Theta has been provided we have to find find out the locus of the midpoint of this cord let's get started with the solution from the given equation we have 2gx = -8x and from here we can say G will be equal to --4 similarly we have 2 FY = -10 y from the given equation then F will be equal to -5 and the constant term is actually POS 5 so we should be able to calculate the radius of the original Circle and that can be written as square root of G2 + F2 - c and and if we plug in the values we can write it as < TK of -4 2 + -5 2 - 5 which can be written as squ < TK of 16 + 25 - 5 that is nothing but < TK of 36 and that is equal to 6 so the radius of the original circle is actually 6 unit and what is the central angle here 2 Pi / 3 so we can say 2 Theta is equal to 2 piun / 3 and from here we can say Theta is equal to < / 3 so what will be the value of r cosine of theta well let's quickly calculate R cosine of theta can be written as 6 cosine of < / 3 and that will Beal to 6 1/2 because cosine of < / 3 which is cosine of 60° that is equal to 1/2 and finally that is equal to just 3 so R cosine of theta is actually 3 Also let's try to interpret the value of H and K H is actually G so we can make a quick note here that H is actually equal to G and that will be equal to posit 4 4 and similarly K is equal to F and that will be equal to POS 5 and next let's try to write down the equation for the locus of the midpoint of this cord the equation is going to look like this x - h² + y - k 2 is = to R cosine of theta squ and now let's plug in the values of h k and R cosine of theta so we can write this equation as x - 4² + y - 5² is = to 3 2 because R Co of Theta has been calculated as 3 H is equal to 4 K is equal to 5 so the equation going to look like this and if we expand it it is going to look like this X2 - 8 x + 16 + y^ 2 - 10 y + 25 is = 9 and we can write it as X2 + y^2 - 8 x - 10 y + 32 is equal 0 and that is the equation of the locus of the midpoint of the cord for this given Circle and that is our answer I hope everything made sense thank you for watching see you in the next video |
190260 | https://tutorial.math.lamar.edu/pdf/calculus_cheat_sheet_integrals.pdf | |
190261 | https://www.scribd.com/document/621672238/Hesitate-no-Longer-hesitation-wounds | Hesitate No Longer Hesitation Wounds | PDF | Wound | Skin
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Hesitate No Longer Hesitation Wounds
This document discusses identifying suicide attempts through the presence of hesitation marks in self-inflicted cuts. Hesitation marks are superficial cuts that result from a suicide attempt…
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1/2 3/2 3, 1 2:2 9 P M H e s i t a t e n o L o n g e r: I d e n t i f y i n g S u i c i d e A t t e m p t s w i t h S e l f-I n f l i c t e d C u t s i n t h e E D ww.nex tleve lemer gency medi cine.com/singl e-pos t/201 8/12/05/h esita te-n o-long er-id entif ying-suici de-at temp ts-w ith-s elf-i nflic ted-c uts-in-th e-ed 1/3
Hesitate no Longer: Identifying Suicide Attempts with Self-Inflicted Cuts in the ED
Case Presentation:
Clinical Question:
Is there a well-defined injury pattern of suicide attempts with self-inflicted cuts that can be identified in patients?
Background:
Summary of Evidence:
50-year-old mal e ambulates to triage with a self-made tourniquet on his right forearm with blo od present. The amount of bleeding cannot be determined, and the decision is made to activate a trauma. The primary survey is negative, except for a 5cm straight laceration with clean edges into the subcutaneous fat with minimal active bleeding on the right anterior forearm, and two superficial parallel abrasions over the patient’s right jugular area. The patient states he was moving a heavy piece of wood furniture and fell. When re-questioned, the patient states glass fell on him while moving furniture. A s the medical student irrigates the right forearm, multiple superficial wounds are seen parallel to the primary wound that were not noted previously. The medical student reports that there is concern for a suicide attempt, as the wound pattern is suggestive of hesitation marks. Among the medical team, there is debate as to whether this wound pattern reflects self-in flicted injury. Forensic Pathology has identified common patterns that can help differentiate nonaccidental vs. accidental injuries. There are over 400,000 annual ED visits for suicide attempts, with 21% attributed to cutting/piercing. (1,2) Completed suicide by cutting/piercing represents 2-3% of the 45,000 annual completed suicides in the US. (3,4) The patient who cuts or stabs himself in a suicide attempt is more likely to present to the ED than the Medical Examiner’s of fice. T erms to describe wounds in clinical medicine ar e often used incorrectly. Revisiting the case, this patient did not have abrasions or lacerations present. Abrasion: A scraping force that rem oves the superficial layer o f skin. (5) Laceration: An impacting that force results in abraded/irreg ular tissue edges, with skin splitting/tissue bridges outside of the directly impacted area. (5) Cut/Incised Wound: T erms are synonymo us. Sharp instrument applied with a force that causes an in jury with length is greater than depth. By definition there is no tissue bridging or irregul ar edges. (5) Hesitation marks are defined as superficial cuts found whe n a patient attempts to cut the skin. Wounds are generally confined to the epidermis, howeve r may involve deeper dermal layers. The “hesitation” results from a the pain associated with cutting, and the patient’s lack of familiarity with the amount of force required to cut through the skin. (3) Hesitation marks are more superficial compared to the primary wound. (6)
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1/2 3/2 3, 1 2:2 9 P M H e s i t a t e n o L o n g e r: I d e n t i f y i n g S u i c i d e A t t e m p t s w i t h S e l f-I n f l i c t e d C u t s i n t h e E D ww.nex tleve lemer gency medi cine.com/singl e-pos t/201 8/12/05/h esita te-n o-long er-id entif ying-suici de-at temp ts-w ith-s elf-i nflic ted-c uts-in-th e-ed 2/3
The Case Revisited:
50-year-old left-hande d male is seen in t he ED with a 5cm horizontal cut through the subcutaneous fat to the right anterior forearm with multiple superficial parallel cuts to the epidermis grouped closely to the primary wound. T wo 3cm superficial parallel cuts to the epidermi s of the right jugular are a are also apprecia ted. The patient’s injury pattern is highly su ggestive of a suicide attempt.
Recommendations:
A special thanks to Jonathan L. Arden, MD for help and guidance with this project. Autopsy photo depicting bilateral hesitation marks courtesy of Dr. Arden.
References:
The presence of hesitation marks is a hallmark of suicide attempts by cutting. Incidence of hesitation marks may be as high as 74% in all suicide cases involving cutting, with similar incidence between genders. (7) The presence of hesitation marks confirms suicide attempt; however, the absence of hesitation marks should not rule this out. (3) The most common locations of hesitation marks (in descending order) are the anterior surface of the upper extremities distal to the antecubital fossa, the neck, and left anterior chest. (7) Hesitation marks are generally group ed adjacent to the location of the primary wound. They overwhelmingl y run parallel to each other, and to the primary wound. The regular pattern of hesitation marks help to differ entiate between suicidal and non-suicidal wounds. (3,7) In general, hesitation marks will be seen on the opposite side of the body fro m the dominant hand. However, hesitation marks may be present on either side or bilaterally. (3,7) Early 40s is the average age of completed suicide by cutting with and without the presence of hesitation marks. Become familiar with the recogni zable injury pattern of suicide attempt by cutting. If you are able to identify hesitation marks, you will be able to identify a suicide attempt reliably. Learn to recognize the difference between cuts and lacerations, which are caused by different mechanisms and therefore have different features. Use correct terminology to describe and diagnose injuries, to communicate correctly the nature and causation of the injuries Recognize that forensic patholo gists are often available for clinical fore nsic consultations. All you need to do is ask,
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1/2 3/2 3, 1 2:2 9 P M H e s i t a t e n o L o n g e r: I d e n t i f y i n g S u i c i d e A t t e m p t s w i t h S e l f-I n f l i c t e d C u t s i n t h e E D ww.nex tleve lemer gency medi cine.com/singl e-pos t/201 8/12/05/h esita te-n o-long er-id entif ying-suici de-at temp ts-w ith-s elf-i nflic ted-c uts-in-th e-ed 3/3
1.Doshi A, Boudreaux E, Wang N, Pelletier A, &Camargo C. National Study of US Emergency Department Visits for Attempted Suicide and Self-Inflicted Injury, 1997-2001. Annals of Emergency Medicine. 2005; 46(4), 369–375. 2.Ting S, Sullivan A, Boudreaux E, Miller I, Camargo C. T rends in US emergency departme nt visits for attempted suicide and self-inflicted injury, 1993–2008. General Hospital Psychiatry. 2012; 34(5), 557–565. 3.Karakasi M, Nastoulis E, Kapetanakis S, Vasil ikos E , Kyropoulos G, Pavlidis P. Hesitation Wounds and Sharp Force Injuries in Forensic Pathology and Psychiatry: Multidisciplinary Review of the Literature and Study of T wo Cases. Journal of Forensic Sciences. 2016; 61(6): 1515–1523.4. Hedegaard H, Curtin SC, Warner M. Suicide rates in the United States continue to increase. NCHS Data Brief, no 309. Hyattsville, MD: National Center for Health Statistics. 2018.5.Spitz WU, Fisher RS. Spitz and Fisher's Medicolegal Investigation of Death: Guidelines for the Application of Pathology to Crime Investigation. 6th edition. Springfield, Il: Charles C. Thomas; 2006. 478, 532-575 6. Arden JL. 2018, Oct 6. Oral Communication.7.Racette S, Kremer C, Desjarlais A, Sauvageau A. Suicidal and homicidal sharp force injury: a 5-year retrospective comparative study of hesitation marks and defense wounds. Forensic Science, Medicine, and Pathology. 2008; 4(4), 221–227.
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190262 | https://en.wikipedia.org/wiki/Relative_change | Jump to content
Search
Contents
(Top)
1 Definition
2 Domain
3 Percentage change
3.1 Percent error
4 Examples
4.1 Valuable assets
4.2 Percentages of percentages
5 Indicators of relative change
6 Logarithmic change
6.1 Additivity
6.2 Uniqueness and extensions
7 See also
8 Notes
9 References
Relative change
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From Wikipedia, the free encyclopedia
Comparisons in quantitative sciences
| |
| This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations. (March 2011) (Learn how and when to remove this message) |
In any quantitative science, the terms relative change and relative difference are used to compare two quantities while taking into account the "sizes" of the things being compared, i.e. dividing by a standard or reference or starting value. The comparison is expressed as a ratio and is a unitless number. By multiplying these ratios by 100 they can be expressed as percentages so the terms percentage change, percent(age) difference, or relative percentage difference are also commonly used. The terms "change" and "difference" are used interchangeably.
Relative change is often used as a quantitative indicator of quality assurance and quality control for repeated measurements where the outcomes are expected to be the same. A special case of percent change (relative change expressed as a percentage) called percent error occurs in measuring situations where the reference value is the accepted or actual value (perhaps theoretically determined) and the value being compared to it is experimentally determined (by measurement).
The relative change formula is not well-behaved under many conditions. Various alternative formulas, called indicators of relative change, have been proposed in the literature. Several authors have found log change and log points to be satisfactory indicators, but these have not seen widespread use.
Definition
[edit]
Given two numerical quantities, vref and v with vref some reference value, their actual change, actual difference, or absolute change is
: Δv = v − vref.
The term absolute difference is sometimes also used even though the absolute value is not taken; the sign of Δ typically is uniform, e.g. across an increasing data series. If the relationship of the value with respect to the reference value (that is, larger or smaller) does not matter in a particular application, the absolute value may be used in place of the actual change in the above formula to produce a value for the relative change which is always non-negative. The actual difference is not usually a good way to compare the numbers, in particular because it depends on the unit of measurement. For instance, 1 m is the same as 100 cm, but the absolute difference between 2 and 1 m is 1 while the absolute difference between 200 and 100 cm is 100, giving the impression of a larger difference. But even with constant units, the relative change helps judge the importance of the respective change. For example, an increase in price of $100 of a valuable is considered big if changing from $50 to 150 but rather small when changing from $10,000 to 10,100.
We can adjust the comparison to take into account the "size" of the quantities involved, by defining, for positive values of vref :
The relative change is independent of the unit of measurement employed; for example, the relative change from 2 to 1 m is −50%, the same as for 200 to 100 cm. The relative change is not defined if the reference value (vref) is zero, and gives negative values for positive increases if vref is negative, hence it is not usually defined for negative reference values either. For example, we might want to calculate the relative change of −10 to −6. The above formula gives (−6) − (−10)/ −10 = 4/ −10 = −0.4, indicating a decrease, yet in fact the reading increased.
Measures of relative change are unitless numbers expressed as a fraction. Corresponding values of percent change would be obtained by multiplying these values by 100 (and appending the % sign to indicate that the value is a percentage).
Domain
[edit]
The domain restriction of relative change to positive numbers often poses a constraint. To avoid this problem it is common to take the absolute value, so that the relative change formula works correctly for all nonzero values of vref:
This still does not solve the issue when the reference is zero. It is common to instead use an indicator of relative change, and take the absolute values of both v and . Then the only problematic case is , which can usually be addressed by appropriately extending the indicator. For example, for arithmetic mean this formula may be used:
Percentage change
[edit]
A percentage change is a way to express a change in a variable. It represents the relative change between the old value and the new one.
For example, if a house is worth $100,000 today and the year after its value goes up to $110,000, the percentage change of its value can be expressed as
It can then be said that the worth of the house went up by 10%.
More generally, if V1 represents the old value and V2 the new one,
Some calculators directly support this via a %CH or Δ% function.
When the variable in question is a percentage itself, it is better to talk about its change by using percentage points, to avoid confusion between relative difference and absolute difference.
Percent error
[edit]
The percent error is a special case of the percentage form of relative change calculated from the absolute change between the experimental (measured) and theoretical (accepted) values, and dividing by the theoretical (accepted) value.
The terms "Experimental" and "Theoretical" used in the equation above are commonly replaced with similar terms. Other terms used for experimental could be "measured," "calculated," or "actual" and another term used for theoretical could be "accepted." Experimental value is what has been derived by use of calculation and/or measurement and is having its accuracy tested against the theoretical value, a value that is accepted by the scientific community or a value that could be seen as a goal for a successful result.
Although it is common practice to use the absolute value version of relative change when discussing percent error, in some situations, it can be beneficial to remove the absolute values to provide more information about the result. Thus, if an experimental value is less than the theoretical value, the percent error will be negative. This negative result provides additional information about the experimental result. For example, experimentally calculating the speed of light and coming up with a negative percent error says that the experimental value is a velocity that is less than the speed of light. This is a big difference from getting a positive percent error, which means the experimental value is a velocity that is greater than the speed of light (violating the theory of relativity) and is a newsworthy result.
The percent error equation, when rewritten by removing the absolute values, becomes:
It is important to note that the two values in the numerator do not commute. Therefore, it is vital to preserve the order as above: subtract the theoretical value from the experimental value and not vice versa.
Examples
[edit]
Valuable assets
[edit]
Suppose that car M costs $50,000 and car L costs $40,000. We wish to compare these costs. With respect to car L, the absolute difference is $10,000 = $50,000 − $40,000. That is, car M costs $10,000 more than car L. The relative difference is, and we say that car M costs 25% more than car L. It is also common to express the comparison as a ratio, which in this example is, and we say that car M costs 125% of the cost of car L.
In this example the cost of car L was considered the reference value, but we could have made the choice the other way and considered the cost of car M as the reference value. The absolute difference is now −$10,000 = $40,000 − $50,000 since car L costs $10,000 less than car M. The relative difference, is also negative since car L costs 20% less than car M. The ratio form of the comparison, says that car L costs 80% of what car M costs.
It is the use of the words "of" and "less/more than" that distinguish between ratios and relative differences.
Percentages of percentages
[edit]
If a bank were to raise the interest rate on a savings account from 3% to 4%, the statement that "the interest rate was increased by 1%" would be incorrect and misleading. The absolute change in this situation is 1 percentage point (4% − 3%), but the relative change in the interest rate is:
In general, the term "percentage point(s)" indicates an absolute change or difference of percentages, while the percent sign or the word "percentage" refers to the relative change or difference.
Indicators of relative change
[edit]
The (classical) relative change above is but one of the possible measures/indicators of relative change. An indicator of relative change from x (initial or reference value) to y (new value) is a binary real-valued function defined for the domain of interest which satisfies the following properties:
Appropriate sign:
R is an increasing function of y when x is fixed.
R is continuous.
Independent of the unit of measurement: for all , .
Normalized:
The normalization condition is motivated by the observation that R scaled by a constant still satisfies the other conditions besides normalization. Furthermore, due to the independence condition, every R can be written as a single argument function H of the ratio . The normalization condition is then that . This implies all indicators behave like the classical one when is close to 1.
Usually the indicator of relative change is presented as the actual change Δ scaled by some function of the values x and y, say f(x, y).
As with classical relative change, the general relative change is undefined if f(x, y) is zero. Various choices for the function f(x, y) have been proposed:
Indicators of relative change
| Name | where the indicator's value is | |
| (Classical) Relative change | x | |
| Reversed relative change | y | |
| Arithmetic mean change | | |
| Geometric mean change | | |
| Harmonic mean change | | |
| Moment mean change of order k | | |
| Maximum mean change | | |
| Minimum mean change | | |
| Logarithmic (mean) change | | |
As can be seen in the table, all but the first two indicators have, as denominator a mean. One of the properties of a mean function is: , which means that all such indicators have a "symmetry" property that the classical relative change lacks: . This agrees with intuition that a relative change from x to y should have the same magnitude as a relative change in the opposite direction, y to x, just like the relation suggests.
Maximum mean change has been recommended when comparing floating point values in programming languages for equality with a certain tolerance. Another application is in the computation of approximation errors when the relative error of a measurement is required.[citation needed] Minimum mean change has been recommended for use in econometrics. Logarithmic change has been recommended as a general-purpose replacement for relative change and is discussed more below.
Tenhunen defines a general relative difference function from L (reference value) to K:
which leads to
In particular for the special cases ,
Logarithmic change
[edit]
See also: Logarithmic derivative
Of these indicators of relative change, arguably the most natural is the natural logarithm (ln) of the ratio of the two numbers (final and initial), called log change. Indeed, when , the following approximation holds:
In the same way that relative change is scaled by 100 to get percentages, can be scaled by 100 to get what is commonly called log points. Log points are equivalent to the unit centinepers (cNp) when measured for root-power quantities. This quantity has also been referred to as a log percentage and denoted L%. Since the derivative of the natural log at 1 is 1, log points are approximately equal to percent change for small differences – for example an increase of 1% equals an increase of 0.995 cNp, and a 5% increase gives a 4.88 cNp increase. This approximation property does not hold for other choices of logarithm base, which introduce a scaling factor due to the derivative not being 1. Log points can thus be used as a replacement for percent change.
Additivity
[edit]
Using log change has the advantages of additivity compared to relative change. Specifically, when using log change, the total change after a series of changes equals the sum of the changes. With percent, summing the changes is only an approximation, with larger error for larger changes. For example:
| Log change 0 (cNp) | Log change 1 (cNp) | Total log change (cNp) | Relative change 0 (%) | Relative change 1 (%) | Total relative change (%) |
--- --- --- |
| 10 | 5 | 15 | 10 | 5 | 15.5 |
| 10 | −5 | 5 | 10 | −5 | 4.5 |
| 10 | 10 | 20 | 10 | 10 | 21 |
| 10 | −10 | 0 | 10 | −10 | −1 |
| 50 | 50 | 100 | 50 | 50 | 125 |
| 50 | −50 | 0 | 50 | −50 | −25 |
Note that in the above table, since relative change 0 (respectively relative change 1) has the same numerical value as log change 0 (respectively log change 1), it does not correspond to the same variation. The conversion between relative and log changes may be computed as .
By additivity, , and therefore additivity implies a sort of symmetry property, namely and thus the magnitude of a change expressed in log change is the same whether V0 or V1 is chosen as the reference. In contrast, for relative change, , with the difference becoming larger as V1 or V0 approaches 0 while the other remains fixed. For example:
| V0 | V1 | Log change (cNp) | Relative change (%) |
--- --- |
| 10 | 9 | −10.5 | −10.0 |
| 9 | 10 | +10.5 | +11.1 |
| 10 | 1 | −230 | −90 |
| 1 | 10 | +230 | +900 |
| 10 | 0+ | −∞ | −100 |
| 0+ | 10 | +∞ | +∞ |
Here 0+ means taking the limit from above towards 0.
Uniqueness and extensions
[edit]
The log change is the unique two-variable function that is additive, and whose linearization matches relative change. There is a family of additive difference functions for any , such that absolute change is and log change is .
See also
[edit]
Approximation error
Errors and residuals in statistics
Relative standard deviation
Logarithmic scale
Notes
[edit]
^ "IEC 60050 — Details for IEV number 112-03-07: "relative"". International Electrotechnical Vocabulary (in Japanese). Retrieved 2023-09-24.
^ a b c d e Törnqvist, Vartia & Vartia 1985.
^ Törnqvist, Vartia & Vartia 1985, p. 11: "We suggest that this indicator should be used more extensively."
^ Vartia 1976, p. 9.
^ Miller, H. Ronald (29 March 2011). Optimization: Foundations and Applications. New York: John Wiley & Sons. ISBN 978-1-118-03118-6.
^ Kazmi, Kumail (March 26, 2021). "Percentage Increase Calculator". Smadent - Best Educational Website of Pakistan. Smadent Publishing. Retrieved March 26, 2021.
^ Bennett & Briggs 2005, pp. 137–139
^ Bennett & Briggs 2005, p.140
^ Bennett & Briggs 2005, p. 141
^ Vartia 1976, p. 10.
^ Vartia 1976, p. 14.
^ a b c Törnqvist, Vartia & Vartia 1985, p. 5.
^ What's a good way to check for close enough floating-point equality
^ Rao, Potluri; Miller, Roger LeRoy (1971). Applied econometrics. Belmont, Calif., Wadsworth Pub. Co. p. 17. ISBN 978-0-534-00031-8.
^ Vartia 1976, pp. 17–18.
^ Tenhunen 1990, p. 20.
^ Békés, Gábor; Kézdi, Gábor (6 May 2021). Data Analysis for Business, Economics, and Policy. Cambridge University Press. p. 203. ISBN 978-1-108-48301-8.
^ a b c d e Karjus, Andres; Blythe, Richard A.; Kirby, Simon; Smith, Kenny (10 February 2020). "Quantifying the dynamics of topical fluctuations in language". Language Dynamics and Change. 10 (1). Section A.3.1. arXiv:1806.00699. doi:10.1163/22105832-01001200. S2CID 46928080.
^ Roe, John; deForest, Russ; Jamshidi, Sara (26 April 2018). Mathematics for Sustainability. Springer. p. 190. doi:10.1007/978-3-319-76660-7_4. ISBN 978-3-319-76660-7.
^ Doyle, Patrick (2016-08-24). "The Case for a Logarithmic Performance Metric". Vena Solutions.
^ Brauen, Silvan; Erpf, Philipp; Wasem, Micha (2020). "On Absolute and Relative Change". SSRN Electronic Journal. arXiv:2011.14807. doi:10.2139/ssrn.3739890. S2CID 227228720.
References
[edit]
Bennett, Jeffrey; Briggs, William (2005), Using and Understanding Mathematics: A Quantitative Reasoning Approach (3rd ed.), Boston: Pearson, ISBN 0-321-22773-5
"Understanding Measurement and Graphing" (PDF). North Carolina State University. 2008-08-20. Archived from the original (PDF) on 2010-06-15. Retrieved 2010-05-05.
"Percent Difference – Percent Error" (PDF). Illinois State University, Dept of Physics. 2004-07-20. Archived from the original (PDF) on 2019-07-13. Retrieved 2010-05-05.
Törnqvist, Leo; Vartia, Pentti; Vartia, Yrjö (1985), "How Should Relative Changes Be Measured?" (PDF), The American Statistician, 39 (1): 43–46, doi:10.2307/2683905, JSTOR 2683905
Tenhunen, Lauri (1990). The CES and par production techniques, income distribution and the neoclassical theory of production (PhD). A. Vol. 290. University of Tampere.
Vartia, Yrjö O. (1976). Relative changes and index numbers (PDF). ETLA A 4. Helsinki: Research Institute of the Finnish Economy. ISBN 951-9205-24-1. Retrieved 20 November 2022.
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190263 | https://www.khanacademy.org/math/class-8-bridge-course-mh/x32413ee700b4a348:direct-and-inverse-variation | Use of cookies
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190264 | https://www.youtube.com/watch?v=CESXLJaq6Mk | Function Transformations: Horizontal and Vertical Translations
Mathispower4u
330000 subscribers
562 likes
Description
311868 views
Posted: 26 Jul 2010
This video explains to graph graph horizontal and vertical translation in the form af(b(x-c))+d. This video looks at how c and d affect the graph of f(x).
19 comments
Transcript:
Introduction - WELCOME TO ANOTHER VIDEO
ON FUNCTION TRANSFORMATIONS. THIS VIDEO DEALS SPECIFICALLY WITH HORIZONTAL AND VERTICAL
TRANSLATIONS, SO THE GOAL OF THE VIDEO
IS TO GRAPH FUNCTIONS USING HORIZONTAL
AND VERTICAL SHIFTS. IF WE ARE LOOKING
AT TRANSFORMATIONS FROM THIS FUNCTION
TO THIS FUNCTION, THIS VIDEO DEALS SPECIFICALLY
WITH HOW C AND D TRANSFORM THE FUNCTION. Horizontal Shift LET'S FIRST TALK ABOUT
A HORIZONTAL SHIFT WHICH MEANS THE GRAPH WILL BE
SHIFTED LEFT OR RIGHT. Y = F OF THE QUANTITY (X + C), WE WILL SHIFT F OF X
LEFT C UNITS AND Y = F OF THE QUANTITY X - C. WE WILL SHIFT F OF X
RIGHT C UNITS. NOW, THIS MAY BE THE OPPOSITE
OF WHAT YOU MIGHT THINK. IF YOU ADD C UNITS TO X, THE
FUNCTION WILL BE SHIFTED LEFT, AND IF YOU SUBTRACT C UNITS
FROM X THE SHIFT WILL BE RIGHT. ONE WAY TO GET A FEEL FOR THIS WOULD BE TO COMPARE A TABLE
OF VALUES FOR F OF X AND F OF THE QUANTITY X - 1. SO LET'S GO AHEAD AND DO THAT. LET'S CHOOSE X = 1, 2, 3, 4. SO TO FIND Y
WE JUST SQUARE X FOR F OF X. SO THIS WOULD BE 1 SQUARED OR 1,
2 SQUARED WHICH IS 4, 3 SQUARED IS 9,
AND 4 SQUARED IS 16. LET'S USE THE SAME X VALUES
FOR F OF X - 1. SO NOW WE ARE GOING TO SUBTRACT
1 FROM THE INPUT AND THEN SQUARE IT. SO 1 - 1 SQUARED WOULD BE 0, 2 - 1 SQUARED WOULD BE 1 SQUARED
OR 1. 3 - 1 SQUARED WOULD BE
2 SQUARED OR 4, AND 4 - 1 SQUARED WOULD BE
3 SQUARED OR 9. SO IF WE COMPARE THE Y VALUES
OF 1, 4, AND 9, NOTICE THAT FOR F OF X - 1
WE HAVE TO INCREASE X BY 1 IN ORDER TO GET
THE SAME Y VALUE. WHEN WE INCREASE X BY 1 WE ARE SHIFTING THE FUNCTION
TO THE RIGHT. SO WHEN WE SUBTRACT A NUMBER
FROM X IT MOVES TO THE RIGHT, AND WHEN WE ADD A VALUE TO X
IT SHIFTS TO THE LEFT. Graph HERE IS THE GRAPH
OF THESE TWO FUNCTIONS, AND WHAT YOU'LL NOTICE IS
FOR ANY CORRESPONDING POINT, LET'S SAY THE VERTEX
ON THE ORIGINAL FUNCTION AND THE VERTEX
ON THE SHIFTED FUNCTION IT'S ONE UNITS TO THE RIGHT. PICK ANY POINT
ON THE ORIGINAL FUNCTION, AND THE TRANSLATED FUNCTION
WILL BE ONE UNIT TO THE RIGHT. LET'S LOOK AT AN ANIMATION
OF THIS. Animation HERE WE HAVE AN ORIGINAL
FUNCTION IN RED, AND AS I CHANGE THE VALUE OF C YOU WILL SEE THE TRANSLATED
FUNCTION IN BLUE AS WELL AS THE FUNCTION NOTATION
FOR THE TRANSLATED FUNCTION. SO NOTICE WHEN IT IS
F OF THE QUANTITY X - 2.5 THE BLUE FUNCTION IS SHIFTED
TO THE RIGHT 2.5 UNITS. WE CAN ALSO COMPARE
CORRESPONDING POINTS, AND WHAT YOU WILL FIND
IS EACH ONE IN BLUE IS SHIFTED 2.5 UNITS
TO THE RIGHT. LET'S GO AHEAD AND SEE
WHAT HAPPENS WHEN WE CHANGE THIS
TO X + A CONSTANT. YOU CAN SEE WHEN WE HAVE
F OF X + 1 THE TRANSLATED FUNCTION
IS SHIFTED TO THE LEFT NOW. LET'S GO AHEAD AND TALK ABOUT
A VERTICAL SHIFT NOW. Y = F OF X + D WILL SHIFT F OF X
UP D UNITS, Y = F OF X - D WILL SHIFT F OF X
DOWN D UNITS, AND THIS TRANSLATION
PROBABLY SEEMS MORE LOGICAL. REMEMBER F OF X = Y, SO IF WE ADD D UNITS TO Y
THE FUNCTION WILL SHIFT UP. IF WE SUBTRACT D UNITS FROM Y
THE FUNCTION WOULD SHIFT DOWN. LET'S GO AHEAD AND DO ANOTHER
COMPARISON USING F OF X AND F OF X - 2. SO AGAIN, FOR F OF X
WE'LL JUST SQUARE THE INPUT. SO WE WILL HAVE 1, 4, 9, 16. FOR F OF X - 2
WE WILL USE THE SAME INPUTS, BUT NOW WE'LL SQUARE THE INPUT
AND THEN SUBTRACT 2. 1 SQUARED - 2 WOULD BE -1. 2 SQUARED - 2,
THAT WOULD BE 4 - 2 OR 2. 3 SQUARED - 2,
THAT WOULD BE 9 - 2 OR 7, AND 4 SQUARED - 2
WOULD BE 16 - 2 OR 14. SO IF WE DO ANOTHER COMPARISON
OF THE Y-VALUES OF THE FUNCTION, NOTICE THAT ALL OF THE Y VALUES
IN RED ARE 2 LESS THAN THE Y-VALUES IN BLUE, THEREFORE THIS FUNCTION
WOULD BE 2 UNITS LOWER THAN THE ORIGINAL. Graphs HERE IS THE GRAPH
OF THOSE 2 FUNCTIONS, AND SO WE CAN PICK ANY POINT
ON THE ORIGINAL BLACK FUNCTION. TO FIND THE CORRESPONDING POINT
ON THE TRANSLATED FUNCTION WE WOULD JUST MOVE THIS POINT
DOWN 2 UNITS. LET'S GO AHEAD AND TAKE A LOOK
AT AN ANIMATION OF THIS AS WELL. SO AS WE CHANGE THE VALUE OF D WE WILL SEE HOW IT AFFECTS
THE GRAPH. THE TRANSLATED GRAPH
WILL BE IN BLUE. AS WE INCREASE D,
THE FUNCTION IS SHIFTED UPWARD. IF WE HAVE F OF X - D THE FUNCTION IS SHIFTED DOWN
FROM THE ORIGINAL. LET'S GO AHEAD AND TAKE A LOOK
AT SOME EXAMPLES. Examples WE WANT TO BE ABLE TO USE
WHAT WE JUST LEARNED IN ORDER TO ACCURATELY
AND QUICKLY GRAPH F OF X = THE ABSOLUTE VALUE
OF THE QUANTITY X + 3 + 2. THE FIRST THING WE NEED TO DO IS RECOGNIZE WHAT THE PARENT
FUNCTION IS, AND IN THIS CASE IT WOULD BE
THE ABSOLUTE VALUE OF X. LET'S GO AHEAD AND CALL IT G
OF X = THE ABSOLUTE VALUE OF X. LET'S GO AHEAD AND GRAPH THIS
FOR REFERENCE. REMEMBER THE ABSOLUTE VALUE
FUNCTION FORMS A V, LOOKS SOMETHING LIKE THIS. LET'S GO AHEAD AND IDENTIFY
A FEW OF THESE POINTS. THIS WOULD BE (2,2). THIS WOULD BE (0,0),
AND THIS WOULD BE (-2,2). THE NEXT THING
WE NEED TO BE ABLE TO DO IS RECOGNIZE HOW TAKING
THE ABSOLUTE VALUE OF X + 3 AND THEN ADDING 2 WOULD
TRANSLATE THE PARENT FUNCTION. SO WHAT WE COULD DO IS SAY
THAT F OF X = G OF X + 3 + 2 IF THAT'S HELPFUL. NOTICE THAT G IS JUST
THE ABSOLUTE VALUE OF X, G OF X + 3 WOULD BE
THIS PART OF THE FUNCTION, AND THEN THE +2 WOULD BE
THE CONSTANT ON THE END. INCREASING X BY 3 HERE WHICH WOULD BE THE SAME
AS THIS X + 3 HERE WOULD SHIFT THE GRAPH LEFT
3 UNITS. THEN ADDING 2 TO THE ABSOLUTE
VALUE FUNCTION HERE OR HERE WOULD SHIFT THE FUNCTION UP
2 UNITS. SO NOW WHAT WE CAN DO
IS TAKE THESE 3 KEY POINTS AND SHIFT THEM LEFT 3 UNITS
AND UP 2 TO GRAPH THE GIVEN FUNCTION. LET'S DO THAT. LET'S START
WITH THE LEFTMOST POINT. WE'RE GOING TO GO LEFT 3 UNITS
AND THEN UP 2, SO WE'D BE OVER TO -5
AND THEN UP TO 4. NEXT, WE TAKE THIS POINT (0,0) AND SHIFT IT LEFT 3 UNITS
AND UP 2. WE WOULD BE HERE AT (-3,2). AND LASTLY, WE'LL TAKE THIS
RIGHTMOST POINT, SHIFT IT LEFT 3 UNITS AND UP 2, AND WE'D BE RIGHT HERE
AT (-1,4). NOW THAT WE KNOW THAT
THE ABSOLUTE VALUE FUNCTION IS A V SHAPE WE CAN FORM THE NEW FUNCTION
HERE IN GREEN USING TRANSLATIONS, AND IT WOULD LOOK
SOMETHING LIKE THAT. LET'S GO AHEAD
AND TRY ANOTHER ONE. AGAIN, THE FIRST STEP
IS GOING TO BE TO RECOGNIZE WHAT THE PARENT FUNCTION
WOULD BE. IF F OF X = THE QUANTITY X - 2
SQUARED - 4 WE SHOULD RECOGNIZE
THE PARENT FUNCTION AS, LET'S CALL IT G OF X
= X SQUARED. SO IF WE WANT TO CREATE F OF X
USING G OF X, F OF X IS GOING TO EQUAL
G OF-- WELL, X - 2 IS BEING SQUARED, SO WE'D HAVE G OF X - 2,
AND THEN WE'RE SUBTRACTING 4. NOW AFTER ALL, YOU MAY NOT HAVE
TO WRITE IT LIKE THIS, BUT IT MAY HELP AT THE BEGINNING
TO RECOGNIZE THAT HERE WE ARE DECREASING
THE INPUT OF X BY 2 WHICH MEANS IT WILL SHIFT IT
RIGHT 2 UNITS. RECOGNIZING THIS X - 2 HERE
IS THE SAME AS THE X - 2 HERE, AND THEN SUBTRACTING 4
FROM THIS FUNCTION VALUE WOULD SHIFT IT DOWN 4 UNITS. SO THIS - 4 IS THE SAME AS THIS
- 4 HERE IN THE GIVEN FUNCTION. SO TO GRAPH THIS GIVEN FUNCTION WE SHOULD FIRST SKETCH THE GRAPH
OF THE PARENT FUNCTION G OF X = X SQUARED. LET'S GO AHEAD AND DO THAT. REMEMBER, THAT IS A PARABOLA
WITH ITS VERTEX AT THE ORIGIN. SO THIS WOULD BE THE VERTEX.
1 SQUARED WOULD BE 1. 2 SQUARED WOULD BE 4, AND THEN WE HAVE A MIRROR IMAGE
ON THE OTHER SIDE OF THE Y-AXIS. SO WE HAVE A POINT HERE
AND A POINT HERE. SO THIS IS THE PARENT FUNCTION, AND WHAT WE'RE GOING TO DO NOW
IS PICK SOME KEY POINTS HERE AND THEN SHIFT EACH OF THEM
RIGHT 2 UNITS AND DOWN 4 UNITS. SO LET'S START WITH THE VERTEX. WE'LL SHIFT IT RIGHT 2 UNITS
AND DOWN 4 UNITS TO HERE. NEXT,
WE WILL TAKE THIS POINT HERE AND SHIFT IT RIGHT 2 UNITS
AND DOWN 4 UNITS. THAT WOULD BE HERE AT (1,-3). THEN WE'LL TAKE THE 0.11, SHIFT
IT RIGHT 2 AND DOWN 4 HERE. MIGHT AS WELL GO AHEAD
AND DO THESE LAST TWO POINTS. LET'S TAKE THIS ONE AND SHIFT IT
TO THE RIGHT 2, DOWN 4. IT WOULD BE HERE ON THE X-AXIS. AND THE SAME HERE,
SHIFT IT RIGHT 2 AND DOWN 4. WE WOULD BE AT THE ORIGIN NOW. SO THE TRANSLATED FUNCTION
WOULD BE THIS GREEN PARABOLA AS WE SEE HERE. THANK YOU FOR WATCHING. |
190265 | https://oeis.org/A000668/internal | A000668 - OEIS
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A000668
Mersenne primes (primes of the form 2^n - 1).
(Formerly M2696 N1080)
641
%I M2696 N1080 #457 Jul 25 2025 15:39:16
%S 3,7,31,127,8191,131071,524287,2147483647,2305843009213693951,
%T 618970019642690137449562111,162259276829213363391578010288127,
%U 170141183460469231731687303715884105727
%N Mersenne primes (primes of the form 2^n - 1).
%C For a Mersenne number 2^n - 1 to be prime, the exponent n must itself be prime.
%C See A000043 for the values of n.
%C Primes that are repunits in base 2.
%C Define f(k) = 2k+1; begin with k = 2, a(n+1) = least prime of the form f(f(f(...(a(n))))). - Amarnath Murthy, Dec 26 2003
%C Mersenne primes other than the first are of the form 6n+1. - Lekraj Beedassy, Aug 27 2004. Mersenne primes other than the first are of the form 24n+7; see also A124477. - Artur Jasinski, Nov 25 2007
%C A034876(a(n)) = 0 and A034876(a(n)+1) = 1. - Jonathan Sondow, Dec 19 2004
%C Mersenne primes are solutions to sigma(n+1)-sigma(n) = n as perfect numbers (A000396(n)) are solutions to sigma(n) = 2n. In fact, appears to give all n such that sigma(n+1)-sigma(n) = n. - Benoit Cloitre, Aug 27 2002
%C If n is in the sequence then sigma(sigma(n)) = 2n+1. Is it true that this sequence gives all numbers n such that sigma(sigma(n)) = 2n+1? - Farideh Firoozbakht, Aug 19 2005
%C It is easily proved that if n is a Mersenne prime then sigma(sigma(n)) - sigma(n) = n. Is it true that Mersenne primes are all the solutions of the equation sigma(sigma(x)) - sigma(x) = x? - Farideh Firoozbakht, Feb 12 2008
%C Sum of divisors of n-th even superperfect number A061652(n). Sum of divisors of n-th superperfect number A019279(n), if there are no odd superperfect numbers. - Omar E. Pol, Mar 11 2008
%C Indices of both triangular numbers and generalized hexagonal numbers (A000217) that are also even perfect numbers. - Omar E. Pol, May 10 2008, Sep 22 2013
%C Number of positive integers (1, 2, 3, ...) whose sum is the n-th perfect number A000396(n). - Omar E. Pol, May 10 2008
%C Vertex number where the n-th perfect number A000396(n) is located in the square spiral whose vertices are the positive triangular numbers A000217. - Omar E. Pol, May 10 2008
%C Mersenne numbers A000225 whose indices are the prime numbers A000043. - Omar E. Pol, Aug 31 2008
%C The digital roots are 1 if p == 1 (mod 6) and 4 if p == 5 (mod 6). [T. Koshy, Math Gaz. 89 (2005) p. 465]
%C Primes p such that for all primes q < p, p XOR q = p - q. - Brad Clardy, Oct 26 2011
%C All these primes, except 3, are Brazilian primes, so they are also in A085104 and A023195. - Bernard Schott, Dec 26 2012
%C All prime numbers p can be classified by k = (p mod 12) into four classes: k=1, 5, 7, 11. The Mersennne prime numbers 2^p-1, p > 2 are in the class k=7 with p=12(n-1)+7, n=1,2,.... As all 2^p (p odd) are in class k=8 it follows that all 2^p-1, p > 2 are in class k=7. - Freimut Marschner, Jul 27 2013
%C From "The Guinness Book of Primes": "During the reign of Queen Elizabeth I, the largest known prime number was the number of grains of rice on the chessboard up to and including the nineteenth square: 524,287 [= 2^19 - 1]. By the time Lord Nelson was fighting the Battle of Trafalgar, the record for the largest prime had gone up to the thirty-first square of the chessboard: 2,147,483,647 [= 2^31 - 1]. This ten-digits number was proved to be prime in 1772 by the Swiss mathematician Leonard Euler, and it held the record until 1867." [du Sautoy] - Robert G. Wilson v, Nov 26 2013
%C If n is in the sequence then A024816(n) = antisigma(n) = antisigma(n+1) - 1. Is it true that this sequence gives all numbers n such that antisigma(n) = antisigma(n+1) - 1? Are there composite numbers with this property? - Jaroslav Krizek, Jan 24 2014
%C If n is in the sequence then phi(n) + sigma(sigma(n)) = 3n. Is it true that Mersenne primes are all the solutions of the equation phi(x) + sigma(sigma(x)) = 3x? - Farideh Firoozbakht, Sep 03 2014
%C a(5) = A229381(2) = 8191 is the "Simpsons' Mersenne prime". - Jonathan Sondow, Jan 02 2015
%C Equivalently, prime powers of the form 2^n - 1, see Theorem 2 in Lemos & Cambraia Junior. - Charles R Greathouse IV, Jul 07 2016
%C Primes whose sum of divisors is a power of 2. Primes p such that p + 1 is a power of 2. Primes in A046528. - Omar E. Pol, Jul 09 2016
%C From Jaroslav Krizek, Jan 19 2017: (Start)
%C Primes p such that sigma(p+1) = 2p+1.
%C Primes p such that A051027(p) = sigma(sigma(p)) = 2^k-1 for some k > 1.
%C Primes p of the form sigma(2^prime(n)-1)-1 for some n. Corresponding values of numbers n are in A016027.
%C Primes p of the form sigma(2^(n-1)) for some n > 1. Corresponding values of numbers n are in A000043 (Mersenne exponents).
%C Primes of the form sigma(2^(n+1)) for some n > 1. Corresponding values of numbers n are in A153798 (Mersenne exponents-2).
%C Primes p of the form sigma(n) where n is even; subsequence of A023195. Primes p of the form sigma(n) for some n. Conjecture: 31 is the only prime p such that p = sigma(x) = sigma(y) for distinct numbers x and y; 31 = sigma(16) = sigma(25).
%C Conjecture: numbers n such that n = sigma(sigma(n+1)-n-1)-1, i.e., A072868(n)-1.
%C Conjecture: primes of the form sigma(4(n-1)) for some n. Corresponding values of numbers n are in A281312. (End)
%C [Conjecture] For n > 2, the Mersenne number M(n) = 2^n - 1 is a prime if and only if 3^M(n-1) == -1 (mod M(n)). - Thomas Ordowski, Aug 12 2018 [This needs proof! - Joerg Arndt, Mar 31 2019]
%C Named "Mersenne's numbers" by W. W. Rouse Ball (1892, 1912) after Marin Mersenne (1588-1648). - Amiram Eldar, Feb 20 2021
%C Theorem. Let b = 2^p - 1 (where p is a prime). Then b is a Mersenne prime iff (c = 2^p - 2 is totient or a term of A002202). Otherwise, if c is (nontotient or a term of A005277) then b is composite. Proof. Trivial, since, while b = v^g - 1 where v is even, v > 2, g is an integer, g > 1, b is always composite, and c = v^g - 2 is nontotient (or a term of A005277), and so is for any composite b = 2^g - 1 (in the last case, c = v^g - 2 is also nontotient, or a term of A005277). - Sergey Pavlov, Aug 30 2021 [Disclaimer: This proof has not been checked. - N. J. A. Sloane, Oct 01 2021]
%D Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 4.
%D John Brillhart, D. H. Lehmer, J. L. Selfridge, Bryant Tuckerman and S. S. Wagstaff, Jr., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.
%D John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 135-136.
%D Graham Everest, Alf van der Poorten, Igor Shparlinski and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
%D Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 76.
%D Marcus P. F. du Sautoy, The Number Mysteries, A Mathematical Odyssey Through Everyday Life, Palgrave Macmillan, First published in 2010 by the Fourth Estate, an imprint of Harper Collins UK, 2011, p. 46. - Robert G. Wilson v, Nov 26 2013
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D Bryant Tuckerman, The 24th Mersenne prime, Notices Amer. Math. Soc., 18 (Jun, 1971), Abstract 684-A15, p. 608.
%H Harry J. Smith, Table of n, a(n) for n = 1..18
%H Peter Alfeld, The 39th Mersenne prime, 2003.
%H Yan Bingze, Li Qiong, Mao Haokun, and Chen Nan, An efficient hybrid hash based privacy amplification algorithm for quantum key distribution, arXiv:2105.13678 [quant-ph], 2021.
%H Andrew R. Booker, The Nth Prime Page
%H John Rafael M. Antalan, A Recreational Application of Two Integer Sequences and the Generalized Repetitious Number Puzzle, arXiv:1908.06014 [math.HO], 2019-2020.
%H W. W. Rouse Ball, Mathematical recreations and problems of past and present times, London, Macmillan and Co., 1892, pp. 24-25.
%H W. W. Rouse Ball, Mersenne's numbers, Messenger of Mathematics, Vol. 21 (1892), pp. 34-40, 121.
%H W. W. Rouse Ball, Mersenne's numbers, Nature, Vol. 89 (1912), p. 86.
%H John Brillhart, D. H. Lehmer, J. L. Selfridge, Bryant Tuckerman and S. S. Wagstaff, Jr., Factorizations of b^n +- 1, Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 3rd edition, 2002.
%H Kevin A. Broughan and Qizhi Zhou, On the Ratio of the Sum of Divisors and Euler's Totient Function II, Journal of Integer Sequences, Vol. 17 (2014), Article 14.9.2.
%H Douglas Butler, Mersenne Primes.
%H C. K. Caldwell, Mersenne primes.
%H C. K. Caldwell, "Top Twenty" page, Mersenne Primes.
%H Luis H. Gallardo and Olivier Rahavandrainy, On (unitary) perfect polynomials over F_2 with only Mersenne primes as odd divisors, arXiv:1908.00106 [math.NT], 2019.
%H Richard K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy]
%H Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 12.
%H Sameen Ahmed Khan, Primes in Geometric-Arithmetic Progression, arXiv preprint arXiv:1203.2083 [math.NT], 2012. - From N. J. A. Sloane, Sep 15 2012
%H Abílio Lemos and Ady Cambraia Junior, On the number of prime factors of Mersenne numbers, arXiv:1606.08690 [math.NT] (2016).
%H Benny Lim, Prime Numbers Generated From Highly Composite Numbers, Parabola (2018) Vol. 54, Issue 3.
%H Math Reference Project, Mersenne and Fermat Primes.
%H Romeo Meštrović, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012. - From N. J. A. Sloane, Jun 13 2012
%H Romeo Meštrović, Goldbach-type conjectures arising from some arithmetic progressions, University of Montenegro, 2018.
%H Romeo Meštrović, Goldbach's like conjectures arising from arithmetic progressions whose first two terms are primes, arXiv:1901.07882 [math.NT], 2019.
%H Landon Curt Noll, Mersenne Prime Digits and Names.
%H Passawan Noppakaew and Prapanpong Pongsriiam, Product of Some Polynomials and Arithmetic Functions, J. Int. Seq. (2023) Vol. 26, Art. 23.9.1.
%H Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
%H Primefan, The Mersenne Primes.
%H Christian Salas, Cantor Primes as Prime-Valued Cyclotomic Polynomials, arXiv preprint arXiv:1203.3969 [math.NT], 2012.
%H Harry J. Smith, Mersenne Primes, 1981-2010.
%H Gordon Spence, 36th Mersenne Prime Found, 1999.
%H Susan Stepney, Mersenne Prime.
%H Thesaurus.maths.org, Mersenne Prime.
%H Bryant Tuckerman, The 24th Mersenne prime, Proc. Nat. Acad. Sci. USA, Vol. 68 (1971), pp. 2319-2320.
%H Samuel S. Wagstaff, Jr., The Cunningham Project.
%H Yunlan Wei, Yanpeng Zheng, Zhaolin Jiang and Sugoog Shon, A Study of Determinants and Inverses for Periodic Tridiagonal Toeplitz Matrices with Perturbed Corners Involving Mersenne Numbers, Mathematics, Vol. 7, No. 10 (2019), 893.
%H Eric Weisstein's World of Mathematics, Mersenne Prime.
%H Eric Weisstein's World of Mathematics, Perfect Number.
%H Wikipedia, Mersenne prime.
%H Marek Wolf, Computer experiments with Mersenne primes, arXiv preprint arXiv:1112.2412 [math.NT], 2011.
%H Chai Wah Wu, Can machine learning identify interesting mathematics? An exploration using empirically observed laws, arXiv:1805.07431 [cs.LG], 2018.
%F a(n) = sigma(A061652(n)) = A000203(A061652(n)). - Omar E. Pol, Apr 15 2008
%F a(n) = sigma(A019279(n)) = A000203(A019279(n)), provided that there are no odd superperfect numbers. - Omar E. Pol, May 10 2008
%F a(n) = A000225(A000043(n)). - Omar E. Pol, Aug 31 2008
%F a(n) = 2^A000043(n) - 1 = 2^(A000005(A061652(n))) - 1. - Omar E. Pol, Oct 27 2011
%F a(n) = A000040(A059305(n)) = A001348(A016027(n)). - Omar E. Pol, Jun 29 2012
%F a(n) = A007947(A000396(n))/2, provided that there are no odd perfect numbers. - Omar E. Pol, Feb 01 2013
%F a(n) = 4A134709(n) + 3. - Ivan N. Ianakiev, Sep 07 2013
%F a(n) = A003056(A000396(n)), provided that there are no odd perfect numbers. - Omar E. Pol, Dec 19 2016
%F Sum_{n>=1} 1/a(n) = A173898. - Amiram Eldar, Feb 20 2021
%p A000668 := proc(n) local i;
%p i := 2^(ithprime(n))-1:
%p if (isprime(i)) then
%p return i
%p fi: end:
%p seq(A000668(n), n=1..31); # Jani Melik, Feb 09 2011
%p # Alternate:
%p seq(numtheory:-mersenne([i]),i=1..26); # Robert Israel, Jul 13 2014
%t 2^Array[MersennePrimeExponent, 18] - 1 ( Jean-François Alcover, Feb 17 2018, Mersenne primes with less than 1000 digits )
%t 2^MersennePrimeExponent[Range] - 1 ( Eric W. Weisstein, Sep 04 2021 )
%o (PARI) forprime(p=2,1e5,if(ispseudoprime(2^p-1),print1(2^p-1", "))) \ Charles R Greathouse IV, Jul 15 2011
%o (PARI) LL(e) = my(n, h); n = 2^e-1; h = Mod(2, n); for (k=1, e-2, h=2hh-1); return(0==h) \ after Joerg Arndt in A000043
%o forprime(p=1, , if(LL(p), print1(p, ", "))) \ Felix Fröhlich, Feb 17 2018
%o (GAP)
%o A000668:=Filtered(List(Filtered([1..600], IsPrime),i->2^i-1),IsPrime); # Muniru A Asiru, Oct 01 2017
%o (Python)
%o from sympy import isprime, primerange
%o print([2n-1 for n in primerange(1, 1001) if isprime(2n-1)]) # Karl V. Keller, Jr., Jul 16 2020
%Y Cf. A000225 (Mersenne numbers).
%Y Cf. A000043 (Mersenne exponents).
%Y Cf. A001348 (Mersenne numbers with n prime).
%Y Cf. A000040, A000203, A000217, A000396, A003056, A007947, A016027, A019279, A023195, A023758, A028335 (lengths), A034876, A046051, A057951-A057958, A059305, A061652, A083420, A085104, A124477, A135659, A173898.
%K nonn,nice,hard
%O 1,1
%A N. J. A. Sloane
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190266 | https://www.ux1.eiu.edu/~addavis/1350/06CirMtn/FlatCurve.html | Circular Motion (and other things)
Circular Motion
Newton's Second Law
applied to a
Flat Curve
What provides the centripetal force necessary for a car to make a curve on a flat road? What happens if that force is not present?
Gravity pulls down on the car with its weight, w = m g. The level road pushes up with a normal force n and horizontally with a friction force Ffrict.
This is the force of static friction! As the car moves and the tires rotate, the tires are momentarily at rest with respect to the road. Otherwise, the tires skid!
Remember, the friction force can be any value from zero up to a maximum of F s = s n when the car is just on the verge of sliding. We will consider this case, when the car is just on the verge of sliding. This means
F frict = F s = s n
The diagram at the right shows all these forces. There is no vertical component to the acceleration so we find that
n = m g
which means
F net = f s = s m g = m v 2/r
s g = v 2/r
s = v 2/g r
That is, we must have a coefficient of static friction of s = v 2/g r to provide the friction force to allow a car, traveling at speed v, to make it around a flat curve of radius r.
Or, we might find the speed in terms of this coefficient of static friction,
v 2 = s g r
v = SQRT[s g r]
This is the maximum speed that a curve of radius r can be taken when the coefficient of static friction between tires and pavement is s.
If the velocity increases, the radius r will also increase! This means the car will not follow the intended curve and may run off the road entirely!
What must be the coefficient of friction between the tires and the level roadway to allow a car to make a curve of radius r = 350 m at a speed of 80 km/h?
UCMnonuniform circular motion
Table of Contents
(c) Doug Davis, 2001; all rights reserved |
190267 | https://danielcameronmd.com/pregnancy-breast-feeding-and-lyme/ | Pregnancy, breast feeding and Lyme - Daniel Cameron MD
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Sep 10
Pregnancy, breast feeding and Lyme
Posted By: Laurie Martin
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Visited 330 Times, 1 Visit today
Welcome to another selection from my book “An Expert’s Guide on Navigating Lyme disease.” The book highlights the findings of my first 600 Lyme disease Science blogs. In this episode, I will discuss pregnancy, breast feeding, and Lyme disease.
Pregnancy and breast feeding are particularly concerning topics for Lyme disease patients and their doctors. Here are few findings to help with discussions.
Pregnancy and Lyme.
If a woman is bit by a deer tick or contracts Lyme disease (LD) while pregnant, she must immediately alert her treating physician. Poor outcomes have been described for newborns whose mothers contracted Lyme disease during pregnancy. Studies have found that stillbirths occurred when LD was contracted during the first trimester. Markowitz et al., 1986, Schlesinger et al., 1985, and MacDonald et al., 1987) Gestational Lyme borreliosis has been described in 5 of 19 pregnancies (26%) resulting in “syndactyly, cortical blindness, intrauterine fetal death, prematurity, and rash” (Markowitz et al., 1986).
A newborn died at 39 hours of life with a left-sided hypoplastic heart and the presence of spirochetes consistent with Bb [Borrelia burgdorferi] “in the spleen, kidneys, and bone marrow” (Schlesinger et al., 1985). Bb was also cultured from fetal liver tissue in 4 stillborn infants (MacDonald 1986). There was insufficient evidence to determine the risk to a child if their mother contracted Lyme disease before pregnancy (Mylonas 2011). A study of 2,000 women with a history of LD did not show an increased risk of fetal death, decreased birth weight, or length of gestation at delivery. There was an increase in the number of congenital defects but the risk may have been by chance alone (Strobino et al., 1993).
Choosing an antibiotic regimen for pregnant women with Lyme disease can be a complex challenge. Amoxicillin, cefuroxime, azithromycin, and IV ceftriaxone have been prescribed for pregnant women (Maraspin et al., 2009).
Author’s note: More studies will be needed to understand pregnancy and breast feeding concerns.
Breast feeding and Lyme.
The CDC addressed the question “Can Lyme disease be transmitted through breast milk?” They announced, “No reports of breast milk spreading Lyme disease to infants exist” (CDC 2022). There is insufficient data to determine if breastfeeding can transmit Bb to the child. Certain antibiotic classes, such as tetracyclines, should not be used in breastfeeding women being treated for Lyme disease to avoid the risk of side effects, such as tooth discoloration. The child’s clinician can help guide treatment options for a breastfeeding mother. Read more.
Diversity of clinical presentations of Lyme and pregnancy.
Doctors followed 11 pregnant women with Lyme disease from 2008 to 2020. “In the present study, we report our case series, which includes 11 pregnant women, 6 of whom developed erythema migrans during pregnancy (between weeks 8 and 34), 3 had monoarticular or neurological symptoms, and 2 had positive serology but did not develop any clinical symptoms” (Trevisan et al., 2020).
All mothers were treated with amoxicillin 1g 3x/ day for 14 days. One child was born prematurely at seven months. Another child was born with angiomatoid patches that regressed spontaneously 18 months later. One of the pregnant women with Lyme disease, confirmed by spinal tap and labs, experienced articular and neurologic involvement and improved with amoxicillin. However, she required treatment with intravenous ceftriaxone because of persistent symptoms. Read more.
Two mothers transmit Lyme to their babies.
Babesia can be contracted from the bite of a deer tick, a blood transfusion, or during pregnancy.This podcast reviews a case in which Babesia was transmitted from mothers to their babies during pregnancy.
Questions raised in the podcast include: • How often do mothers contract Babesia from a tick bite during pregnancy? • Is there an effective and safe treatment for Babesia in pregnant women? • How does a mother or doctor recognize Babesia in a pregnant mother? • Should doctors follow pregnant mothers with a tick bite or Lyme disease for Babesia and what evidence should be investigated? • Will the mothers develop complications of Babesia in the future if not treated? • Should the two mothers have been treated for Babesia? Read more.
[bctt tweet=”Two mothers transmit Lyme to their babies.” username=”DrDanielCameron”]
A baby girl with Lyme disease.
Slovenian researchers investigated whether Borrelia burgdorferi bacteria, the pathogen causing Lyme disease, might impact pregnancy outcomes. Pregnancy outcome was unfavorable in 13.8% (42/304) of patients. They found that the outcome of pregnancy in Lyme disease patients was not significantly worse.
There were 22 pre-term births, 10 fetal/perinatal deaths, and/or 15 anomalies. Several mothers had potential explanations for their unfavorable pregnancy outcomes. The poor outcome for Lyme disease patients was not significantly different compared to the general population (Maraspin et al., 2020).
Author’s note: The study did not follow the 262 women who gave birth with a favorable outcome for any long-term problems. Nor did the authors describe the outcome for women who were not treated for early Lyme disease. Read more.
Little information on treatment of tick bites during pregnancy.
Smith et al., (2020) argue that “high-level evidence” supports using a single 200 mg dose of doxycycline for tick bites during pregnancy. The evidence they cited is not high-level. Instead, they focused on a small Meta-Analysis study. Regrettably, there is no evidence that a single 200 mg dose of doxycycline prevents other manifestations of Lyme disease, such as heart block, 7th nerve palsy, Lyme arthritis, Lyme encephalopathy, or Neuropsychiatric Lyme disease.Read more.
Congenital transmission of Babesia to a 5-year-old twin.
A baby girl was born to a mother who showed no evidence of Lyme or a related tick-borne illness during her pregnancy (Walker et al., 2022). The 5-week-old female diamniotic dichorionic twin was born at 36 5/7 weeks by C-section and diagnosed with Babesia. Her twin brother was asymptomatic. The mother described several trips to Cape Cod, Massachusetts, an area endemic to Lyme disease.
“The patient’s mother had one febrile illness during pregnancy, occurring at approximately 23-24 weeks of gestation, associated with a maculopapular rash that resolved spontaneously” (Walter et al., 2022). The daughter was treated with a blood transfusion, intravenous atovaquone twice daily, and azithromycin daily. The authors of the article pictured a blood smear with intraerythrocytic inclusions consistent with Babesia microti. Read more.
Delayed onset Babesia in two newborns.
A study from the Mayo clinic described two newborn infants diagnosed with Babesia several weeks after the mothers were treated for Lyme disease (Hoversten and Bartlett, 2018). Infant 1: A baby boy was diagnosed with Babesia at 4-1/2 weeks. His mother had been diagnosed and treated for early Lyme disease at 32 weeks gestation.Infant 2: A baby girl was diagnosed with Babesia at 18-days-old. Her mother had been diagnosed and treated for early Lyme disease at 37 weeks gestation. Neither mother was treated for Babesia during their pregnancy. Read more.
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2 thoughts on “Pregnancy, breast feeding and Lyme”
Jen Miklavcic September 11, 2023 at 2:56 pm
Interesting, that is not what I was told when I was pregnant. I was being treated by one of the top Lyme Doctors(He is no longer practicing but may still be doing research) and he was very clear that breastfeeding was not an option. He also indicated that most women feel great during pregnancy and had a horrible time after. My kids are 22, 19, and 15 and I was told I could not breastfeed. During my first pregnancy, I was not on antibiotics and after the pregnancy, I had a horrible relapse. In my other two pregnancies, I was on antibiotics and did not have nearly the same reaction. I would be curious as to what has warranted the change of thinking esp since the number of mothers and studies seem rather low. Overall, I have trouble trusting the CDC when it comes to anything related to Lyme and tick borne diseases. I hope the experts in the field are not relying only on the CDC
Reply
1. Dr. Daniel Cameron September 12, 2023 at 11:40 am
I have not seen published data to advise my patients not to breast feed. I have colleagues who continued to advise their patients not to breast feed. There advise could be out of cases they have seen or an abundance of concern. I choose to discuss options with the mother and father until data is available.
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190268 | https://www.amplivox.com/education/knowledge-hub/identifying-age-related-hearing-loss | Identifying age related hearing loss | Amplivox
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Identifying age related hearing loss (Presbycusis)
Reading Time: 10 minutes
Written by Amplivox on 26 March 2024
What is Presbycusis?
As people get older, age related hearing loss (or presbycusis) may gradually affect their hearing. Essentially, it's the slow decline of high-pitched hearing in both ears. It’s one of the most common conditions accompanying ageing, affecting about one in three people ages 65–74 and nearly half of those 75 and older.1
Hearing problems are common in older adults and are most commonly the result of the natural ageing process, but other factors such as working environment, past activities, and health issues can also impact a person's hearing as they age. For instance, a person's previous social life can impact their hearing in the future.2
Age related hearing loss usually has an equal affect on both ears but people with presbycusis may not notice they are losing their hearing because it tends to happen gradually over time.
Common causes of hearing loss with old age
As we get older, our hearing might naturally deteriorate. However, experts believe that various factors, not just ageing, can impact hearing loss. As mentioned, this includes things like genetics, long-term exposure to loud noises, and certain medical conditions like diabetes, cardiovascular disease and high blood pressure.
Complex changes in the nerve pathways from the ear to the brain can worsen hearing loss, as can abnormalities in the middle ear (however these are less common).
Certain drugs, such as those used for chemotherapy can also be harmful to the sensory cells in your ears, which can also lead to hearing loss.3
Symptoms of age related hearing loss
Presbycusis tends to occur within the highest frequency sounds first. This results in people having difficulty with certain tones, including:
| Speech | Alarms and warnings |
--- |
| Consonants provide clarity of speech. People with age-related hearing loss often struggle to hear and distinguish these kinds of sounds, which can lead to speech sounding unclear, muffled, or distorted. | Technological devices such as oven timers and alerts have high frequency sounds. Those with presbycusis will struggle to hear them. |
| Difficulty hearing in noisy environments | Needing higher volumes |
| Background noise can make it difficult for people with age-related hearing loss to grasp conversations. This is partly because they have difficulty distinguishing between different sounds. | They may need to increase the volume on their TV, radio, or other audio devices to hear better. |
| Tinnitus | Social isolation |
| Tinnitus is a noise in the ears, like ringing or buzzing, that some older people with hearing loss can experience. Age-related hearing loss doesn't exclusively cause tinnitus, but it can be an associated symptom. | People who have difficulty hearing may struggle to participate in social activities. As a result, they can choose to distance themselves from others, which can lead to feelings of loneliness and sadness. |
It's important to remember that age-related hearing loss can be different for each person, and not everyone will have the same symptoms.
How is age related hearing loss treated?
Once an audiologist has identified and diagnosed a type of hearing loss, they can determine a suitable rehabilitation method. This will vary depending on the situation. An audiologist will talk about appropriate treatment options.
One possible method of rehabilitation is through the prescription of a hearing aid. The type of hearing aid can vary in output performance, size, style, and features, depending on the nature and severity of the hearing loss.
In severe cases a cochlear implant might be required, for example when hair cells are damaged or missing, or where sound isn't able to properly reach the hearing nerve.
Industry-leading audiometers
It's important to routinely assess the hearing function to identify any hearing loss. Hearing checks can take place in the community, at work, or by visiting a doctor or audiologist.
Our extensive range of dependable, PC-based screening audiometersprovide a customised and flexible mobile or static screening solution. They also provide data management tailored to the user’s specific requirements, guaranteeing accurate and efficient testing.
Before choosing the correct treatment, it's important to accurately diagnose the cause and type of hearing loss.
We've developed a suite of easy-to-use diagnostic audiometers. Our devices can perform a wide range of hearing loss tests including AC, BC, speech, and special tests like ABLB, Stenger, SISI, Tone decay, HLS, and MHA.
To learn more about our audiometers, you can visit our audiometerswebpage. Or contact our customer support team on +44 (0)1865 880 846 or via email.
References
1 Johns Hopkins Medicine. Age-Related Hearing Loss (Presbycusis). Accessed at:
2 National Institute on Aging. Hearing Loss: A Common Problem for Older Adults. Accessed at:
3 National Institute on Deafness and Other Communication Disorder. Age-Related Hearing Loss (Presbycusis). Accessed at:
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190269 | https://www.youtube.com/watch?v=wKOTYNfw3DM | Geometry - Perpendicular Bisectors and Angle Bisectors, Circumcenter and Incenter
iteachalgebra
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Posted: 11 Dec 2020
Join me as I explain what perpendicular bisectors look like, the theorem, and how they meet at the circumcenter, as well as explain angle bisectors and how they meet at the incenter.
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Transcript:
Intro hi everyone welcome to this lesson where today we're talking about perpendicular bis sectors and angle bis sectors so let's take a look so first of all we have What's called the perpendicular bis sector theorem and it says if a point is on the perpendicular bis sector of a segment then it is equidistant from the end points of the segment so basically what this is saying is that if I have a point if a point is on the perpendicular bis sector so first of all let's say I have this segment here AB and this line that's going through it let's say line M is my perpendicular bis sector so perpendicular bis sector means two things it means I am bisecting the segment so though I have two uh congruent segments created and it's also at a right angle so the bis sector and that segment create um right angles so it's saying if a point is on the perpendicular bis sector so line M is my perpendic bis sector so let's say I mark off a point here a point on the perpendicular bis sector then it is equidistant from the end points of the segment so if I was to connect that any point on my perpendicular bis sector to those sides there it's basically just telling me that these two segments that I've now created are congruent to each other which in theory should also make total sense because I did just create two right triangles and by side side angle side I could prove that the two triangles are congruent and then by cpctc prove that those two other segments of the triangles are congruent to each other the converse says if a point is equidistant from the end points on a segment then it would mean that there is this is a perpendicular bis sector so if I tell you there's a perpendicular bis sector then I know I would know that these two segments are congruent or the converse if I just tell you that hey these two segments are congruent well that must mean that there is a perpendicular by sector there circum Center so the circum Center is where the perpendicular bis sectors intersect um at this one point of um concurrency so it's that point is called the circum Center and you're going to see in just a moment what that actually means and that circumcenter is equidistant from the vertices and again in just a moment you're going to see what that means so here is just a full Perpendicular Bisector out explanation of a perpendicular bis sector so when you see th that term perpendicular bis sector it means you are working with um right angles and it also means that you're working with congruent segments and just for sake of time I'm not going to write out the full word so perpendicular bis sector perpendicular means we're working with right angles bis sector means that we're working with congruent segments and here it says the circum Center is where the perpendicular bis sectors meet so I want you to look at this triangle ABC and what I want us to be able to understand from this diagram is if I look at segment AB this line here is the perpendicular bis sector it is creating two congruent segments and a right angle if I look at segment BC okay BC is now split up into two congruent segments by that line at a right angle and then segment AC is by bisected into two congruent segments and I just need to go back actually for a moment and add in another marker here at a right angle and now when you construct perpendicular bis sectors in a triangle where those points all meet where those lines all meet rather is called the circum Center so this point here is called the circum Center and something special about the circum Center that was mentioned in the previous uh slide rather is that the circum Center is equidistant to all of the vertices so all of the segments I'm making in yellow that those are actually all congruent to each other the circum Center um so it's pretty interesting it's like as if you know there were three towns towns a b and c and you wanted to put a new store and you wanted the store to be equally distant from town a town B Town C if you actually on a map constructed the circum Center you would find the location that's actually directly the same length from all three towns so no matter what town you were from you would travel the same distance and get to that store all right so we're going to take a look at this diagram which is super similar but you're going to see I actually marked up some of the letters here um if I wanted to understand that point D is the circum Center so again that does mean and I'm going to go ahead um mark up some things here that these are all right angles okay and then my segments are all marked up as congruent my sides of my triangle Are all uh bisected it says be is equal to I'm sorry congruent to so segment be I would know is then congruent to segment AE right because the entire side of the triangle has been B Ed segment AF is then congruent to segment CF segment CG is congruent to segment um BG okay the distance from the circum Center to the vertices is congruent so that's what I marked up in yellow on the previous screen so from a to d would be congruent from B to D and c2d okay so where the circum Center is to those vertices those segments are all congruent to each other you also have quite a few congruent triangles created in this diagram now so if I was to look at triangle a I'm sorry AF D what I should notice is that triangle AF D this triangle here is congruent to Triangle cfd right because they share a common side a common right angle and then they share this common side and all three sets of sides are congruent because they share the side from reflexive property and then these segments here are congruent to each other so then therefore those two triangles are congruent triangle a d would then be congruent to Triangle b d so you've got this little mini triangle drawn and then triangle b g d is congruent to Triangle CG D okay now actually constructing the Circumcenter circum Center and this is going to be a little challeng in for me here um virtually to do but I'm going to show you to the best of my ability I have a ruler here and that was going to help me pretty much figure out where the bis sector is so I'm going to show you what that would look like and then I may put the ruler away just to save myself some time so let me um create a line here okay okay so if I wanted to figure out what the where the perpendicular bis sector would be I would measure the side of a triangle figure out where my halfway point would be which is approximately right here and then from that point I would go ahead and I would construct a line so that I have officially bisected the segment at a right angle and I know it's a right angle only because I was lining it up with my ruler to make sure that if it's perpendicular to each other it's a vertical line to my horizontal line and then what you would need to do is you would need to um you know rotate your ruler whoops you would need to move your ruler I'm having a hard time with that measure another side this is obviously not going to be the best thing for me so I'm just going to eyeball it right now for the sake of this lesson but I would go ahead and I would measure this side here on the left figure out what the where the midpoint is and then construct my body B sector and then my third side here again find the midpoint of that side BCT that side and make a right angle a perpendicular line rather and you can see my sketch is not perfect perfect but you can see that these two lines in these three lines intersect each other now this triangle guys is acute this triangle is acute and what you're going to notice is that the inter section is actually inside of the triangle whereas when I go over to a right triangle if I go to construct some perpendicular Bice sectors just make sure my pen is all good I go to construct this side I find the midpoint of this side and set up a line perpendicular to that side and then my bottom of my triangle here if I construct that and I really should move that one over a bit to really bict it and then if I bict my hypotenuse of this triangle what I'm going to notice is look at that if I create the perpendicular bis sectors for all three sides of a right triangle what we're going to notice is that the circumcenter is actually on the hypotenuse okay so if you have an acute triangle the circumcenter is inside of the triangle if you have a right triangle the circum Center is actually on the triangle or on on the hypotenuse and now the last example I want to show you is if I have an obtuse triangle and so I'm going to try my best here to create a line that's perpendicular to my segment that's on the left create a perpendicular somewhat Midway so that Midway wasn't great I'm going to scooch that over a bit so again I'm trying to bict the side and make a perpendicular line and then if I go ahead and I BCT my third side and create a perpendicular to it you're going to notice here that if your triangle is obtuse then the circum Center actually intersects outside of the triangle right so that's a pretty interesting um series of events rather and it looks like Angle Bisector that okay the next part of this lesson is the angle bis sector theorem so it says if a point is on the bis sector of an angle so I'm going to mark this up if a point is on the bis sector of an angle so here we have an angle that's in Black we have a red angle bis sector if a point is on the angle bis sector then it is equidistant from the sides of an angle so the distance from this point to um one of the sides of an angle here and remember distance is always created at a perpendicular line distance is always the shortest distance uh created at a perpendicular then those lengths would be equidistant to each other okay so any point on an angle bis sector is then equidistant to the sides of that angle which then do create two little congruent right triangles for sure the converse would be the opposite then it says if the point in the interior is equidistant from both sides then that means that the red line would be an angle bis sector so the regular theorem is if you have an angle bis sector then any point on the angle bis sector is congruent the distance is congruent to the sides of the angle and the converse is if I tell you that the two sides um these two segments are congruent and that must mean that that was an angle bis sector in Center so angle bis sectors meet at the in Center okay so just like before perpendicular bis sectors meet at the circumcenter angle bis sectors meet at the in Center so it says the angle bis sectors of a triangle intersect at a point called the in center that is equidistant from the sides of a triangle so perpendicular bis sectors meet at the circum Center which is equidistant to the vertices the in Center where all the angle bis sectors meet is equidistant from the sides okay so an angle bis sector means you are working with congruent angles okay an angle bis sector just simp means you're working with congruent angles so here an in center so if I these are all angle bis sectors here so angle a is split into two congruent angles angle B is split into two congruent angles and angle C is split into two congruent angles and then the point where they meet is called the in center now the in Center it says um I said before is equidistant from the sides so if I was to constru constuct the length here the distance to the sides remember distance is always perpendicular that distance is all the same now in my previous example I talked about there being three towns and then the circum Center would be the location that's closest to all three towns in this case if you're talking about a side then it's kind of like this would be the best location so let's say if you're playing tag and there was three walls that were safe this would be the point where it wouldn't matter which wall you went to wall AB wall AC or wall BC um it's the same distance from any wall and you could go to any wall to basically be safe on home base I just think of my um times as a kid just playing tag and you know going to the wall that was safe and that would kind of be that same thing so that position here the in Center is the same distance to any of let's say the three walls in a room okay so Point D is the in center of triangle ABC so again in Center is going to mean that it's the um the intersection of all three angle bis sectors so I'm just going to mark up my diagram so again we know what we're working with and then remember these segments here that I'm going to draw in yellow those distances which are all at a right angle are congruent to each other so angle e a d is obviously congruent to angle f a angle FCD is congruent to angle G CD and angle GB D is congruent to angle EBD the perpendicular distance from the in Center to the sides is congruent which I marked up in yellow so um H which is a new point that I created so this point here I'm going to call it H the perpendicular distance I'm going to call I here this perpendicular distance I'm going to call J so H to D is congruent from I to D which is congruent from J to D okay now the in Center it says here is always in the interior of a triangle so the best thing we can always do guys is grab a protractor actually line up the protractor with the vertex of the triangle measure the angle so this one I'm measuring is approximately let's say 58 cut that angle in half so if I take half of 58 I'm at 39 I'm going to make a little dot there move my protractor over and then construct that angle bis sector so um measuring an angle and of course this is just a sketch of what it would look like but actually measuring the angle and then marking up where halfway would be and then going through and create doing that for every single side is how you would construct your incenter so if I was to go ahead and then measure whoops measure this angle here on the right I would measure that this angle is about 70 uh 71 if I go halfway that's going to be at 35 so if I make a marking where 35 would be approximately and of course pencil and paper is going to be the easiest way to do this for sure me doing it on my screen here and I know my screen's shaking um and creating those angle bis sectors here for each one and I can mark up you know what the whole point is that these two angles are congruent these two angles are supposed to be congruent you can see my red line is not the greatest but the whole point of showing you this is that it's not like perpendicular bis sectors anytime you have angle bis sectors the angle bis sectors will always intersect in the interior of the triangle so the in Center is always in the interior even over here if I was to just visually try to um bict these angles here if I BCT my 90° angle if I BCT that angle there and if I BCT this angle here again you can see the in Center would be in the interior of the triangle last one if I BCT my angle on the left left just eyeballing it BCT my angle on the right again just eyeballing it for the sake of the lesson and bisecting this angle here my drawing again is not the best but notice it's never going to be on the outside it's never going to be inside I'm sorry on the line of the triangle it will always be in the interior well that's it for today just showing you and that's obviously a much much better diagram thank you so much for following along about perpendicular bisc and angle bis sectors bye |
190270 | https://www.quora.com/How-many-right-angled-triangles-can-be-formed-by-joining-the-vertices-of-cuboid | How many right angled triangles can be formed by joining the vertices of cuboid? - Quora
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How many right angled triangles can be formed by joining the vertices of cuboid?
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Mihir Choudhary
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Originally Answered: How many right angled triangles can be formed by joining the vertices of a unit cube? ·
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William Harris
someone who wishes he knew more about mathematics than he does. · Author has 1K answers and 1.1M answer views
·7y
Originally Answered: How many right angled triangles can be formed by joining the vertices of a unit cube? ·
A cube has face-diagonals (across each face) and body-diagonals (through the cube from one corner to the opposite corner). Consider a face-diagonal. It is the hypotenuse of 2 triangles on the same face. It can also be the side (non-hypotenuse) of 2 triangles of which a body-diagonal is the hypotenuse. Thus each face-diagonal is part of 4 triangles. There are 2 face-diagonals on each side. Thus the face-diagonals of each side are part of 8 triangles. The cube has 6 sides. Hence there are 8x6 that is 48 triangles of which a face-diagonal is part.
We have considered only the face-diagonals. What a
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A cube has face-diagonals (across each face) and body-diagonals (through the cube from one corner to the opposite corner). Consider a face-diagonal. It is the hypotenuse of 2 triangles on the same face. It can also be the side (non-hypotenuse) of 2 triangles of which a body-diagonal is the hypotenuse. Thus each face-diagonal is part of 4 triangles. There are 2 face-diagonals on each side. Thus the face-diagonals of each side are part of 8 triangles. The cube has 6 sides. Hence there are 8x6 that is 48 triangles of which a face-diagonal is part.
We have considered only the face-diagonals. What about an edge on the side? It is part of triangles but each has a side that is a face-diagonal and so has already been counted. Thus there are no more triangles.
Hence the total number of right-angled triangles formed by the vertices of a cube is 48.
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Doug Dillon
Ph.D. Mathematics · Author has 12.4K answers and 11.4M answer views
·7y
Originally Answered: How many right angled triangles can be formed by joining the vertices of a unit cube? ·
Consider cases bases on the number of cube edges which are used to make a right triangle.
Case 1: 2 edges.
In this case, the two edges meet at a corner to make a right angle. This happens 4 4 times per faces and accounts for 24 right triangles.
Case 2: 1 edge.
In this case, the right angle is formed by one edge and one face diagonal of the cube. The hypo...
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Adithya Lanka
Worked at Bloomberg (company) · Author has 92 answers and 469.2K answer views
·9y
Originally Answered: How many right angled triangles can be made out of a cuboid by joining its vertices? ·
Let us first count the triangles formed on the faces.
On each face, we can draw four right-angled triangles by choosing one vertex to serve as the right angled and two adjacent vertices to complete the triangle. Since there are six faces, you can form 64=24 right-angled triangles on the faces.
In addition to these, depending on the dimensions of the particular cube, you might be able to form right-angled triangles by joining the vertices which are on opposite faces. These can't be formed in every cuboid. But, these is a possibility that they can also be formed - depending on the dimensions.
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To determine how many right-angled triangles can be formed by joining the vertices of a cuboid, we first note the following:
A cuboid has 8 vertices. A right-angled triangle can be formed by selecting 3 vertices such that one angle is 90∘90∘.
Right-Angled Triangle Formation
Understanding the Vertices: The vertices of a cuboid can be denoted as:
A(0,0,0)A(0,0,0)
B(a,0,0)B(a,0,0)
C(a,b,0)C(a,b,0)
D(0,b,0)D(0,b,0)
E(0,0,c)E(0,0,c)
F(a,0,c)F(a,0,c)
G(a,b,c)G(a,b,c)
H(0,b,c)H(0,b,c)
Identifying Right Angles: A right-angled triangle can be formed if two sides of the triangle are perpendicular. In a cuboid, t
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To determine how many right-angled triangles can be formed by joining the vertices of a cuboid, we first note the following:
A cuboid has 8 vertices. A right-angled triangle can be formed by selecting 3 vertices such that one angle is 90∘90∘.
Right-Angled Triangle Formation
Understanding the Vertices: The vertices of a cuboid can be denoted as:
A(0,0,0)A(0,0,0)
B(a,0,0)B(a,0,0)
C(a,b,0)C(a,b,0)
D(0,b,0)D(0,b,0)
E(0,0,c)E(0,0,c)
F(a,0,c)F(a,0,c)
G(a,b,c)G(a,b,c)
H(0,b,c)H(0,b,c)
Identifying Right Angles: A right-angled triangle can be formed if two sides of the triangle are perpendicular. In a cuboid, this can happen in the following ways:
The right angle can be formed by selecting one vertex and two vertices that are on the edges meeting at that vertex.
Counting the Right-Angled Triangles
For each vertex of the cuboid, there are three edges meeting at that vertex, and we can select any two of these edges to form a right angle.
Number of vertices: 8
Choosing 2 edges from 3: The number of ways to choose 2 edges from 3 is given by (3 2)=3(3 2)=3.
Thus, for each of the 8 vertices, we can form 3 right-angled triangles.
Total Calculation
The total number of right-angled triangles is:
Total Right-Angled Triangles=8(vertices)×3(triangles per vertex)=24 Total Right-Angled Triangles=8(vertices)×3(triangles per vertex)=24
Conclusion
Therefore, 24 right-angled triangles can be formed by joining the vertices of a cuboid.
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L Viswanathan
Loves Math, Physics. Enjoys Sanskrit · Author has 3K answers and 1.4M answer views
·7y
Originally Answered: How many right angled triangles can be formed by joining the vertices of a unit cube? ·
From each of the four vertices at the top, one can form 3 rt angled triangles totaling 12 rt angled triangles. Again, from each of the four vertices at the bottom, 12 rt angled triangles may be formed. Thus in total 24 rt angled triangles may be formed.
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Gary Ward
MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views
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What is the dimensions of the cuboid, if diagonal long 53 dm has an angle with one edge 42° and with another edge 64°?
What is the dimensions of the cuboid, if diagonal long 53 dm has an angle with one edge 42° and with another edge 64°?
tan 64∘=2.05 1 a n d tan 42∘=0.9 1 tan64∘=2.05 1 a n d tan42∘=0.9 1
k=tan 64∘tan 42∘=2.277 1 k=tan64∘tan42∘=2.277 1 making the ratio of the sides 1:2.05:2.277 1:2.05:2.277
53 2=u 2+(2.05 u)2+(2.277 u)2=(1+4.2+5.18)u 2=10.38 u 2 53 2=u 2+(2.05 u)2+(2.277 u)2=(1+4.2+5.18)u 2=10.38 u 2
u=√53 2 10.38=16.45 u=53 2 10.38=16.45
Dimensions are u:2.05 u:2.277 u or 16.45 d m:33.72 d m:37.46 d m u:2.05 u:2.277 u or 16.45 d m:33.72 d m:37.46 d m
The square root of the sum of the squares of these values is 53.01785 dm due to rounding.
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What is the dimensions of the cuboid, if diagonal long 53 dm has an angle with one edge 42° and with another edge 64°?
tan 64∘=2.05 1 a n d tan 42∘=0.9 1 tan64∘=2.05 1 a n d tan42∘=0.9 1
k=tan 64∘tan 42∘=2.277 1 k=tan64∘tan42∘=2.277 1 making the ratio of the sides 1:2.05:2.277 1:2.05:2.277
53 2=u 2+(2.05 u)2+(2.277 u)2=(1+4.2+5.18)u 2=10.38 u 2 53 2=u 2+(2.05 u)2+(2.277 u)2=(1+4.2+5.18)u 2=10.38 u 2
u=√53 2 10.38=16.45 u=53 2 10.38=16.45
Dimensions are u:2.05 u:2.277 u or 16.45 d m:33.72 d m:37.46 d m u:2.05 u:2.277 u or 16.45 d m:33.72 d m:37.46 d m
The square root of the sum of the squares of these values is 53.01785 dm due to rounding.
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Allen Ries
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How many different triangles can be formed by joining the vertices of a Pentagon if the vertices of each triangle is a Pentagon?
35
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35
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James Buddenhagen
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How many triangles can be formed by joining the vertices of a nonagon?
How many triangles can be formed by joining the vertices of a nonagon?
Here is what it looks like:
According to this site the total number of triangles formed is 1302.
However, if you require that each of the three vertices of each triangle is one of the vertices of the given nonagon, then there are far fewer, namely “9 take 3”, that is:
(9 3)=9×8×7 3×2×1=3×4×7=84(9 3)=9×8×7 3×2×1=3×4×7=84.
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How many triangles can be formed by joining the vertices of a nonagon?
Here is what it looks like:
According to this site the total number of triangles formed is 1302.
However, if you require that each of the three vertices of each triangle is one of the vertices of the given nonagon, then there are far fewer, namely “9 take 3”, that is:
(9 3)=9×8×7 3×2×1=3×4×7=84(9 3)=9×8×7 3×2×1=3×4×7=84.
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Doug Dillon
Ph.D. Mathematics · Author has 12.4K answers and 11.4M answer views
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How many right-angled triangles can be formed by joining three vertices of a given regular 8-gon?
Assuming that the vertices of the triangle are also vertices of the octagon, each right triangle must have hypotenuse a longest diagonal of the octagon. There are 4 such diagonals. For each diagonal, there are 6 right triangle, the are 6 right triangles wit it as hypotenuse. In the diagram 3 are shown and there are their mirror images through the diagonal. 4×6=24 4×6=24
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Assuming that the vertices of the triangle are also vertices of the octagon, each right triangle must have hypotenuse a longest diagonal of the octagon. There are 4 such diagonals. For each diagonal, there are 6 right triangle, the are 6 right triangles wit it as hypotenuse. In the diagram 3 are shown and there are their mirror images through the diagonal. 4×6=24 4×6=24
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Henry Burek
M.Phil, B.Sc. in Ophthalmic Optics, University of Bradford (MDIS) (Graduated 1977) · Author has 2.2K answers and 3.3M answer views
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How many right triangles can be formed by connecting three points in a 4x4 unit grid?
How many right triangles can be formed by connecting three points in a 4x4 unit grid?
It’s easier to think of the grid comprising a 3x3 squares array and count the number of rectangles to be found:
1x1 = 9
1x2 = 12
1x3 = 6
2x2 = 4
2x3 = 4
3x3 = 1
Making a total of 36 rectangles.
Each rectangle can be split into two right triangles in two different ways so the total number of right triangles obtained in this manner is 4 times 36 = 144. Additionally each 2x2 rectangle can be divided along both its diagonals into four right triangles thereby increasing the total by a further 16, making the grand total 160
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How many right triangles can be formed by connecting three points in a 4x4 unit grid?
It’s easier to think of the grid comprising a 3x3 squares array and count the number of rectangles to be found:
1x1 = 9
1x2 = 12
1x3 = 6
2x2 = 4
2x3 = 4
3x3 = 1
Making a total of 36 rectangles.
Each rectangle can be split into two right triangles in two different ways so the total number of right triangles obtained in this manner is 4 times 36 = 144. Additionally each 2x2 rectangle can be divided along both its diagonals into four right triangles thereby increasing the total by a further 16, making the grand total 160.
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Haresh Sagar
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Given that a cuboid ABCDEFGH with side AE=20 cm,AD=15 cm,DC=15 cm,AB=15 cm ,what is the angle between the planes FAC&ABCD?
On x y z x y z plane plot points of cuboid as,
A=(0,15,0)A=(0,15,0), B=(15,15,0)B=(15,15,0), C=(15,0,0)C=(15,0,0)
D=(0,0,0)D=(0,0,0), E=(0,15,20)E=(0,15,20), F=(15,15,20)F=(15,15,20)
G=(15,0,20)G=(15,0,20), H=(0,0,20)H=(0,0,20)
Centre point of plane A B C D A B C D as M=(7.5,7.5,0)M=(7.5,7.5,0)
→M F=<7.5,7.5,20>M F→=<7.5,7.5,20>
→M B=<7.5,7.5,0>M B→=<7.5,7.5,0>
→M F∙→M B=112.5 M F→•M B→=112.5
||→M F||=√512.5||M F→||=512.5
||→M B||=√112.5||M B→||=112.5
θ=a r c c o s(3√41)=62.06°θ=a r c c o s(3 41)=62.06°
→D H=<0,0,20>D H→=<0,0,20> is normal to plane A B C D A B C D.
→A F=<15,0,20>A F→=<15,0,20>
→C F=<0,15,20>C F→=<0,15,20>
→A F×→C F=<−300,−300,225>A F→×C F→=<−300,−300,225>
Dot product of both the normals is 4500 4500
Magnitude of D H D H is 20 20
Magnitude of A F×C F A F×C F is 75√41 75 41
\Huge{\theta=arccos(\frac{4500}{2075\sqrt{41}})=(\frac{3}{\sqrt{4\Huge{\theta=arccos(\frac{4500}{2075\sqrt{41}})=(\frac{3}{\sqrt{4
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On x y z x y z plane plot points of cuboid as,
A=(0,15,0)A=(0,15,0), B=(15,15,0)B=(15,15,0), C=(15,0,0)C=(15,0,0)
D=(0,0,0)D=(0,0,0), E=(0,15,20)E=(0,15,20), F=(15,15,20)F=(15,15,20)
G=(15,0,20)G=(15,0,20), H=(0,0,20)H=(0,0,20)
Centre point of plane A B C D A B C D as M=(7.5,7.5,0)M=(7.5,7.5,0)
→M F=<7.5,7.5,20>M F→=<7.5,7.5,20>
→M B=<7.5,7.5,0>M B→=<7.5,7.5,0>
→M F∙→M B=112.5 M F→•M B→=112.5
||→M F||=√512.5||M F→||=512.5
||→M B||=√112.5||M B→||=112.5
θ=a r c c o s(3√41)=62.06°θ=a r c c o s(3 41)=62.06°
→D H=<0,0,20>D H→=<0,0,20> is normal to plane A B C D A B C D.
→A F=<15,0,20>A F→=<15,0,20>
→C F=<0,15,20>C F→=<0,15,20>
→A F×→C F=<−300,−300,225>A F→×C F→=<−300,−300,225>
Dot product of both the normals is 4500 4500
Magnitude of D H D H is 20 20
Magnitude of A F×C F A F×C F is 75√41 75 41
θ=a r c c o s(4500 20∗75√41)=(3√41)=62.06°θ=a r c c o s(4500 20∗75 41)=(3 41)=62.06°
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Bhargav Prajapati
B.E in Product Design of Physical Goods&Current Events, Lalbhai Dalpatbhai College of Engineering (Graduated 2015)
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190271 | https://www.youtube.com/watch?v=8hixh8gQ_II | Differentiate Ln functions using the product rule
Magic Monk
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Posted: 8 Sep 2017
In this tutorial we will go through how to differentiate Ln functions where it's multiplied by another functon using the product rule.
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Transcript:
hey guys it's Eddie the magic monk today I want to show you how to do some mathematical problems where you have to differentiate the natural log function R multiplied by some other functions so let's give you guys an example so we want to differentiate this function y equals 4x cubed times Ln X now before we do this let's just type it into geogebra and see what the answer is so the graph is y equals 4x cubed times x ln x okay so there is our line and what we want to do is we want to try and differentiate this line so you can see that the line has a minimum point at x equals about 0.7 so if we differentiate it the derivative should have a y-value of 0 at this point because the gradient at the minimum is 0 so let's differentiate it derivative of f and this looks right because you can see at this point here okay where the maximum point is let's see if we can get that point up so let's click on tools' you can see some of the image hasn't loaded yet because I'm using a web version but if you go to points and you select the point at the maximum you can see that point a is at about 0.7 3 and if you select the point where it touches the x-axis ok and you can see B is also at about the same point now you can probably drag it left and right to try and get to the right to get them to match up but it's pretty obvious that the x-coordinate off the minimum is the same as the x-coordinate after derivative I'm just gonna try and match it up better so yep you can see that they match okay so this is the answer for the derivative y - equals 12x squared ln x + 4 x squared so let's talk about how to do that on the exam because you're probably not going to have geogebra in the exam otherwise it'll be too easy so let's try and differentiate this so let's go y equals F times G right that's what the product rule is this formula is basically a product of two functions we got the function 4x cubed where there's a variable X and we got another function Ln X and it's the product of these two functions so let f be 4x cubed and if we want to differentiate that if - how do we differentiate this let's move the 3 to the front so 3 times 4x let's - 1 from the original power so 3 - 1 so that's 12x squared and let's do the same thing to the G function Ln X how do we differentiate that well if you guys remember that differentiates into 1 over X ok so now let's try and put that back into the product rule so we have Y - equals F times G - plus G times F - that's the product rule so let's try and put all that in F X cubed times 1 over x plus lnx times 12x squared okay so that's times as well the little dot okay Sonia how do I simplify this well 4x cubed times 1 over X - X cubed and the X at the bottom you can simplify that into 4x squared plus 12x squared Ln X now you can simplify that further because you can see there is an x squared in both of these functions so let's take out the x squared let's take out the 4 so then it becomes 4x square bracket 1 plus 3 Ln X okay and let's try and match that up with what we have for X we're in geogebra 12x squared and X plus so that's basically the same right they didn't factorize it like we have it's actually good practice to factorize things okay so that is our answer let's try into another question okay we want to differentiate y equals e to the power of minus 4 x times Ln bracket 5x okay so I want you guys to have a go and come back for the solution when you aren't done okay so as always let's try and get the solution in geogebra first so we know that we are doing things correctly let's just go to a new one and let's get rid of the calculator thing on the bottom okay so let's type the same y equals e to the power of negative 4x so in case you're also typing it in to the power of is just shift 6 on your keyboard ok and then once you typed it in you want to press the right key to bring it back down ellen:oh sorry times Ln bracket 5x ok there is our flat graph you probably have to zoom in a bit to see what's going on okay so now we want to differentiate it so just type ter I derivative press Enter and now let's differentiate the F function and you can see that it has come up with the red line and if you go to where the maximum point is on the blue line it's where the red line touches the x-axis so that proves that it's the derivative of the original function ok so this is the solution we're trying to get looks a bit complicated let's see if we can do it so again let's use the product rule y equals F times G F equals e negative 4x f - so when you to friend share exponential functions it's basically the same thing except you have to differentiate the power and multiply at the front so what's negative 4x differentiate into differentiating a negative force and multiply by that at the front ok now you have g equals ln v x g - equals so you're gonna turn the Ln function into 1 over the same thing in the bracket and you're gonna multiply by the derivative of the inside which is 5 so the whole thing becomes 5 over 5 X which is just 1 over X again because 5 divided by 5 is 1 so let's put everything inside the product rule formula F times G - plus G times F - so f is e to the power negative 4x G - is 1 over X plus G is Ln 5x + f - is negative 4 times e negative 4 X a lot of stuff there but hopefully we can simplify it so it's e to the power of negative 4 X over X plus negative 4 e to the power of negative 4 x times Ln X now in 5 X and for simplicity normally you would move actually let's just take out the common factors so let's take out eetu the power of negative 4 X bracket 1 over X plus negative 4 Ln 5 X so we take out all of common factors so you can see 1 over X and negative 4 Allen 5 X they have nothing in common so let's move then negative 4 X to the bottom of the fraction because normally we want to keep the powers positive and then 1 over X + let's just do negative - 4 Ln 5 X so that is my final answer let's see it in geogebra again see if they match I've already forgotten what it looks like ok so looks like this one now the format is a little bit different but in order to check that these two are the same thing you can simply just type what you differentiate it so let's type in y equals 1 over e to the power of 4x all right press the right key times 1 over X right key - 4 ln 5x and you can see does this line so press enter at the end do those two lines come up with the same line so that's let's get rid of the blue line so these two lines are the exact same line all right exact same line so that's why both equations are correct so again this is a correct solution okay thank you for watching guys see you next time |
190272 | https://www.chem.tamu.edu/rgroup/hughbanks/courses/102/slides/slides15.pdf | Introduction to Chemical Kinetics CHEM 102 T. Hughbanks Chemical Kinetics Reaction rates – “How fast?” Reaction mechanisms – “How?” Answers to these questions depend on the path taken from reactants to products. Reaction Rates αA + βB → γC + δD Follow progress by measuring any one concentration: Rates of change related by coefficients from balanced equation. −1 α Δ[A] Δt , −1 β Δ[B] Δt , 1 γ Δ[C] Δt , 1 δ Δ[D] Δt 2 NO2 → 2 NO + O2 concentration time NO O2 NO2 rate = −1 2 Δ[NO2] Δt = 1 2 Δ[NO] Δt = Δ[O2] Δt Factors Which Influence Rates Identity & form of reactants, products – H2 + I2 vs. H2 + Br2 – solution vs. gas phase, etc. Concentrations of various species – usually reactants – sometimes products, other species Temperature – usually, faster at higher T – strong dependence Catalysts Concentration Effects: Rate Laws αA + βB → Products Empirically, usually find that Rate = k [A]n[B]m n = “order of reaction with respect to A” m = “order of reaction with respect to B” n + m= “overall order of reaction” k = rate constant = k (T) Example: rate of a redox reaction Reaction Orders Order of a reaction can NOT be found by looking at a balanced equation! αA + βB → Products Rate = k [A]n[B]m In general: α & n, β & m are not necessarily equal — because this isn’t an elementary step Reaction order can only be found by experiments Examples 2 N2O5 → 4NO2 + O2 rate = k [N2O5] BUT 2 NO2 → 2NO + O2 rate = k [NO2]2 CAN’T predict these from equations! More Examples H2 + I2 → 2HI rate = k [H2][I2] BUT H2 + Br2 → 2HBr rate = k [H2][Br2]1/2 1 + ′ k [HBr][Br2]−1 Finding rate laws, rate constants “Method of Initial Rates” – combine known amounts of reactants – determine rate by measuring change in some concentration over a “short” time – repeat with different initial concentrations – find experimental rate law Problem A + 2B → products Expt. [A]0 [B]0 Initial Rate 1 0.10 0.10 0.0032 2 0.10 0.20 0.0032 3 0.20 0.30 0.0128 find rate law & rate constant, k (concentrations in M, rates in M/min) Rates & Mechanisms Experiments → Rate Law Rate Law → Mechanism (?) MECHANISM: “The detailed molecular processes by which a chemical reaction proceeds.” A series of “elementary steps” which combine to give an observed net reaction. Rate laws & mechanisms Start with overall reaction Guess some mechanism(s) Derive corresponding rate laws Compare with experiments Repeat as needed We need to relate rates of individual steps to the overall, observable rate laws. A reaction profile example reactants products intermediate A + B → C → D + E A + B C D + E Energy “Reaction Coordinate” 1st step is rate- determining Elementary Steps ELEMENTARY STEP: A chemical equation or reaction that describes a process as it occurs at the molecular level. A single reaction event which occurs in one simple atomic or molecular collision. Most reactions do not occur in a single elementary step. Reactions vs. Elementary Steps Normal chemical eqs. tell us the overall stoichiometry of a reaction. 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O Eq. for an elementary step looks just like a “normal” eq., but actually describes a simple molecular event. NO2 + NO2 → N2O4 Reactions vs. Elementary Steps Not always easy to tell an elementary step from a (slightly) more complicated reaction 2 NO2 → N2O4 2 NO2 → 2 NO + O2 The first one is an elementary step, the second is not. You can’t really tell this from the equations. Types of Elementary Reactions Unimolecular decomposition: one molecule falls apart: A → Product(s) Bimolecular reaction: two reactant molecules collide: A + B → Product(s) Termolecular reaction: three reactant molecules: A + B + C → Product(s) (such steps rare in gas-phase and soln. rxns.) NO examples of more complex elementary reactions are known. Rates of Elementary Steps For an elementary step, the rate law can be written from the equation. A → Product(s) rate = k [A] A + B → Product(s) rate = k [A][B] 2A → Product(s) rate = k [A]2 A + B + C → Product(s) rate = k [A][B][C] (not for gas phase reactions) Can ONLY do this for an elementary step! Rate Determining Steps If a single step in a reaction mechanism is much slower than the other steps, then the rate of the slow step is crucial in determining overall rate. The rate determining step (RDS) can be thought of as a “bottleneck” in the formation of products. Steps that follow the RDS have negligible effect on the overall rate of reaction. A reaction profile example reactants products intermediate A + B → C → D + E A + B C D + E Energy “Reaction Coordinate” 1st step is rate- determining Example: rates & mechanisms 2 NO2 → 2 NO + O2 Consider 2 mechanisms for this: NO2 → NO + O (slow) O + NO2 → O2 + NO (fast) 2 NO2 → NO3 + NO (slow) NO3 → NO + O2 (fast) Example: rates & mechanisms experimental rate law is: rate = k [NO2]2 NO2 → NO + O (slow) O + NO2 → O2 + NO (fast) rate = ? 2 NO2 → NO3 + NO (slow) NO3 → NO + O2 (fast) rate = ? Rate Laws Can Prove a Mechanism is Wrong but Can’t Prove one Right! 2 NO + O2 → 2NO2 rate = k [NO]2[O2] A single step mechanism? Rate Law is consistent. Rate Laws ... Proof? 2 NO + O2 → 2NO2 rate = k [NO2]2[O2] Two-step mechanism? (1) NO + NO ⇌ N2O2 (fast equilibrium) (2) N2O2 + O2 → 2NO2 (slow) Rate Law for this mechanism? Mechanism & Rate (1) NO + NO ⇌ N2O2 (fast equilibrium) (2) N2O2 + O2 → 2NO2 (slow) rate = rate of slow step = k 2[N2O2][O2] N2O2 is a reactive intermediate, NOT a reactant or a product. We should eliminate it from the rate law. Reversible Step: 2 NO ⇌ N2O2 Rates of forward and backward reactions will quickly become equal. Set: rate forward = rate backward k f [NO]2 = k r [N2O2] From this: [N2O2] = (k f /k r ) [NO]2 2 NO ⇌ N2O2" "(fast) " 2 NO2 + O2 → 2 NO2 , cont... rate = rate of slow step = k 2[N2O2][H2] [N2O2] = (k f /k r ) [NO]2 So: rate = k 2 [N2O2][O2] = k 2 (k f /k r )[NO]2[O2] = k observed [NO]2[O2] 2 NO ⇌ N2O2 " "(fast) N2O2 + O2 → 2NO2 "(slow) " Equilibrium: 2 NO ⇌ N2O2 The interconversion of products and reactants are an example of equilibrium Set: rate forward = rate backward k f [NO]2 = k r [N2O2] The Equilibrium Constant, Keq, is defined as: k f k r = [N2O2] [NO]2 = constant = Keq Reversible Step: 2 NO ⇌ N2O2 Rates of forward and backward reactions will quickly become equal. Set: rate forward = rate backward k f [NO]2 = k r [N2O2] From this: [N2O2] = (k f /k r ) [NO]2 2 NO ⇌ N2O2"(fast) " 2 NO2 + O2 ⇌ 2 NO2 , cont... rate = rate of slow step = k 2[N2O2][O2] [N2O2] = (k f /k r )[NO]2 So: rate = k 2 [N2O2][O2] = k 2 (k f /k r)[NO]2[O2] = k observed [NO]2[O2] = k 2 Keq[NO]2[O2] 2 NO ⇌ N2O2 " "(fast) N2O2 + O2 → 2NO2 "(slow) " |
190273 | https://byjus.com/maths/nth-term-of-an-ap/ | In Mathematics, an arithmetic progression (AP) is defined as the list or sequence of numbers, in which each term in the sequence is obtained by adding a fixed number to the preceding term. The fixed number is called the common difference of the arithmetic progression. The fixed number can be positive or negative or zero. Generally, the first term of an AP is represented by a1, the second term by a2, … and the nth term by an. The common difference of an AP is denoted by “d”. In this article, let us discuss how to find the nth term of an AP with many solved examples.
nth Term of an AP Formula
Assume that a1, a2, a3,… be an arithmetic progression (AP), in which first term a1 is equal to “a” and the common difference is taken as “d”, then the second term, third term, etc can be calculated as follows:
Second term, a2 = a+d
Third term, a3 = (a+d)+d = a+2d,
Fourth term, a4 = (a+2d)+d = a+3d, and so on.
Therefore, the nth term of an AP (an) with the first term “a” and common difference “d” is given by the formula:
nth term of an AP, an = a+(n-1)d.
(Note: The nth term of an AP (an) is sometimes called the general term of an AP, and also the last term in a sequence is sometimes denoted by “l”.)
Also, read: Sum of N terms of AP
nth Term of an AP Examples
Go through the following examples to understand the procedure for finding the nth term of an AP.
Example 1:
Determine the 10th term of an AP 2, 7, 12, ….
Solution:
Given arithmetic progression (AP) is 2, 7, 12, …
Here, the first term, a = 2.
Common difference, d = 7-2 = 5
n=10.
The formula to find the nth term of an AP, an = a+(n-1)d
Now, substitute the values in the formula, we get
a10 = 2 + (10-1)5
a10 = 2 + (9)5
a10 = 2+45
a10 = 47.
Hence, the 10th term of an AP 2, 7, 12, … is 47.
Example 2:
The third term of an AP is 5 and the 7th term of an AP is 9. Find the arithmetic progression (AP).
Solution:
Given that, Third term of AP = 5
Seventh term of AP = 9
(i.e) a3 = a+(3-1)d = a+2d = 5 …(1)
a7 = a+(7-1)d = a+6d = 9 …(2)
Now, solve the equations (1) and (2), we get
a=3 and d = 1.
Therefore, the first term is 3 and the common difference is 1.
Therefore, the arithmetic progression (AP) is 3, 4, 5, 6, 7, 8, 9, ….
Example 3:
How many two-digit numbers are divisible by 3?
Solution:
The sequence of two-digit numbers which are divisible by 3 are:
12, 15, 18, 21, …, 99.
To find whether the given sequence is an AP, find the common difference.
The common difference (d) of the above-given sequence is 3, and hence, the given sequence is an Arithmetic progression (AP).
Hence, a = 12, d = 3, an = 99.
Now, we have to find the value of “n”.
Now, substitute the values in the formula, an = a+(n-1)d, we get
99 = 12+(n-1)3
99 = 12+3n-3
99-12+3 = 3n
3n = 90
n= 90/3
n=30.
Hence, there are 30 two-digit numbers that are divisible by 3.
| |
| Also, read: Geometric Progression Harmonic Progression |
Practice Problems
Solve the following problems:
Find the 30th term of an AP: 10, 7, 4, …..
Determine the 31st term of an AP, if its 11th term is 38 and its 16th term is 73.
Find the missing terms for the following AP: -4, __, __, ___, ___, 6.
Stay tuned with BYJU’S – The Learning App and download the app to learn all Maths-related concepts by exploring more exciting videos.
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190274 | https://www.imaios.com/en/e-anatomy/anatomical-structures/genitofemoral-nerve-1557860936 | View the module
Genitofemoral nerve
Nervus genitofemoralis
Definition
Muhammad A. Javaid
Origin
The genitofemoral nerve arises from the anterior rami of the first and second lumbar spinal nerves (L1 and L2). It is a mixed nerve, comprising both sensory and motor fibers.
Course
The genitofemoral nerve forms within the substance of the psoas major muscle. It emerges on the anterior surface of the muscle while remaining retroperitoneal. Along its course, it descends deep to the psoas fascia and crosses posterior to the parietal peritoneum, as well as structures like the ureter and gonadal vessels. Near the inguinal region, it pierces the psoas fascia and divides into two terminal branches—the genital branch and the femoral branch—just above the inguinal ligament.
Branches and Innervation
1. Genital Branch
Course:
This branch enters the deep inguinal ring and traverses the inguinal canal. In males, it accompanies the spermatic cord. In females, it follows the round ligament of the uterus.
Innervations:
Males: Provides motor innervation to the cremaster muscle and sensory innervation to the skin of the upper anterior scrotum. spermatic fasica and tunica vaginalis of testis.
Females: Supplies sensory innervation to the skin of the mons pubis and labium majus.
2. Femoral branch
Course:
The femoral branch runs along the lateral side of the external iliac artery, passing beneath the inguinal ligament. It enters the femoral sheath lateral to the femoral artery, then pierces the sheath and the fascia lata.
Innervation:
Supplies sensory fibers to the skin covering the upper anterior thigh, specifically in the femoral triangle region.
Fiber types and neural pathways
The genitofemoral nerve comprises both motor and sensory fiber types:
1. General Somatic Afferent (GSA)
Sensory fibers transmit information from the skin of the upper anterior thigh and external genitalia to the spinal cord.
Pathway: Sensory impulses travel through the genitofemoral nerve → anterior rami of L1 & L2 → spinal nerves → posterior roots → dorsal gray horns of spinal segments L1 and L2.
2. General Somatic Efferent (GSE)
Motor fibers innervate the cremaster muscle.
Pathway: Signals originate in the anterior gray horns of spinal segments L1 and L2 → anterior rami of L1 & L2 → genital branch of the genitofemoral nerve → cremaster muscle.
References
Drake, R.L., Vogl, A.W., and Mitchell, A.W.M. (2010). ‘Chapter 4: Abdomen’, in Gray’s anatomy for students. (2nd ed.) Churchill Livingstone Elsevier, Philadelphia PA 19103, pp.379-381.
Singh O, Al Khalili Y. Anatomy, Back, Lumbar Plexus. [Updated 2023 Aug 7]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Available from:
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190275 | https://www.quora.com/What-is-the-difference-between-the-addition-of-halogens-to-alkenes-and-alkanes-in-terms-of-Markovnikovs-rule | What is the difference between the addition of halogens to alkenes and alkanes in terms of Markovnikov's rule? - Quora
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Chemical Reactions
Anti Markonikov Rule
Organic Chemistry
Alkanes
Nucleophilic Addition
Halogenation
Organic Chemistry Basics
Reaction of Alkenes
Organic Reactions
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What is the difference between the addition of halogens to alkenes and alkanes in terms of Markovnikov's rule?
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John Britto
BS/MS Organic Chemistry in Organic Chemistry&Biochemistry, Lehigh University · Author has 125 answers and 17.5K answer views
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First alkanes - the only addition of a halogen to alkanes is via radical propagation and it only works with bromine or chlorine. Alkanes are typically considered to be unreactive.
For alkenes, Markovnikov’s Rule doesn’t necessarily apply for the addition of halogens to alkenes. Halogens (bromine and chlorine) react via a halonium ion intermediate which forms a bridge-type structure at the site of the double bond with the halogen acquiring a + charge. The other portion of the diatomic halogen ( X-) which was released when the halonium ion formed then opens the halonium ion from the opposite side
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First alkanes - the only addition of a halogen to alkanes is via radical propagation and it only works with bromine or chlorine. Alkanes are typically considered to be unreactive.
For alkenes, Markovnikov’s Rule doesn’t necessarily apply for the addition of halogens to alkenes. Halogens (bromine and chlorine) react via a halonium ion intermediate which forms a bridge-type structure at the site of the double bond with the halogen acquiring a + charge. The other portion of the diatomic halogen ( X-) which was released when the halonium ion formed then opens the halonium ion from the opposite side of where it formed. This is designated anti addition and results in a trans 1,2 dihalide.
Markovnikov addition would apply in the addition of a hydrogen halide (HX) or halohydrin (HOX) in which the nucleophilic portion of the halide or halohydrin would add at the more substituted carbon of the double bond and the electrophilic portion would add to the least substituted carbon. Be alert to potential carbocation rearrangement to give a more stable carbocation.
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What is the difference between the halogenation of alkanes and those of alkenes?
What is the addition of halogens to alkenes?
What is the comparison between the preparation of alkanes and alkenes?
How do you halogenate an alkene? What is Markovnikov's rule?
How do you halogenate an alkene using an anti-Markovnikov approach?
Kumaraswamy Sathiavasan
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What is the difference between the halogenation of alkanes and those of alkenes?
Alkanes are saturated hydrocarbons and so their hologenation involves substitution of one or more hydrogen atoms by halogens.
Example: Reaction of ethane with chlorine to form chloroethane
C2H6 + Cl2 → C2H5Cl + HCl
Thus, halogenation of an alkane is a substitution reaction as one hydrogen in the ethane molecule gets replaced by a chlorine atom. Hydrochloride is the other product of this reaction.
But alkenes are unsaturated hydrocarbons and therefore their halogenation results in addition of halogens to the alkene molecules.
Example: Reaction of ethylene with bromine to form dibromoethane
C2H4 + Br2
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Alkanes are saturated hydrocarbons and so their hologenation involves substitution of one or more hydrogen atoms by halogens.
Example: Reaction of ethane with chlorine to form chloroethane
C2H6 + Cl2 → C2H5Cl + HCl
Thus, halogenation of an alkane is a substitution reaction as one hydrogen in the ethane molecule gets replaced by a chlorine atom. Hydrochloride is the other product of this reaction.
But alkenes are unsaturated hydrocarbons and therefore their halogenation results in addition of halogens to the alkene molecules.
Example: Reaction of ethylene with bromine to form dibromoethane
C2H4 + Br2 → C2H4Br2
Thus, halogenation of an alkene is an addition reaction as one molecule of bromine gets entirely consumed by the ethylene molecule to form ethylene dibromide as the single product.
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Donald Montecalvo
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Bob Mouk
, PhD Organic Chemistry (1970) · Author has 1.5K answers and 1.4M answer views
·4y
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What happens when an Alkenes gets additioned by Halogen Acid if both carbon atoms have equal number of hydrogens using Markovnikov's Rule? Should we consider the neighboring carbon atoms? Or the hydrogen atom can bond with any carbon atoms?
I think that we can consider a molecule like cis-2-pentene to illustrate my answer.
Each carbon of the double bond has one hydrogen and one alkyl group. Adding the initial hydrogen atom from, for example, HCl to either carbon would create a carbocation of approximately equal stability (both being secondary), and both cations would be created at, roughly, the same rate. Therefore, there will be approximately equal amounts of 2-chloropentane and 3-chloropentane.
Note that both cations are not exactly the same, despite both being labeled as secondary. Here the neighboring carbon atoms may come into
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I think that we can consider a molecule like cis-2-pentene to illustrate my answer.
Each carbon of the double bond has one hydrogen and one alkyl group. Adding the initial hydrogen atom from, for example, HCl to either carbon would create a carbocation of approximately equal stability (both being secondary), and both cations would be created at, roughly, the same rate. Therefore, there will be approximately equal amounts of 2-chloropentane and 3-chloropentane.
Note that both cations are not exactly the same, despite both being labeled as secondary. Here the neighboring carbon atoms may come into play, but only in a minor way. So we may expect a small deviation from 50/50 products.
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Ayush Sharma
Class 12th Science stream student. · Author has 77 answers and 510K answer views
·9y
Related
What's the difference between Markovnikov and Anti-Markovnikov addition?
Markownikoff's and Anti-Markownikoff's rule help us to predict the product formed on addition of H-X (where x is usually Cl or Br) to alkene.
Markownikoff's rule follows "rich become rich, poor becomes poorer" analogy. It states that "on addition of H-X to alkene, the negative part of addendum (that is X¯ or Cl¯/Br¯) goes to the carbon which has less number of hydrogen attached to it.
Example :
But hey!Don't be too much dependent on this rule, be careful, it's not applicable to all alkenes. Since it involves formation of carbocation, rearrangement may take place.
Example :
So, before u
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Markownikoff's and Anti-Markownikoff's rule help us to predict the product formed on addition of H-X (where x is usually Cl or Br) to alkene.
Markownikoff's rule follows "rich become rich, poor becomes poorer" analogy. It states that "on addition of H-X to alkene, the negative part of addendum (that is X¯ or Cl¯/Br¯) goes to the carbon which has less number of hydrogen attached to it.
Example :
But hey!Don't be too much dependent on this rule, be careful, it's not applicable to all alkenes. Since it involves formation of carbocation, rearrangement may take place.
Example :
So, before using this rule justcheck if 2° carbocation is rearrange-able (look if there is any adjacent tertiary carbon).
Anti-Markownikoff's rule is just the opposite, it follows " rich becomes poor, and poor becomes rich" analogy. It is applicable only when you perform the reaction in PRESENCE OF PEROXIDE and only when you add H-Br to alkene. In this case, the negative part of addendum goes to the carbon which has more number of carbon attached to it.
Example :
Feel free to use this rule anywhere. Since it involves free radicals, rearrangement does not take place.
But, again be careful, it is only applicable to H-Br. Suppose if I were to take H-Cl in presence of peroxide, product will be formed according to Markownikoff's rule.
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More answers below
When halogens are added to symmetrical alkenes at low temperature or room temperature, do they show as anti-Markovnikov products?
Why are hydrogen halides and not halogen acid preferred for preparing halogen alkanes from alkenes?
What is the difference between the reaction of an alkene with a halogen acid and the reaction of an alkane with a halogen acid?
What is the condition necessary for the reaction between alkanes and halogens?
Can light be used instead of peroxides in the anti-Markovnikov's rule for the addition of HBr to alkenes?
Vishnuthirtha Madaksira
Professor in Chemistry at Sri Guru Krupa Coaching Centre (2009–present) · Author has 10.6K answers and 2.7M answer views
·1y
Related
What is the difference between the halogenation of alkanes and those of alkenes?
The halogenation of alkanes is an example for free radical substitution reaction. ( Free radical substitution reaction involves : Chain Initiation, Chain propagation and Chain termination steps ).
Example :
CH4 + Cl2 /sunlight → CH3Cl + HCl ( Free radical is Cl• which gets substituted ).
The halogenation of Alkenes is an example for electrophilic addition reaction. ( Electrophilic addition reaction involves addition of an electrophile as the rate determining step ).
Example :
H2C=CH2 + Cl2 /CCl4 → ClCH2-CH2Cl ( Electrophile is Cl+ which gets added up ) or
H2C=CH2 + HCl → CH3-CH2Cl ( Electrophile is
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The halogenation of alkanes is an example for free radical substitution reaction. ( Free radical substitution reaction involves : Chain Initiation, Chain propagation and Chain termination steps ).
Example :
CH4 + Cl2 /sunlight → CH3Cl + HCl ( Free radical is Cl• which gets substituted ).
The halogenation of Alkenes is an example for electrophilic addition reaction. ( Electrophilic addition reaction involves addition of an electrophile as the rate determining step ).
Example :
H2C=CH2 + Cl2 /CCl4 → ClCH2-CH2Cl ( Electrophile is Cl+ which gets added up ) or
H2C=CH2 + HCl → CH3-CH2Cl ( Electrophile is H+ which gets added up ).
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Ray Menon
Studied at Rutgers University · Upvoted by
Daniel James Berger
, Ph.D. in organic chemistry · Author has 3.4K answers and 10.9M answer views
·6y
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Does the halogenation of alkene with anti-Markovnikov by ROOR need HV?
hv (electromagnetic energy; UV light) is not essential, but it is preferred.
UV Light energy is used to excite and ultimately break the O-O bond in the peroxide initiator, in order to produce the alkyloxy free radical(s).
R’-O-O-R’ —-> 2R’O
This is the step that sets up the entire chain of reaction.
RO is not a very stable radical, and tries to steal an electron from whatever is around it that can provide it one.
R’O + H-Br → R’-OH + Br
Now, Br (bromine atom a.k.a bromine radical) will look for any source of electron density it can find, in order to stabilize its extra electron. Clearly the alk
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hv (electromagnetic energy; UV light) is not essential, but it is preferred.
UV Light energy is used to excite and ultimately break the O-O bond in the peroxide initiator, in order to produce the alkyloxy free radical(s).
R’-O-O-R’ —-> 2R’O
This is the step that sets up the entire chain of reaction.
RO is not a very stable radical, and tries to steal an electron from whatever is around it that can provide it one.
R’O + H-Br → R’-OH + Br
Now, Br (bromine atom a.k.a bromine radical) will look for any source of electron density it can find, in order to stabilize its extra electron. Clearly the alkene is around it in abundance, so it attacks the alkane.
RCH= CH2 + Br → RCH-CH2Br.
The reason bromine attacks the primary carbon (=CH2) rather than the secondary carbon (RCH=) of the alkene is because the resulting secondary alkyl radical is more stable than the primary radical.
Once this step is complete, the mechanism involves a hydrogen abstraction from H-Br, thereby generating a Br atom and setting up a continuation of the chain reaction.
RCH-CH2Br + H-Br → RCH2-CH2-Br + Br
As can be seen, the bromine substituent ends up on the LESS substituted carbon due to the initiation step of the peroxide radical, and thereby effecting Anti-Markownickoff addition results.
If you didn’t use UV light in this reaction, the R’-O-O-R’ can break down THERMALLY to produce R’O radicals all by itself. However, depending on the peroxide reagent (dibenzoyl peroxide is frequently used C6H5-COO-OOC-C6H5), the temperature of decomposition is quite high - at 92 deg. C, the half-life of Benzoyl Peroxide decomposition is about 1 hour. When you run any reaction at a higher temperature, you invariably sacrifice selectivity, and you will end up with a mixture of Markownickof and Anti-Markownickof addition products. This is why using UV light, at room temperature (25 deg. C), is preferred in this reaction.
reference: hydrogen bromide and alkenes
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Rohan Naik
Data Scientist at Wipro (Indian company) (2016–present) · Author has 1.4K answers and 4.3M answer views
·6y
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What is the halogenation of alkanes?
Alkanes: Halogenation
The reaction of a halogen with an alkane in the presence of ultraviolet (UV) light or heat leads to the formation of a haloalkane (alkyl halide). An example is the chlorination of methane.
Experiments have shown that when the alkane and halogen reactants are not exposed to UV light or heat, the reaction does not occur. However, once a reaction is started, the light or heat source can be removed and the reaction will continue. The mechanism of the reaction explains this phenomenon.
Halogenation mechanism. In the methane molecule, the carbon‐hydrogen bonds are low‐polarity cov
Continue Reading
Alkanes: Halogenation
The reaction of a halogen with an alkane in the presence of ultraviolet (UV) light or heat leads to the formation of a haloalkane (alkyl halide). An example is the chlorination of methane.
Experiments have shown that when the alkane and halogen reactants are not exposed to UV light or heat, the reaction does not occur. However, once a reaction is started, the light or heat source can be removed and the reaction will continue. The mechanism of the reaction explains this phenomenon.
Halogenation mechanism. In the methane molecule, the carbon‐hydrogen bonds are low‐polarity covalent bonds. The halogen molecule has a nonpolar covalent bond. UV light contains sufficient energy to break the weaker nonpolar chlorine‐chlorine bond (∼58 kcal/mole), but it has insufficient energy to break the stronger carbon‐hydrogen bond (104 kcal/mole). The fracture of the chlorine molecule leads to the formation of two highly reactive chlorine free radicals (chlorine atoms). A free radical is an atom or group that has a single unshared electron.
The bond that is ruptured is said to have broken in a homolytic fashion; that is, each of the originally bonded atoms receives one electron. This initial reaction is called the initiation step of the mechanism. The chlorine free radicals that form are in a high‐energy state and react quickly to complete their octets and liberate energy. Once the high‐energy chlorine free radicals are formed, the energy source (UV light or heat) can be removed. The energy liberated in the reaction of the free radicals with other atoms is sufficient to keep the reaction running.
When a chlorine free radical approaches a methane molecule, homolytic fission of a carbon‐hydrogen bond occurs. The chlorine free radical combines with the liberated hydrogen free radical to form hydrogen chloride and a methyl free radical. This is called a propagation step, a step in which both a product and a reactive species, which keeps the reaction going, are formed.
A second propagation step is possible. In this step, a methyl free radical reacts with a chlorine molecule to form chloromethane and a chlorine free radical.
When a reaction occurs between free radicals, product forms, but no new free radicals are formed. This type of reaction is called a termination step because it tends to end the reaction. There are several termination steps in the chlorination of methane.
A methyl free radical reacts with a chlorine free radical to form chloromethane.
Two methyl free radicals react to form ethane.
Two chlorine free radicals react to form a chlorine molecule.
To summarize, this free‐radical chain reaction initially contains few free radicals and many molecules of reactants. As the reaction proceeds, the number of free radicals increases, while the number of reactant molecules decreases. Near the end of the reaction, many more free radicals exist than reactant molecules. At this stage of the overall reaction, termination steps become the predominant reactions. All of the halogenation mechanism reactions occur very rapidly, and the formation of the products takes only microseconds.
For more information, you can also watch the below video.
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Daniel Iyamuremye
Former Senior Lecturer (Retired) (2000–2018) · Author has 12.1K answers and 2M answer views
·1y
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What is the difference between the halogenation of alkanes and those of alkenes?
Halogenation of alkanes consists into a substitution of hydrogen by halogen.
Halogenation of alkenes consists in an addition reaction on the double bond.
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Rahul Dey
Studied at Medical College and Hospital, Kolkata
·9y
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What's the difference between halogenation of benzene and addition of chlorine?
Benzene can undergo both halogenation and addition of chlorine depending on the conditions of reactions and other factors like catalyst and all.
Benzene undergo Friedel Crafts chlorination in presence of Lewis Acids(like AlCl3 and halogen abstractor/electron carrier) to give substituted products (via Electrophilic substitution mechanism).
C6H6 —-(Cl2/AlCl2)——→ C6H5Cl (a substituted product)
However if you add excess of Cl2 you will get highly substituted Hexachloro Benzene
C6H6 ——(Cl2 excess/AlCl3)—→ C6Cl6
Now the same reagents(without the catalyst) may undergo Halogenation to give Benzene Hexa
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Benzene can undergo both halogenation and addition of chlorine depending on the conditions of reactions and other factors like catalyst and all.
Benzene undergo Friedel Crafts chlorination in presence of Lewis Acids(like AlCl3 and halogen abstractor/electron carrier) to give substituted products (via Electrophilic substitution mechanism).
C6H6 —-(Cl2/AlCl2)——→ C6H5Cl (a substituted product)
However if you add excess of Cl2 you will get highly substituted Hexachloro Benzene
C6H6 ——(Cl2 excess/AlCl3)—→ C6Cl6
Now the same reagents(without the catalyst) may undergo Halogenation to give Benzene Hexachloride (which is an addition product). This takes place in diffused sunlight/UV/heat (being a Free Radical reaction) with the help of free radical generation.
C6H6 ——(Cl2/UV)—→ C6H6Cl6 (addition product)
[note in substitution product aromatic character remains while in addition one, it changes to non aromatic]
So the major reason is the conditions, as long as no condition is stated either of them is right. However if any is specific you know now what to write and balance.
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Vishnuthirtha Madaksira
Professor in Chemistry at Sri Guru Krupa Coaching Centre (2009–present) · Author has 1.7K answers and 1.7M answer views
·6y
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How do you do anti-Markovnikov's rule to alkynes?
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Ravi Teja Gannavarapu
Studied at International Institute of Information Technology, Bhubaneswar (IIIT-BH) · Author has 81 answers and 347.8K answer views
·9y
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What's the difference between Markovnikov and Anti-Markovnikov addition?
When an alkyl halide is added to an unsymmetrical alkene, an addition reaction takes place. To predict the product of the reaction, we use the markovnikovs rule. Markovnikovs rule says that "The electronegative part (halogen) of the alkyl halide gets attached to the least substituted carbon (carbon having the lowest number of hydrogens attached to it) out of the double bonded carbons." Anti-Markovnikov addition (otherwise known as Peroxide effect or Kharasch effect) just says the opposite, "The halogen of the alkyl halide gets attached to the more substituted carbon out of the double bonded ca
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When an alkyl halide is added to an unsymmetrical alkene, an addition reaction takes place. To predict the product of the reaction, we use the markovnikovs rule. Markovnikovs rule says that "The electronegative part (halogen) of the alkyl halide gets attached to the least substituted carbon (carbon having the lowest number of hydrogens attached to it) out of the double bonded carbons." Anti-Markovnikov addition (otherwise known as Peroxide effect or Kharasch effect) just says the opposite, "The halogen of the alkyl halide gets attached to the more substituted carbon out of the double bonded carbons.
But remember, the reagent plays a important role here and the mechanism is different in both the cases. The intermediate formed in Markovnikovs addition is a carbocation which can undergo rearrangement which can also be expained via 1,2-hydride shift. The stability of carbocations increases as 3°>2°>1°, so rearrangement occurs whenever possible. In case of Anti Markovnikovs addition, the intermediate formed is a free radical which doesn't undergo rearrangement.
Also note that the Anti Markovnikov addition only takes place in case of highly selective halide like HBr. If you add HCl or HI in presence of Peroxide, then the reaction will still proceed via the Markovnikov addition.
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Allen Erickson
PhD. in Organic Chemistry, Purdue University
·3y
Related
What is the difference between a haloalkane and an alkene or an alkyne?
A halocompound contains a halogen i.e. Flourine, chlorine bromine or iodine (or astatine if you like radioactive halogens). An alkene contains a double bond (olefin) while an alkyne contains a triple carbon carbon bond. The term alkene or alkyne refers to the bond between two carbon atoms. There are alkenes and alkynes that contain halogens (vinyl chloride for example the monomer for PVC plastic).
The difference is that the term alkane indicate there is no unsaturation (double or triple bonds i.e. no alkene or alkynes) while the term halo indicates there is a halogen in the molecule. Therefore
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A halocompound contains a halogen i.e. Flourine, chlorine bromine or iodine (or astatine if you like radioactive halogens). An alkene contains a double bond (olefin) while an alkyne contains a triple carbon carbon bond. The term alkene or alkyne refers to the bond between two carbon atoms. There are alkenes and alkynes that contain halogens (vinyl chloride for example the monomer for PVC plastic).
The difference is that the term alkane indicate there is no unsaturation (double or triple bonds i.e. no alkene or alkynes) while the term halo indicates there is a halogen in the molecule. Therefore a haloalkane is an organic compound with no double or triple bonds but with a halogen atom someplace in the molecule.
Hope this helps.
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Sirat Yadav
Studied at DAV Public School, Cheeka
·6y
Related
How do you do anti-Markovnikov's rule to alkynes?
We can do anti markonikov ' s rule with alkynes by the help of addition of peroxide . In this redical mechanism takes place .
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We can do anti markonikov ' s rule with alkynes by the help of addition of peroxide . In this redical mechanism takes place .
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Related questions
What is the difference between the halogenation of alkanes and those of alkenes?
What is the addition of halogens to alkenes?
What is the comparison between the preparation of alkanes and alkenes?
How do you halogenate an alkene? What is Markovnikov's rule?
How do you halogenate an alkene using an anti-Markovnikov approach?
When halogens are added to symmetrical alkenes at low temperature or room temperature, do they show as anti-Markovnikov products?
Why are hydrogen halides and not halogen acid preferred for preparing halogen alkanes from alkenes?
What is the difference between the reaction of an alkene with a halogen acid and the reaction of an alkane with a halogen acid?
What is the condition necessary for the reaction between alkanes and halogens?
Can light be used instead of peroxides in the anti-Markovnikov's rule for the addition of HBr to alkenes?
Can LAH and SBH reduce alkenes to alkane?
Does the addition of hydrogen halides to alkenes rule by the Markovnikov rule?
Does the addition of hydrogen halides to alkenes lead to the formation of halogen derivatives of alkanes?
How do you name the alkane formed when it is reacted with a halogen?
What happens when an alkene reacts with a halogenation reagent that has bromine and chlorine in it?
Related questions
What is the difference between the halogenation of alkanes and those of alkenes?
What is the addition of halogens to alkenes?
What is the comparison between the preparation of alkanes and alkenes?
How do you halogenate an alkene? What is Markovnikov's rule?
How do you halogenate an alkene using an anti-Markovnikov approach?
When halogens are added to symmetrical alkenes at low temperature or room temperature, do they show as anti-Markovnikov products?
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June 2013 - Volume 40 - Issue 6
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Case Report
Secondary Syphilis Presenting as a Generalized Lymphadenopathy
Clinical Mimicry of Malignant Lymphoma
Park, Seong Yeon MD; Kang, Jung Hyun MD; Roh, Ji Hyeon MD†; Huh, Hee Jin MD‡; Yeo, Jeong Seok MD§; Kim, Do Yeun PhD, MD
Author Information
From the Departments of Internal Medicine; †Pathology; ‡Laboratory Medicine; and §Nuclear Medicine, Dongguk University Ilsan Hospital, University of Dongguk College of Medicine, Goyang-si, Republic of Korea.
The polymerase chain reaction assay for Treponema pallidum was performed in Samkwang Medical Laboratories.
Conflict of interest: There are no potential conflicts of interest for any authors. No author received any financial support.
Correspondence: Do Yeun Kim, PhD, MD, Department of Internal Medicine, Divisions of Hematology and Oncology, Dongguk University Ilsan Hospital, University of Dongguk College of Medicine, Siksa-dong, Ilsandong-gu, Goyang-si, Republic of Korea. E-mail: smdkdy@hanmail.net.
Received for publication July 29, 2012, and accepted January 23, 2013.
Sexually Transmitted Diseases 40(6):p 490-492, June 2013. | DOI: 10.1097/OLQ.0b013e3182897eb0
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Abstract In Brief
The diagnosis of syphilis remains challenging. The absence of classical features of the disease, such as the rash of secondary syphilis or genital lesion, may pose diagnostic difficulties. In this article, we report a case of secondary syphilis in which the clinical syndrome and pattern of fluorodeoxyglucose uptake mimicked malignant lymphoma. This case highlights the importance of thorough history taking including sexual contact. Clinicians should be alert for syphilis-underlying unexplained lymphadenopathy, even in the absence of typical rash or genital lesion.
The authors report a case of secondary syphilis with generalized lymphadenopathy that mimicked lymphoma on clinical syndrome and fluorodeoxyglucose–positron emission tomography imaging. This is a first case report of secondary syphilis with fluorodeoxyglucose–positron emission tomography findings.
© Copyright 2013 American Sexually Transmitted Diseases Association
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Secondary Syphilis Presenting as a Generalized Lymphadenopathy: Clinical Mimicry of Malignant Lymphoma
Sexually Transmitted Diseases40(6):490-492, June 2013.
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190277 | https://simpop.org/ohms-law/ohms-law.htm | Ohm's Law Simulation | SimPop
☰Categories
PhysicsChemistryBiology
Ohm's Law Simulation
Grades 10th - 12th by Ankush Naskar
1 Ω
2 Ω
3 Ω
∼ 0 Ω
12 V
0 A
0 A
0 A
0 A
Don't forget to turn on the switch! Current flows at speed of light. It's slowed down here.
Instructions : Drag and drop the resistors into the drop areas (dashed grey boxes) and find out the relation between voltage (V), current (I) and resistance (R) of circuit.
❮ ❯ Embed Code
What is Resistance?
Resistance is the measure of obstruction offered to the flow of electric current through a material. The SI unit of electrical resistance is ohm (Ω). It depends upon the nature of material, length of the conductor and cross-section area of the wire.
What is Ohms Law?
Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
So, I = V/R where V is voltage, I is current and R is resistance of circuit. Or, rearranging, V = I × R.
What is a Series Circuit?
When the components are connected end to end in a sequence, we call it a series circuit. The current flows along a single path so the current through each component remains the same. There is a different voltage drop across each component depending upon its resistance (as per Ohm's law, Vn = Icircuit × Rn). The sum of voltages across each component is equal to the total voltage provided by the power source (Vcircuit = V1+V2+...+Vn).
What is a Parallel Circuit?
When each component is individually connected to the power source such that the current can take multiple paths, we call it a parallel circuit. Each component gets the same voltage and draws a different current depending upon its resistance (as per Ohm's law, In = Vcircuit/Rn). The sum of currents through each component is equal to the total current flowing through the circuit (Icircuit = I1+I2+...+In).
To learn more, watch this video: Ohm's Law
Related SimPop Simulations:
Conductivity ------------
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190278 | https://www.youtube.com/watch?v=dtXLJ7v1LII | Average and Marginal Costs | Production, Cost and Growth | A Level Economics 9708
A Level Economics | Alt Academy
1170 subscribers
Description
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Posted: 15 Mar 2024
Welcome to our comprehensive tutorial on production, growth, and costs, designed specifically for Cambridge A2 Level economics students! In this chapter, we delve deep into the intricacies of average and marginal cost, essential concepts for understanding production processes and economic decision-making.
Throughout this video, learners will embark on a journey to unravel the relationship between marginal cost and average cost, gaining insights into their significance in cost analysis. By examining the U-shaped curve of the marginal cost, students will explore its implications on production efficiency and resource allocation.
Furthermore, we delve into the impact of diminishing marginal returns on marginal cost, highlighting its implications for businesses and producers. Through graphical analysis, learners will also discover how to derive average variable cost and average total cost, enhancing their understanding of cost structures in different production scenarios.
Whether you're aiming to excel in your economics exams or seeking a deeper understanding of production and cost concepts, this video provides clear explanations and practical examples to support your learning journey. Join us as we demystify average and marginal cost and empower you to analyze economic decisions with confidence.
Don't miss out on this essential chapter in your economics education – press play and unlock the secrets of production costs today!
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in the last video we looked at both the average cost and the marginal cost now is the time to bring them all together and see the relationship between them if we draw a marginal cost curve our marginal cost curve is u-shaped and let's draw that to see how the relationship with average cost look like we saw for the first two workers or quantity of 90 in our spotless Car Wash uh our marginal cost was going down and then our marginal cost starts to increase as we saw more and more workers were hired this happens primarily because of again the law of diminishing return if you remember for the first two workers our marginal product was going up so for the first two workers on 90 units of output when the marginal product was going up our marginal cost is going down while marginal cost starts to rise Beyond 190 units of output and we can attribute this to our increasing MP for the first 90 units and then later on we will say there is diminishing MP we can also look at it from the perspective that as your MP is rising and I would like to draw here simultaneously my mp for the first two workers when MP was Rising you can see this modal cost is going down and then for after the two workers MP starts to fall or modal cost starts to rise now we can say this that each worker hired at more at is adding more to production than the worker before when we are looking at marginal cost going down or MP going up but that also means the that fewer additional workers are needed to produce an additional unit of output because workers are very productive beyond the two workers however Things become different because then after the first two workers MP starts to fall or diminishing Mar return sets in and MP Falls which also means MC starts to rise Mar cost starts to rise which means each additional unit of output cost more and more to produce because the worker productivity is going down so as long as mp uh is falling marginal cost must be rising and vice versa as long as MP is rising MC is falling so they have an inverse relationship between MP and MC now on this marginal cost if I now plot my average cost so my average fixed cost will of course declines that's uh something which we may not need to plot here but let's look at average variable cost our average variable cost looks something like this it falls and but as soon as it meets MC it starts to rise on the other hand ATC also falls but it will be above ABC because ATC includes AFC so it will be also a u-shape curve starts to fall but as soon as it meets MC it starts to rise again but the gap between ATC and uh ABC will be shrinking we did talk about why the gap between ATC and ABC is shrinking and the reason primarily for that is that that this Gap here is basically my average fixed cost and this average fixed cost is declining as uh quantity goes up and that's why this Gap is shrinking as we produce more and more now when we look at margin cost the marginal cost relationship seems to be such that when MC is below uh our ATC and ABC both ATC and ABC are falling look at this when the yellow line is below ABC the Blue Line the blue line is going down as soon as MC is above ABC ABC starts to rise and similarly when MC is below ATC ATC Falls but then ATC starts to rise when model cost is below our average cost we know that the cost of producing one more unit of output which is your marginal cost is less than the average cost of all units produced so far so if your marginal cost is 20 while the average is simply 25 then clearly average will go down because producing one more unit will bring this average down because margin is below the average so when the marginal cost is below average cost average cost will come down and this applies both to the variable cost and the total cost so when marginal is below average the average goes down when margin is above average the average goes up we can say this that um when you do better than the average the average goes up and when you do worse than the average the average goes down so when you do better than the average let's say when margin is above average you can pull up the average and that's why averages are going up when the margin is above the average when you do worse than the average when margin is below the average of course you bring down the average the average goes down interestingly graphically the ABC at each level of output is deriv from the slope of a line drawn from the region to the point on the TVC curve corres responding to the particular level of output what do I mean by this if you look closely to this diagram and let's say choose any point let's say we choose a point like uh Q2 which is uh this quantity and let's go and find Q2 here so this is the total variable cost and this is the quantity so if I look at the value of average variable cost at Q2 what I need to do is that I can find out the TVC which is this distance divided by your quantity which is Q2 and therefore I can get myself my abc at that point so in short that ABC at let's say each point for example at Point uh Q2 for quity Q2 is simply a line uh drawn from the uh origin to the point on the TVC curve so the slope of this remember the slope the formula for slope is also change in y over change in X so change in y will be this distance change in X will be this distance so the slope of of the line which is drawn from origin towards the Curve will give you a the value of the ABC and you can see that and we can see also that the point where your TVC is minimum this quantity Q2 is the the flattest possible line which is coming from the origin then touching the TVC curve in other words here TVC uh it's tangent to the TVC curve and uh if I draw any more lines which are in lower in slope or more flatter than this line I will not have any line touching the TVC so in order to find the lowest point uh of the ad ABC from the TVC curve all we need to draw is uh a line from the origin all the way touching the TVC at the lowest possible Point similarly we can do the exercise for uh total cost the slope of uh the line which is coming from the region and touching total cost Curve will give you the value of uh uh ATC and the lowest point of uh ATC where ATC is the minimum will be the line that is the flattest line possible touching the total cost C so look at this quantity Q3 which is this one at this quantity Q3 or Point C your uh ATC is minimum primarily because this line that is coming from the origion and touching the total cost curve is the flattest possible line so if if you if somebody says what's the ATC at Q3 all I need to do is find out the vertical distance which is my total cost right which is this point my total cost curve and then look at my uh my quantity which is this Q3 and I'll get myself my ATC so ATC simply is uh TC over Q so we can say this information is pretty useful because uh I can from this graph figure out uh all my points where MC TC and TVC is minimum for example point a is when the TC starts to rise at a faster rate is when my marginal cost is minimum point B with the slope uh that is drawn from the origin to TVC um is is the flattest is the minimum is the point where a C is minimum and similarly a line that is drawn from origin to TVC the which is the flattest one is the one where ATC will be minimum and the flattest means that the both B and C are forming a tangent to your TC and TVC |
190279 | https://mindyourdecisions.com/blog/2023/06/01/number-minus-reverse-is-divisible-by-9/ | Number Minus Reverse Is Divisible By 9 – Mind Your Decisions
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About Me: Presh Talwalkar
I run the MindYourDecisions channel on YouTube, which has over 2 million subscribers and 400 million views. I am also the author of The Joy of Game Theory: An Introduction to Strategic Thinking, and several other books which are available on Amazon.
(As you might expect, the links for my books go to their listings on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.)
By way of history, I started the Mind Your Decisions blog back in 2007 to share a bit of math, personal finance, personal thoughts, and game theory. It's been quite a journey! I thank everyone that has shared my work, and I am very grateful for coverage in the press, including the Shorty Awards, The Telegraph, Freakonomics, and many other popular outlets.
I studied Economics and Mathematics at Stanford University.
Feel free to send me an email presh@mindyourdecisions.com. I get so many emails that I may not reply, but I save all suggestions for puzzles/video topics.
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If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.
Book ratings are from January 2025.
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Mind Your Decisions is a compilation of 5 books:
(1) The Joy of Game Theory: An Introduction to Strategic Thinking
(2) 40 Paradoxes in Logic, Probability, and Game Theory
(3) The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias
(4) The Best Mental Math Tricks
(5) Multiply Numbers By Drawing Lines
The Joy of Game Theory shows how you can use math to out-think your competition. (rated 4.2/5 stars on 564 reviews)
40 Paradoxes in Logic, Probability, and Game Theory contains thought-provoking and counter-intuitive results. (rated 4.2/5 stars on 81 reviews)
The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias is a handbook that explains the many ways we are biased about decision-making and offers techniques to make smart decisions. (rated 4.2/5 stars on 55 reviews)
The Best Mental Math Tricks teaches how you can look like a math genius by solving problems in your head (rated 4.3/5 stars on 148 reviews)
Multiply Numbers By Drawing Lines This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 4.5/5 stars on 57 reviews)
Mind Your Puzzles is a collection of the three “Math Puzzles” books, volumes 1, 2, and 3. The puzzles topics include the mathematical subjects including geometry, probability, logic, and game theory.
Math Puzzles Volume 1 features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. Volume 1 is rated 4.4/5 stars on 138 reviews.
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Math Puzzles Volume 3 is the third in the series. (rated 4.3/5 stars on 38 reviews)
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Number Minus Reverse Is Divisible By 9
If you buy from a link in this post, I may earn a commission. This does not affect the price you pay. As an Amazon Associate I earn from qualifying purchases. Learn more.
Posted June 1, 2023 By Presh Talwalkar. Read about me, or email me.
Thanks to Trần for the suggestion! Here’s a nice little mathematical “magic” trick.
Take any number, say 8,675,309. Reverse the digits to get 9,035,768. The difference between the two numbers is 360,459 and this is evenly divisible by 9 since 360,459/9 = 40,051. But there’s even something more interesting. You can re-arrange (or permute) the digits in any random order. Let’s say the new number is 3,580,697. The difference with the original number 8,675,309 is 5,094,612, and that is evenly divisible by 9 since 5,094,612/9 = 566,068.
This will always be true!
“Magic” trick
Number
n
Re-arrange digits/permutation
σ(n)
Always divisible by 9
n – σ(n)
σ(n) – n
But why is this true? Let’s work it out with some number theory!
As usual, watch the video for a solution.
Number Minus Reverse Is Divisible By 9
Or keep reading.
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Answer To Number Minus Reverse Is Divisible By 9
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
Let’s go back to something you probably learned in school. A number is divisible by 9 if the sum of its digits is divisible by 9. Why is that true? Let’s work out a pattern with powers of 10.
1 = 1 + 0
10 = 1 + 9
100 = 1 + 99
1000 = 1 + 999
…
10 k = 1 + 99…9
Clearly a number with only 9s is a multiple of 9 since 99…9 = 9×11…1. So every power of 10 is 1 more than a multiple of 9. This is written mathematically as a modulo equation:
10 k = 1 (mod 9)
Just like a regular equation, you can multiply both sides by the same number. So let’s multiply both sides by x to get:
x(10 k) = x (mod 9)
So how does this result help us explain the divisibility rule? Let’s say a number n has digits d k-1 d k-2…d 0. We can write this in expanded form:
n
= d k-1 d k-2…d 0
= d k-1 10 k-1 + d k-2 10 k-2 + … + d 0 10 0
When we take modulo 9 on both sides, every power of 10 becomes 1, so we have:
n = d k-1 + d k-2 + … + d 0 (mod 9)
So a number is equal to the sum of its digits modulo 9.
But we also know n is divisible by 9 is equivalent to saying n = 0 (mod 9). Therefore, a number is divisible by 9 if and only if the sum of its digits is divisible by 9.
Explaining the magic trick
Let s(n) = sum of the digits of n. As shown above, this is equivalent to n modulo 9.
If we re-arrange the digits, the new number clearly has the same sum of digits. That is s(σ n) = s(n). Therefore, we have:
n – σ(n) (mod 9)
= σ(n) – σ(n) (mod 9)
= 0 (mod 9)
A number minus any permutation of its digits will result in a new number that is divisible by 9. Amazing!
References
Math StackExchange
Published by
PRESH TALWALKAR
I run the MindYourDecisions channel on YouTube, which has over 1 million subscribers and 200 million views. I am also the author of The Joy of Game Theory: An Introduction to Strategic Thinking, and several other books which are available on Amazon.
(As you might expect, the links for my books go to their listings on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.)
By way of history, I started the Mind Your Decisions blog back in 2007 to share a bit of math, personal finance, personal thoughts, and game theory. It's been quite a journey! I thank everyone that has shared my work, and I am very grateful for coverage in the press, including the Shorty Awards, The Telegraph, Freakonomics, and many other popular outlets.
I studied Economics and Mathematics at Stanford University.
People often ask how I make the videos. Like many YouTubers I use popular software to prepare my videos. You can search for animation software tutorials on YouTube to learn how to make videos. Be prepared--animation is time consuming and software can be expensive!
Feel free to send me an email presh@mindyourdecisions.com. I get so many emails that I may not reply, but I save all suggestions for puzzles/video topics.
MY BOOKS
If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.
Book ratings are from January 2025.
(US and worldwide links)
Mind Your Decisions is a compilation of 5 books:
(1) The Joy of Game Theory: An Introduction to Strategic Thinking
(2) 40 Paradoxes in Logic, Probability, and Game Theory
(3) The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias
(4) The Best Mental Math Tricks
(5) Multiply Numbers By Drawing Lines
The Joy of Game Theory shows how you can use math to out-think your competition. (rated 4.2/5 stars on 564 reviews)
40 Paradoxes in Logic, Probability, and Game Theory contains thought-provoking and counter-intuitive results. (rated 4.2/5 stars on 81 reviews)
The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias is a handbook that explains the many ways we are biased about decision-making and offers techniques to make smart decisions. (rated 4.2/5 stars on 55 reviews)
The Best Mental Math Tricks teaches how you can look like a math genius by solving problems in your head (rated 4.3/5 stars on 148 reviews)
Multiply Numbers By Drawing Lines This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 4.5/5 stars on 57 reviews)
Mind Your Puzzles is a collection of the three "Math Puzzles" books, volumes 1, 2, and 3. The puzzles topics include the mathematical subjects including geometry, probability, logic, and game theory.
Math Puzzles Volume 1 features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. Volume 1 is rated 4.4/5 stars on 138 reviews.
Math Puzzles Volume 2 is a sequel book with more great problems. (rated 4.2/5 stars on 45 reviews)
Math Puzzles Volume 3 is the third in the series. (rated 4.3/5 stars on 38 reviews)
KINDLE UNLIMITED
Teachers and students around the world often email me about the books. Since education can have such a huge impact, I try to make the ebooks available as widely as possible at as low a price as possible.
Currently you can read most of my ebooks through Amazon's "Kindle Unlimited" program. Included in the subscription you will get access to millions of ebooks. You don't need a Kindle device: you can install the Kindle app on any smartphone/tablet/computer/etc. I have compiled links to programs in some countries below. Please check your local Amazon website for availability and program terms.
US, list of my books (US)
UK, list of my books (UK)
Canada, book results (CA)
Germany, list of my books (DE)
France, list of my books (FR)
India, list of my books (IN)
Australia, book results (AU)
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190280 | https://edtech506finalproject.files.wordpress.com/2012/07/mhm-point-slope-form-lesson.pdf | Lesson Plan Topic: Point-Slope Form Subject Area: Mathematics - Algebra 1 Grade Level: Grade 7 - Accelerated Time Frame: Approximately two 45-minute class periods Lesson Goals: Algebra 1 is the building block for accelerated mathematics. This lesson will give students detailed instruction on writing linear equations. At the conclusion of this lesson, students will be able to: Use the point-slope form to write an equation of a line. Common Core State Standards: Grade 7: Solve real-life and mathematical problems using numerical and algebraic expressions and equations. 7.EE.3. Solve multi-step real-life and mathematical problems posed with positive and negative rational numbers in any form, using tools strategically. Grade 8: Use functions to model relationships between quantities. 8.F.4. Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values. 8.F.5. Describe qualitatively the functional relationship between two quantities by analyzing a graph. Sketch a graph that exhibits the qualitative features of a function that has been described verbally. Credits: Student Characteristics: This lesson is intended for seventh grade gifted students in Algebra 1. These students are on the accelerated, college-bound, mathematics track. This course used to grant students a high school credit for the completion of Algebra 1 in the middle school. Students should expect daily homework assignments, weekly formative assessments, group collaboration, and a summative assessment. Required accommodations will be made for English Language Learners (ELL) and learning disabled students. The prerequisite for this course is Pre-Algebra. Point-Slope Form Lesson Megan Hoopes Myers EDTECH 506 Summer 2012 The prerequisite for this lesson are mastery of the following: Find the intercepts of the graph of a linear equation. Find the slope of a line using two of its points. Write a linear equation in slope-intercept form. Student's Present Level of Performance and Knowledge: Students need to have successfully completed Pre-Algebra before entering Algebra 1. Eligible students for Algebra 1 are those who also have scored advanced on the Mathematics PSSA Exams. Proficient skills with the TI-83 Plus and TI-84 Plus Graphing Calculators are required. Students must have basic knowledge with computers and Internet. Classroom Layout: The student desks are arranged in five rows consisting of six desks, allowing for individual work space. Class size averages 28 students hindering productive large group work. (Optional: Grouping desks into pairs and groups of four will coordinate with the appropriate lesson plans.) Lesson Procedures: DAY 1: "Trick of the Day" Solve the equation for y. Approximately 5 minutes "Point-Slope Form Scavenger Hunt" Each student will independently complete the scavenger hunt while viewing the unit's website. Approximately 40 minutes DAY 2: "Trick of the Day" Write an equation for the line in point-slope form. The line through (-1, 3) with slope -2. Approximately 5 minutes "Point-Slope Form Scavenger Hunt" Each student will independently complete the scavenger hunt while viewing the unit's website. Approximately 10 minutes "Point-Slope Form Assessment" Each student will independently complete the assessment. The use of a graphing calculator is allowed. Approximately 30 minutes. Required Materials: Teacher and Student textbook editions of McDougal Littell Algebra 1 ©2004 TI-83 Plus or TI-84 Plus Graphing Calculator (students are highly encouraged to have their own technology) Promethean Board and accessories (allows for interactive lessons and instruction) TI-SmartView software (graphing calculator presentation) Handouts (electronic and printable versions) Internet (to access Unit's website) Writing utensils Visuals: "At a Glance: Writing Linear Equations" image [this is the introductory image providing the learner a quick overview of the unit's goals] "Point-Slope Form Overview" image [concept map including the point-slope form definition, equation, and example] "Point-Slope Form Equation" image [consistent graph explaining how to find an equation of a line] "Slope-Intercept Form vs. Point-Slope Form" image [demonstrates how an equation of a graphed line can be written in more than one form] Assessment: The students will be assessed daily (i.e. "trick of the day", independent and collaborative assignments). Independent lesson assessments will be given upon completion of main topics. Outcome: The variety of assessments will demonstrate the students' mastery of the goals in the unit. Achievement of each goal will become a helpful building block for the next topic, section, chapter, and mathematical course. Teacher Resources: Point-Slope Form Lesson Point-Slope Form Scavenger Hunt Point-Slope Form Assessment Adapted from Unit Plan at |
190281 | https://proofwiki.org/wiki/Definition:Multiplicative_Group_of_Real_Numbers | Definition:Multiplicative Group of Real Numbers - ProofWiki
Definition:Multiplicative Group of Real Numbers
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The multiplicative group of real numbers(R≠0,×)(R≠0,×) is the set of real numbers without zero under the operation of multiplication.
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Non-Zero Real Numbers under Multiplication form Abelian Group
Thus real multiplication is:
Well-defined on R≠0 R≠0
Closed on R≠0 R≠0
Associative on R≠0 R≠0
Commutative on R≠0 R≠0
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The identity of (R≠0,×)(R≠0,×) is 1 1
Each element of (R≠0,×)(R≠0,×) has an inverse.
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190282 | https://pdfs.semanticscholar.org/089d/8ff0d1da59303e4d2c32e59e34c0dfbaf49a.pdf | ORIGINAL ARTICLES Epidemiology Biostatistics and Public Health - 2019, Volume 16, Number 2 Mini-review on continuity correction of Pearson’s x2 Continuity correction of Pearson’s chi-square test in 2x2 Contingency Tables: A mini-review on recent development Nicola Serra (1), Teresa Rea (1), Paola Di Carlo (2), Consolato Sergi (3) (1) Department of Public Health, University Federico II of Naples, Italy (2) Department of Sciences for Health Promotion, Mother & Child Care, Univ. of Palermo, Italy (3) Department of Lab. Medicine and Pathology, Univ. of Alberta, Edmonton, AB, Canada, Stollery Children’s Hospital, Univ. of Alberta, Edmonton, AB, Canada CORRESPONDING AUTHOR: Nicola Serra Ph.D., Department of Public Health, School of Medicine and Surgery, University Federico II of Naples, Italy. E-mail: nicola.serra@unina.it DOI: 10.2427/13059 Accepted on April 16, 2019 ABSTRACT The Pearson’s chi-square test represents a nonparametric test more used in Biomedicine and Social Sciences, but it introduces an error for 2x2 contingency tables, when a discrete probability distribution is approximated with a continuous distribution. The first author to introduce the continuity correction of Pearson’s chi-square test has been Yates F. (1934). Unfortunately, Yates’s correction may tend to overcorrect of p-value, this can implicate an overly conservative result. Therefore many authors have introduced variants Pearson’s chi-square statistic, as alternative continuity correction to Yates’s correction. The goal of this paper is to describe the most recent continuity corrections, proposed for Pearson’s chi-square test. Key words: Pearson’s x2 statistic; continuity correction; 2x2 contingency table; Yates’s continuity correction, Serra’s continuity correction INTRODUCTION Pearson’s chi-square test or c2 test is the nonparametric test commonly used by researchers in Biology, Medicine and Social Sciences. This test is based on the calculation of Pearson’s c2 statistic, introduced by Pearson K. , considering a sample of a population characterized by two o more dichotomous variables. For two dichotomous variables, it is possible to define a 2x2 contingency table, with the frequencies of occurrence of all combinations of their levels, considering a sample size equal to N, as it is shown in Table 1 In a 2x2 contingency table, Pearson’s c2 statistic is used to test the association between dichotomous variables, for example to individualize a possible e13059-1 ORIGINAL ARTICLES Epidemiology Biostatistics and Public Health - 2019, Volume 16, Number 2 Mini-review on continuity correction of Pearson’s x2 association between variables such as sex (Male/Female) and smoke (Yes/No). For this scope Pearson introduce the chi-square statistic to evaluate the discrepancy between observed (Oi,j ) and expected frequencies (Ei,j ), where the observed frequencies are a, b, c and d of Tables 1. Instead the expected frequencies are defined for every cell such as: , , , 1,2 = = i j i j rc E i j N where i and j indicate the row and column index respectively. The formula to compute Pearson’s c2 statistic is described by Pearson K. (1900): where r1, r2, c1 and c2 i.e. the totals across rows and columns are generally called marginal totals.
Using the c2 distribution to interpret Pearson’s c2 statistic requires one to assume that the discrete probability of observed binomial frequencies of 2x2 contingency table, can be approximated by the continuous c2 distribution. This assumption is not entirely correct and introduces some error. To reduce the error in approximation, many authors introduced a continuity correction or variants of Pearson’s c2 test.
To reduce the error introduced by Pearson’s c2 statistic, Yates F. suggested a correction for continuity that adjusts the formula for Pearson’s c2 by subtracting the value 0.5, from the difference between each observed value and its expected value for 2x2 contingency table. This correction reduces the c2 value obtained and consequently increases its p-value. The formula to compute Yates’s c2 statistic in a 2x2 contingency table is: Unfortunately, Yates’s correction may tend to overcorrect of p-value; this can implicate an overly conservative result, as reported by several authors [3-7].
The goal of this study is with literature review, to describe the most recent development about the continuity corrections by variants of Pearson’s c2 test defined for 2x2 contingency tables. METHODS In this section we introduce the most recent study about continuity correction of Pearson’s c2 statistic in 2x2 contingency tables. Serra’s continuity correction Recently Serra N. introduces a significant minimized of Pearson’s c2 statistic as a continuity correction of Pearson’s c2 test, for small samples (sample size ≤ 25). This approach is based on the observation that the denominator r1 r2 c1 c2 of (1), can be interpreted as a geometric mean. The formula to compute minimize Pearson’s c2 statistic in a 2x2 contingency table is: Serra N., showed with a statistical approach, that for small samples (≤25), the minimized Pearson’s c2 statistic in 2x2 contingency tables, represents a continuty correction for Pearson’s c2 statistic more effective in comparison to Yates’continuity correction. Particularly in this study the author verify that, the Fisher’s exact test [9,10], actually considered the “gold test” used when c2 test is not appropriate, i.e. when the sample size is small and the expected values in any of the cells of a 2x2 contingency table are below 5, had performance statistically equal to c2 Serra test.
Kajita Matchita et al.’s continuity correction Kajita Matchita et al. proposed a continuity correction to maintain a continuity value to be used when small expected cell frequencies on Pearson’s c2 test for independence exist in the research data. This correction method is used to control the type I error and obtained using a developed correction in more condition. For this scope the authors used a simulation study. The simulations were performed with Monte Carlo method, to evaluate the performance of their method in comparison to other continuity corrections such as Yates’s correction and Williams’s correction . It shows an outperformed control of type I error, considering a pattern of data set at a significant level of 0.05 and 0.01, simulated contingency tables between 2x2 and 4x4 (2x2, 2x3, 2x4, 3x3, 3x4 TABLE 1. 2x2 contingency table form.
Column variable (X) Row variable (Y) State 1 State 2 Row totals State 1 a b a + b = r1 State 2 c d c + d = r2 Column totals a + c = c1 b + d = c2 N = a + b + c + d e13059-2 ORIGINAL ARTICLES Epidemiology Biostatistics and Public Health - 2019, Volume 16, Number 2 Mini-review on continuity correction of Pearson’s x2 and 4x4), a number of small expected cell frequencies up to 30% of the total cell used, a sample size between 5 and 10 times that total cell, and using 10,000 data set simulated by Monte Carlo method for each pattern. The type I error (number rejection of null hypothesis divided by 10,000) was evaluated by Pearson’s c2 test, i.e. by classical c2 test without continuity correction.
In the case of 2x2 contingency tables, where the type I error is greater than the significant level, the c2 test equation to be used is as follows: instead, where the type I error is less than the significant level, the c2 test equation is where Oi,j and Ei,j represent the observed and expected frequencies respectively, instead C is the developed correction value. It was computed in two cases as follows, if the type I error is higher than the significant level, the authors try to replace the value C into equation (4) start from 0.01, 0.02, 0.03, ..., . If the type I error is less than the significant level, they try to replace the value C into equation (5) start from 0.01, 0.02 , 0.03 ..., . After they replaced value C and computed type I error then to compared with significant level. Developed correction value (C) is the value which gets very similar values between type I error and significant level.
CONCLUSION In this paper we described the most recent studies of continuity correction of Pearson’s c2 test. Since the first continuity correction proposed by Yates (1934), produced an overcorrection of the p-value, many authors are discouraging its use. Instead other authors [13-18], have followed Yates (1934) in claiming that the use of Pearson’s c2 in the case of 2x2 contingency tables tends to generate too many type I errors, especially with small samples, therefore they defined different continuity corrections of Pearson’s c2 statistic, to reduce the type I error, and simultaneously to reduce the type II error that Yates’s correction introduces Unfortunately, the study of continuity correction of Pearson’s c2 statistic is very limited in the recent statistical literature, only two recent studies are dedicated at this problem (Serra N., 2018 and Kajita Matchita et al., 2018), showing of the variants of c2 statistic as continuity correction of Pearson’s c2 test.
Funding statement This research did not receive any specific grant from funding agencies in the public, commercial, or not for profit sectors.
Competing interests statement There are no competing interests for this study.
References 1. Pearson K. (1900), On the criterion that a given system of deviations from the probable in the case of a correlated system of variables is such that it can be reasonably supposed to have arisen from random sampling. Philosophical Magazine Series 5; 50(302):157–175.
2. Yates, F. (1934). Contingency tables involving small numbers and the c2 test. Supplement to the Journal of the Royal Statistical Society, 1(2), 217-235.
3. Camilli, G., & Hopkins, K. D. (1978). Applicability of chi-square to 2×2 contingency tables with small expected cell frequencies. Psychological Bulletin, 85(1), 163.
4. Campbell, I. (2007). Chi-squared and Fisher–Irwin tests of two-by-two tables with small sample recommendations. Statistics in Medicine, 26(19), 3661-3675.
5. Haber, M. (1982). The continuity correction and statistical testing. International Statistical Review/Revue Internationale de Statistique, 135-144.
6. Richardson, J. T. (1990). Variants of chi-square for 2×2 contingency tables. British Journal of Mathematical and Statistical Psychology, 43(2), 309-326.
7. Richardson, J. T. (2011). The analysis of 2×2 contingency tables— Yet again. Statistics in Medicine, 30(8), 890-890.
8. Serra, N. (2018). A significant minimization of Pearson’s c2 statistics in 2x2 contingency tables: preliminary results for small samples. Epidemiology, Biostatistics and Public Health, 15(3).
9. Agresti, A. (2001). Exact inference for categorical data: recent advances and continuing controversies. Statistics in medicine, 20(17-18), 2709-2722.
10. Fisher, R.A. (1934), Statistical Methods for Research Workers. Chapter 12. 5th Ed., Oliver & Boyd.
11. Matchima, K., Vongprasert, J., & Chutiman, N. (2018). The Development of a Correction Method for Ensuring a Continuity Value of The Chi-square Test with a Small Expected Cell Frequency. Naresuan University Journal: Science and Technology (NUJST), 26(1), 98-105.
12. Mcdonald, J.H. (2014), Handbook of Biological Statistics. Maryland: Sparky House Publishing.
13. Cochran WG. (1954), Some methods for strengthening the common c2 tests. Biometrics; 10(4):417–451.
14. Cox, D.R. (1970). The continuity correction. Biometrika 57: 217-219.
15. Feller, W. (1968). An Introduction to Probability Theory and Its Applications. Volume I, 3rd ed. John Wiley & Sons, Inc. New York.
16. Mantel, N., & Haenszel, W. (1959). Statistical aspects of the analysis of data from retrospective studies of disease. Journal of the National Cancer Institute, 22(4), 719-748.
17. Maxwell, E. A. (1976). Analysis of contingency tables and further e13059-3 ORIGINAL ARTICLES Epidemiology Biostatistics and Public Health - 2019, Volume 16, Number 2 Mini-review on continuity correction of Pearson’s x2 reasons for not using Yates correction in 2× 2 tables. The Canadian Journal of Statistics/La Revue Canadienne de Statistique, 277-290.
18. Upton, G. J. G. (1982). A comparison of alternative tests for the 2× 2 comparative trial. Journal of the Royal Statistical Society. Series A (General), 86-105.
e13059-4 |
190283 | http://home.miracosta.edu/dlr/qnum.htm | Quantum Numbers and Electronic Structure
Quantum Numbers and Electronic Structure
Quantum Numbers Table | Atomic Structure Slideshow | Quantum Chemistry Quizzes
To account for the behavior of an electron in an atom, it is not sufficient to describe the electron simply as a negatively charged particle; its wavelike properties must also be considered. The first such description of an electron came in 1926 with the development of Schrodinger's wave equation. The branch of science that deals with the solution of wave equations is called quantum mechanics (or wave mechanics). Each solution to a wave equation is characterized by three integers called quantum numbers. Each solution corresponds to a discrete energy and defines a region of space about the nucleus (called an orbital) where an electron having that energy is generally found. A fourth quantum number is also necessary for a unique description of an equation.
According to the quantum mechanical model, the allowed energy levels of an electron are composed of one or more orbitals, and the distribution of electrons about the nucleus is determined by the number and kinds of energy levels that are occupied. Therefore, in order to understand the way electrons are distributed, we must first examine the energy levels. This is best accomplished through a discussion of the four quantum numbers. The most important aspects of each quantum number are presented below.
The principal quantum number, n.The electronic energy levels in an atom are arranged roughly into principal levels (or shells) as specified by n. The value of n gives an indication of the position of an electron in the energy level relative to the nucleus; the larger the value of n, the greater the average distance of an electron from the nucleus and the higher its energy. The principal quantum number may have values as follows: n = 1, 2, 3, 4, …..
The azimuthal quantum number (also called subsidiary or secondary), l. Each principal energy level may be split into closely spaced sublevels (or sublevels) as specified by l. This quantum number may be more aptly named the orbital shape quantum number, since each orbital in a given type of sublevel (i.e., a given value of l) has the same "electron cloud" shape. For example, when l = 0, the orbital is spherical. For each principal energy level (designated by n) there are n sublevels (i.e., n values of l): l=0,1,2,3,…(n -1). Sublevels are commonly given letter designations. The l = 0,1,2,3,4,5, … sublevels are designated as s, p, d, f, g,…. sublevels, respectively. For known elements no value of l higher than 3 (f sublevel) is necessary.
Two quantum numbers (n and l) are required to specify a particular energy sublevel.
The magnetic quantum number, m l. Each orbital within a particular sublevel is distinguished by its value of m l. This quantum number may be more aptly named the orbital orientation quantum number. In each energy sublevel (designated by l) there are 2 l+1 possible independent orientations of the electron cloud. Each orientation is defined by a value of ml and is called an orbital.
m l = l, (l -1), (l -2), … 0 … -(l -2), -(l -1), -l or m l = 0, ±1, ±2, ±3, … ±1
All orbitals in a given sublevel are of equal energy (they are degenerate). In the presence of a magnetic field their different orientations cause them to have different energies.
Three quantum numbers (n, l, and m l) are required to specify a particular orbital.
The spin quantum number, m s. An electron spins on its own axis as characterized by m s. There are two possible directions of spin: m s = +1/2 or -1/2. Since a spinning charge generates a magnetic field, an electron has a magnetic field associated with it. Two electrons in the same orbital are most stable when they have opposite spins (+1/2 and -1/2) due to a magnetic attraction. Such electrons are said to be paired electrons or each other's magnetic fields, but an unpaired electron may be detected by magnetic measurements. In fact, elements with unpaired electron are attracted by magnetic fields; such elements are called paramagnetic. Magnetic measurements have shown that electrons are distributed among the orbitals of a sublevel in a way that gives the maximum number of unpaired electrons with parallel spins (all m s values have the same sign).
Four quantum numbers (n, l, m l, and m s) are required to specify a particular electron.
The following list is a condensation of some of the most useful information relevant to quantum numbers and electronic structure.
The four quantum numbers:
n = 1, 2, 3, …
l = 0, 1, 2, 3, …. (n-1)
m l = 0, ±1, ±2, ±3, … ±l
m s = +1/2 or -1/2
Sublevel Types
QN lTypeOrbitalsTotal ElectronsQN lTypeOrbitalsTotal Electrons
0 s 1 2 3 f 7 14
1 p 3 6 4 g 9 18
2 d 5 10 5 h 11 22
3. Principal energy level n contains: (a) n sublevels (b) n 2 orbitals (c) 2 n 2 electrons maximum (see table above)
4. Sublevel l contains: (a) 2 l+1 orbitals (b) 2(2 l+1) electrons maximum
5. In the ground state electrons fill orbitals so that the total energy of the atom is minimized.
6. Sublevel energies increase as: (a) n increases: 1 s<2 s<3 s…; 2 p<3 p<4 p ….. etc. (b) l increases: 2 s<2 p; 3 s<3 p<3 d; 4 s<4 p<4 d<4 f; etc.
7. Each orbital can hold a maximum of two electrons; they must be paired.
8. In a given sublevel electrons are distributed among the orbitals in a way that yields the maximum number of unpaired electrons with parallel spins.
9. For a given value of l, orbital shape remains the same. For example, orbitals with l = 0 (1 s, 2 s, 3 s, etc. sublevels) are all spherical.
10. A given value of n designates a specific principal energy level. Given values of n and l designate a specific energy sublevel. Given values of n, l, and m l designate a specific orbital. Given values of n, l, m l, and m s designate a specific electron.
To review the definitions and interrelationships of the 4 quantum numbers, study this Table. Careful study will show that the table is consistent with the statements listed above.
| n | l | m l | Subshell | Number of orbitals in subshell | Number of electrons in subshell |
--- ---
| 1 | 0 | 0 | 1s | 1 | 2 |
| 2 | 0 | 0 | 2s | 1 | 2 |
| 2 | 1 | -1, 0, +1 | 2p | 3 | 6 |
| 3 | 0 | 0 | 3s | 1 | 2 |
| 3 | 1 | -1, 0, +1 | 3p | 3 | 6 |
| 3 | 2 | -2, -1, 0, +1, +2 | 3d | 5 | 10 |
| 4 | 0 | 0 | 4s | 1 | 2 |
| 4 | 1 | -1, 0, +1 | 4p | 3 | 6 |
| 4 | 2 | -2, -1, 0, +1, +2 | 4d | 5 | 10 |
| 4 | 3 | -3, -2, -1, 0, +1, +2, +3 | 4f | 7 | 14 |
There are two common methods of indicating the arrangement of electrons in an atom. Unless otherwise stated, the lowest energy state (ground state) is given. These methods are electron configurations and atomic orbital diagrams.
An electron configuration shows the electrons distribution by sublevel using the quantum numbers n and l, where the notation for l is by its letter designation (s, p, d, f, etc.). For example, the notation 3 d 4 indicates 4 electrons in the d sublevel (l=2) of the n =3 principal energy level. You are referred to Chapter 7 in the textbook where a complete discussion of electron configuration may be found.
An atomic orbital diagram shows the electron distribution in an atom by means of a diagram which accounts for the distribution by all four quantum numbers.
An orbital is shown by a box, circle or line.
An electron is shown by an arrow.
The arrows also show the spin of the electron so that when two electrons are in the same orbital, the arrows point in opposite directions to represent their opposing spins.
Sublevels are shown by a designation under the appropriate orbitals (see examples below).
In the figure below, an atomic orbital diagram is used to illustrate the order of filling for the first ten electrons as shown by the numbers entered in the boxes. As an example, consider the electronic structure of sulfur.
Since sulfur has 16 electrons, its electron configuration is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 4
Atomic orbital diagram for sulfur with
two unpaired electrons:
The chart below shows the order of orbital fill up by electrons. Remember that electrons ar put into the lowest available energy orbital before filling up a higher energy orbital. In addition, you always put one electron in each of the degenerate orbitals of a particlar energy level before putting a second electron in any of the orbitals of the same energy (Hund's Rule). |
190284 | https://stats.stackexchange.com/questions/575393/survival-vs-hazard-when-to-use-which | regression - "Survival" vs. "Hazard" : When to Use Which? - Cross Validated
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"Survival" vs. "Hazard" : When to Use Which?
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When dealing with Survival Analysis, we create models that estimate two properties:
Survival: Survival Probabilities tend to be more straightforward to understand. Survival Probabilities estimate the probability of surviving some "event" past a certain time, given that you have survived up until that time.
Hazard: On the other hand, Hazard Rate is said to represent the "instantaneous rate of experiencing the event" at any given time.
I have the following question: When we work with on Survival Analysis data, what kinds of problems are better answered using "Survival Probabilities" and what kinds of problems are better answered using "Hazard Rates"?
For instance, I could use Survival Probabilities to estimate out which cohorts of patients are more likely to survive a certain "event" (e.g. some disease) - and I could then use Hazard Rates to estimate which of these same cohorts are more likely to experience this "event" at different times. I am not a biomedical researcher by background, but both of these quantities seem equally important to me. But I was hoping someone could help me better understand this question.
Usually in biomedical studies, which of these two quantities tends to be "more important" - are we ever able to interpret both of these quantities simultaneously? In short, which kinds of problems are better answered using "Survival Probabilities" and what kinds of problems are better answered using "Hazard Rates"?
Thanks!
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asked May 15, 2022 at 20:00
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The survival function and hazard function are directly related. In continuous time:
h(t)=−S′(t)S(t).h(t)=−S′(t)S(t).
So "answering a problem" in terms of survival probabilities is equivalent to doing so in terms of hazard rates. Which you work with first depends on the type of model.
If a proportional hazards (PH) assumption holds, then you can work first in terms of hazards with a semi-parametric Cox PH survival model. Under the PH assumption only the relative hazards are affected by covariate values; the underlying shape of the baseline survival curve doesn't matter. A Cox PH regression model thus doesn't model the survival curve directly. It estimates the relative hazards as functions of covariates under a PH assumption. Hazard ratios then provide compact summaries of the associations of covariates with outcome, without considering the survival curve per se. My impression is that Cox models are predominant in clinical survival analysis. Once the Cox model is fit, however, you can estimate corresponding survival curves.
With a fully parametric survival model that involves censored or truncated survival-time data, you could say that model fitting is done more in terms of the survival function. This page shows the forms of the likelihoods associated with each type of data that are used to fit the model, expressed in terms of the survival function and the event density, f(t)=−S′(t)f(t)=−S′(t). If the parametric functional form isn't maintained under a PH assumption, then relative hazards aren't constant over time and interpretation of results in terms of hazards is less straightforward. Such models (e.g., log-normal models) can instead follow an accelerated failure time interpretation, in which covariates stretch or compress the time axis of a single underlying survival curve.
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answered May 16, 2022 at 12:11
EdMEdM
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Could you clarify what S' is? how does it differ from S?Wojty –Wojty 2023-11-08 13:37:35 +00:00 Commented Nov 8, 2023 at 13:37
@Wojty the ' is a standard symbol for the first derivative of a function of a single variable. In this case, S′(t)=d S(t)/d t S′(t)=d S(t)/d t.EdM –EdM 2023-11-08 14:42:53 +00:00 Commented Nov 8, 2023 at 14:42
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Hazard models are highly flexible methods that produce summary measures of differences in the time to event.
Using the hazard function, we can create multivariable regression models with independent variables of any form. Such models can be extended to accommodate situations that include repeated measures, stratification, competing outcomes, or time-dependent covariates. The resulting models can be used to create plots of any form (eg, cumulative hazard, instantaneous hazard, or survival) or to generate probabilities or rates that can be used in an array of applications.
The cumulative hazard can exceed a value of 1 (it’s a rate), and some people find it unsettling or nonintuitive to see graphs that exceed a rate of 1.0 event/time.
Survival models depict the empiric time to event data and can provide a test of the difference in group survival.
Survival analysis (traditional Kaplan-Meier analysis) is a much more restrictive approach to modeling time to event data, but it models or displays the difference between distributions as they are (ie, does not “transform” the time to event data). Traditional Kaplan-Meier analysis can only compare groups and can only make a determination on the probability of seeing the observed data assuming all groups were sampled from the same distribution (ie, provides a p-value, but does not provide measures of relative risk). When all of the work to reduce bias and confounding is done with the experiment's design, as in a randomized trial, traditional survival analysis can provide an excellent measure of the differences between survival at some time point along with confidence intervals and a p-value.
Survival estimates are bounded by 1 and 0; most people find these measures intuitive and easy to interpret in graphical form.
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answered May 17, 2022 at 15:30
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190285 | https://www.healthline.com/health/superfetation | Superfetation
Medically reviewed by Holly Ernst, PA-C — Written by Jacquelyn Cafasso on March 26, 2018
Superfetation is when a second, new pregnancy occurs during an initial pregnancy. It’s rare in humans and observed in animals much more frequently.
Superfetation is a rare phenomenon where a second pregnancy occurs alongside an existing one.
Another ovum (egg) is fertilized by sperm and implanted in the womb days or weeks later than the first one. Babies born from superfetation are often considered twins since they may be born during the same birth on the same day.
Superfetation is common in other animal speciesTrusted Source, like fish, hares, and badgers. Its likelihood of occurring in humans is controversial. It’s considered extremely rare.
There are only a few cases of supposed superfetation in the medical literature. Most cases occurred in woman undergoing fertility treatments such as in vitro fertilization (IVF).
How does superfetation happen?
In humans, a pregnancy occurs when an ovum (egg) is fertilized by sperm. The fertilized ovum then implants itself in a woman’s uterus. For superfetation to happen, another completely different ovum needs to be fertilized and then implanted separately in the womb.
For this to happen successfully, threeTrusted Source very unlikely events need to take place:
Ovulation (release of on ovum by an ovary) during an ongoing pregnancy. This is incredibly unlikely because hormones released during pregnancy function to prevent further ovulation.
The second ovum must be fertilized by a sperm cell. This is also unlikely because once a woman is pregnant, their cervix forms a mucus plug that blocks the passage of sperm. This mucus plug is the result of elevations of hormones produced in pregnancy.
The fertilized egg needs to implant in an already pregnant womb. This would be difficult because implantation requires the release of certain hormones that wouldn’t be released if a woman were already pregnant. There is also the issue of having enough space for another embryo.
The chances of these three unlikely events occurring simultaneously seem nearly impossible.
This is why, of the few cases of potential superfetation reported in the medical literature, most have been in women undergoing fertility treatmentsTrusted Source.
During a fertility treatment, known as in vitro fertilization, fertilized embryos are transferred into a woman’s uterus. Superfetation might happen if the woman also ovulates and the egg becomes fertilized by sperm a few weeks after the embryos are transferred into her uterus.
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Are there any symptoms that superfetation has occurred?
Because superfetation is so rare, there are no specific symptoms associated with the condition.
Superfetation may be suspected when a doctor notices that twin fetuses are growing at different rates in the womb. During an ultrasound test, a doctor will see that the two fetuses are different sizes. This is called growth discordance.
Still, a doctor probably won’t diagnose a woman with superfetation after seeing that the twins are different in size. This is because there are several more common explanations for growth discordance. One example is when the placenta isn’t able to sufficiently support both fetuses (placental insufficiency). Another explanation is when blood is unevenly distributed between the twins (twin-to-twin transfusion).
Are there any complications of superfetation?
The most important complication of superfetation is that the babies will be growing at different stages during the pregnancy. When one baby is ready to be born, the other fetus might not be ready yet. The younger baby would be at risk of being born prematurely.
Premature birth puts the baby at a higher risk of having medical problems, such as:
trouble breathing
low birth weight
movement and coordination problems
difficulties with feeding
brain hemorrhage, or bleeding in the brain
neonatal respiratory distress syndrome, a breathing disorder caused by underdeveloped lungs
In addition, women carrying more than one baby are at increased risk of certain complications, including:
high blood pressure and protein in the urine (preeclampsia)
gestational diabetes
The babies may need to be born via Cesarean section (C-section). The timing of the C-section depends on the difference in the development of the two babies.
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Is there any way to prevent superfetation?
You can decrease your chances of superfetation by not having sexual intercourse after you’ve already become pregnant. Still, superfetation is extremely rare. It’s incredibly unlikely that you would become pregnant for a second time if you did have sex after you’ve already become pregnant.
Of the few cases of potential superfetation reported in the medical literature, most have been in women undergoing fertility treatments. You should be tested to make sure you’re not already pregnant before undergoing these treatments, and follow all recommendations from your fertility doctor if undergoing IVF, including certain times of abstinence.
Are there any known cases of superfetation?
Most reports of superfetation in humans are in women who have undergone fertility treatments to become pregnant.
A case reportTrusted Source published in 2005 discusses a 32-year-old woman who had undergone in vitro fertilization and became pregnant with twins. Around five months later, the woman’s doctor noticed during an ultrasound that she was actually pregnant with triplets. The third fetus was much smaller in size. This fetus was found to be three weeks younger than its siblings. The doctors concluded that another fertilization and implantation took place naturally weeks after the in vitro fertilization procedure.
In 2010, there was another case report of a woman with superfetation. The woman was undergoing an artificial insemination (IUI) procedure and was taking medications to stimulate ovulation. It was later found out that she was already pregnant with an ectopic (tubal) pregnancy. Doctors didn’t know the woman was already pregnant with an ectopic pregnancy when they performed the IUI procedure.
In 1999, there was a reportTrusted Source of a woman who is believed to have experienced superfetation spontaneously. The fetuses were found to be four weeks apart. The woman went through a normal pregnancy and both babies were born healthy. Twin one was a female born at 39 weeks and twin two was a male born at 35 weeks.
Takeaway
Superfetation is often observed in other animals. The possibility of it happening naturally in a human remains controversial. There have been a few case reports of superfetation in women. Most had been undergoing assisted reproduction techniques, like in vitro fertilization.
Superfetation results in two fetuses with different ages and sizes. Despite this, it’s possible for both babies to be born fully developed and completely healthy.
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How we reviewed this article:
Healthline has strict sourcing guidelines and relies on peer-reviewed studies, academic research institutions, and medical journals and associations. We only use quality, credible sources to ensure content accuracy and integrity. You can learn more about how we ensure our content is accurate and current by reading our editorial policy.
Baijal N, et al (2007). Discordant twins withthe smaller baby appropriate for gestational age – Unusual manifestation ofsuperfoetation: A case report. DOI:
Harrison A, et al. (2005). Superfetation as acause of growth discordance in a multiple pregnancy. DOI:
Lantieri T, et al. (2010). Superfetation afterovulation induction and intrauterine insemination performed during an unknownectopic pregnancy. DOI:
Pape O, et al. (2008). Superfetation: Casereport and review of the literature. DOI:
Roellig K, et al. (2010). Superconception inmammalian pregnancy can be detected and increases reproductive output perbreeding season. DOI:
Roellig K, et al. (2011). The concept ofsuperfetation: A critical review on a ‘myth’ in mammalian reproduction. DOI:
Tuppen GD, et al. (1999). Spontaneoussuperfetation diagnosed in the first trimester with successful outcome.
Twins, triplets, and other multiples. (2018).
Our experts continually monitor the health and wellness space, and we update our articles when new information becomes available.
Current Version
Mar 26, 2018
Written By
Jacquelyn Cafasso
Edited By
Kelly Morrell
Medically Reviewed By
Holly Ernst, PA-C
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190287 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_22?srsltid=AfmBOooLCqCLjFc5O-7vGyNEubNoT5PkjxBASlwPmYFFIiQyn4C_Pv_w | Art of Problem Solving
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2001 AMC 10 Problems/Problem 22
Contents
1 Problem
2 Solutions
3 Video solution 1
3.1 Solution 1
3.2 Solution 2
3.3 Really Easy Solution
4 Systems of Equations
5 Video Solution 2
6 See Also
Problem
In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by , , , , and . Find .
Solutions
Video solution 1
-DaBob
Solution 1
We know that , so we could find one variable rather than two.
The sum per row is .
Thus .
Since we needed and we know , .
Solution 2
The magic sum is determined by the bottom row. .
Solving for :
.
To find our answer, we need to find . .
Really Easy Solution
A nice thing to know is that any numbers that go through the middle form an arithmetic sequence.
Using this, we know that , or because would be the average.
We also know that because is the average the magic sum would be , so we can also write the equation using the bottom row.
Solving for x in this system we get , so now using the arithmetic sequence knowledge we find that and .
Adding these we get .
-harsha12345
Systems of Equations
Create an equation for every row, column, and diagonal. Let be the sum of the rows, columns, and diagonals. .
Notice that and both have . Equate them and you get that . Using that same strategy, we use instead. is good for our purposes. It turns out that . Since we already know those numbers, and , We can say that will be . We are now able to solve: , , , and . Respectively, , , , , and . We only require The sum of , which is . We get that the sum of and respectively is
-OofPirate
Video Solution 2
~savannahsolver
See Also
2001 AMC 10 (Problems • Answer Key • Resources)
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Problem 23
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190288 | https://cs.uwaterloo.ca/journals/JIS/VOL12/Osgood/osgood6.pdf | 23 11 Article 09.7.8 Journal of Integer Sequences, Vol. 12 (2009), 2 3 6 1 47 Falling Factorials, Generating Functions, and Conjoint Ranking Tables Brad Osgood and William Wu Department of Electrical Engineering Stanford University Stanford, CA 94305 USA osgood@stanford.edu willywu@stanford.edu Abstract We investigate the coefficients generated by expressing the falling factorial (xy)k as a linear combination of falling factorial products xlym for l, m = 1, . . . , k. Algebraic and combinatoric properties are discussed, some in relation to Stirling numbers.
1 Introduction Let xk denote the falling factorial power, xk = x(x −1) . . . (x −k + 1) , k ∈N, k ≥1 .
We think of x ∈R, but the definition can make sense in more general settings. Problems from discrete Fourier analysis – distant from the topics considered here – led us to falling factorial powers of products expressed as (xy)k = k X l,m=1 c(k) l,mxlym .
(1) There is the obvious symmetry c(k) l,m = c(k) m,l. Since c(1) 1,1 = 1 the interest begins when k ≥2.
For example, (xy)2 = x1y2 + x2y1 + x2y2 , (xy)3 = x1y3 + x3y1 + 6x2y2 + 3x2y3 + 3x3y2 + x3y3 .
1 Note that c(2) 1,1 = 0 c(3) 1,1 = 0 , c(3) 1,2 = c(3) 2,1 = 0 .
and that the coefficients that do appear are positive.
Here are the values for c(9) l,m displayed as a symmetric matrix: 0 0 0 0 0 0 0 0 1 0 0 0 0 15120 40320 24192 4608 255 0 0 10080 544320 1958040 1796760 588168 74124 3025 0 0 544320 6108480 12267360 7988904 2066232 218484 7770 0 15120 1958040 12267360 18329850 9874746 2229402 212436 6951 0 40320 1796760 7988904 9874746 4690350 965790 85680 2646 0 24192 588168 2066232 2229402 965790 185766 15624 462 0 4608 74124 218484 212436 85680 15624 1260 36 1 255 3025 7770 6951 2646 462 36 1 .
(2) The numbers c(k) l,m have a number of interesting properties that are the subject of the present paper. We found a recurrence relation, several closed-form expressions (which appear rather different from each other), a natural combinatorial interpretation in terms of conjoint ranking tables, and we can extend these results to products of more than two variables.
The combinatorics presented here have been considered before in other contexts. To make the present approach self-contained and more readable we have rederived (briefly) some of these earlier results, with references. We also mention several questions that we were unable to answer.
2 Stirling Numbers and Falling Factorial Powers One can solve for the coefficients c(k) l,m in any particular case, but that there should generally be such an expansion emerges from the connection between falling factorial powers, ordinary powers, and Stirling numbers; see Knuth , whose notation we follow.
In combinatorics one defines the (unsigned) Stirling numbers of the first kind by k l = the number of permutations of k letters which consist of l disjoint cycles.
In particular k k = 1 , k l = 0 if k < l.
(3) For us, the important fact is xk = n X l=1 (−1)k−l k l xl .
(4) 2 Stirling numbers of the second kind, denoted by k l , are defined by k l = the number of partitions of a set of k elements into l nonempty subsets.
Here k 1 = k k = 1 , k l = 0 if k < l.
(5) The special values in (3) and (5) will come up in adjusting summation indices.
Corresponding to (4) one has xk = n X l=1 k l xl .
(6) We included the phrase ‘Generating Functions’ in the title of the paper because the sequences k l , k l , and c(k) l,m each count something and each appears in an expansion in powers, (4), (6), and (1), much like classical generating functions.
Proceeding with the derivation of (1), from (4) and (6) we have (xy)k = k X p=1 (−1)k−p k p (xy)p = k X p=1 (−1)k−p k p xpyp = k X p=1 (−1)k−p k p p X l=1 p l xl ! p X m=1 p m ym !
= k X p=1 p X l=1 p X m=1 (−1)k−p k p p l p m xlym = k X l=1 k X m=1 k X p=1 (−1)k−p k p p l p m xlym = k X l,m=1 c(k) l,mxlym , where the second-to-last equality holds since by convention a b = 0 when a < b, and c(k) l,m = k X p=1 (−1)k−p k p p l p m .
(7) This establishes the expansion (1) and provides a formula for the coefficients.
It is not clear from this expression that the c’s are nonnegative. We will deduce this from a combinatorial characterization in Section 4.
3 When l = k, c(k) k,m = k X p=1 (−1)k−p k p p k p m = k k k k k m = k m .
Thus the Stirling numbers of the second kind appear in the last row (or column) of the matrix of the c’s, as we see in (2).
There are a few more arithmetic properties of the c(k) l,m that we wish to note, expressed most easily in terms of the symmetric matrix C(k) whose (l, m)-entry is c(k) l,m. Let Σ(k) be the k × k diagonal matrix whose entries are Σ(k) lm = (−1)k−l k l δlm .
Let S(k) 2 be the upper triangular matrix with nonzero entries (S(k) 2 )lm = m l , l ≤m .
One can check that C(k) = S(k) 2 Σ(k)(S(k) 2 )T .
In turn it follows that C(k) is invertible and has the same signature as Σ(k).
Since the diagonal entries in Σ(k) alternate sign, C(k) has ⌈k/2⌉positive eigenvalues and k −⌈k/2⌉ negative eigenvalues. Moreover, the diagonal entries of S(k) 2 are 1’s so det S(k) 2 = 1, and then det C(k) = k Y l=1 (−1)k−l k l .
For all k, it appears that the rows (columns), the diagonals, and the anti-diagonals of C(k) are all unimodal; this is quite visible for k = 9 in (2). Stronger than unimodality, numerical evidence suggests that the rows, diagonals, and anti-diagonals are log-concave. In fact, numerical evidence suggests that the associated polynomials for these sequences have negative real roots, satisfying Newton’s sufficient test for log-concavity .
3 Recurrence Relation Falling factorial powers satisfy ∆xn = nxn−1 , 4 where ∆is the forward difference operator. A little more generally, using the two-variable analog ∆f(x, y) = f(x + 1, y + 1) −f(x, y), one can verify that ∆(xlym) = (∆xl)ym + xl(∆ym) + (∆xl)(∆ym) = lxl−1ym + mxlym−1 + lmxl−1ym−1 .
Then, on the one hand, ∆(xy)k = k X l,m=1 c(k) l,m(lxl−1ym + mxlym−1 + lmxl−1ym−1) = k−1 X l=0 k X m=1 c(k) l+1,m(l + 1)xlym + k X l=1 k−1 X m=0 c(k) l,m+1(m + 1)xlym + k−1 X l,m=0 c(k) l+1,m+1(l + 1)(m + 1)xlym .
On the other hand, apply the identity vk = 1 v(vk+1 + kvk) with v = (x + 1)(y + 1) to write ∆(xy)k = ((x + 1)(y + 1))k −(xy)k = 1 (x + 1)(y + 1) ((x + 1)(y + 1))k+1 + k((x + 1)(y + 1))k −(xy)k , and then ∆(xy)k = k+1 X l,m=1 c(k+1) l,m (x + 1)l x + 1 (y + 1)m y + 1 + k k X l,m=1 c(k) l,m (x + 1)l x + 1 (y + 1)m y + 1 − k X l,m=1 c(k) l,mxlym = k+1 X l,m=1 c(k+1) l,m xl−1ym−1 + k k X l,m=1 c(k) l,mxl−1ym−1 − k X l,m=1 c(k) l,mxlym = k X l,m=0 c(k+1) l+1,m+1xlym + k k−1 X l,m=0 c(k) l+1,m+1xlym − k X l,m=1 c(k) l,mxlym .
Comparing the two expressions for ∆(xy)k, and rearranging terms, we have shown the following: Theorem 1. The coefficients c(k) l,m satisfy the recurrence c(k+1) l+1,m+1 = c(k) l,m + (l + 1)c(k) l+1,m + (m + 1)c(k) l,m+1 + ((l + 1)(m + 1) −k)c(k) l+1,m+1 .
We remark that while this recurrence generally has four terms, exceptions occur at the edges of the matrix.
Careful bookkeeping shows that in the last row when l = k, the recurrence reduces to the two term recurrence c(k+1) k+1,m+1 = c(k) k,m + (m + 1)c(k) k,m+1 which is precisely the recurrence relation describing Stirling numbers of the second kind.
The c(k) l,m coefficients can thus be viewed as a generalization of Stirling numbers.
Lastly, we comment that other proofs of the recurrence are possible, for example one that uses the recurrence relations for Stirling numbers.
5 4 Combinatorial Characterizations Conjoint analysis is a method in marketing that assigns weightings to independent attributes of a product. It is only the first step that we consider, that of setting up a conjoint ranking table. Suppose there are l ≥1 choices for one attribute (e.g., price) and m ≥1 for the other (e.g., color). Form an l×m table, in which each cell represents the two attributes considered jointly; hence the contraction ‘conjoint.’ Let k ≤lm and rank the conjoint preferences from 1 to k. Since we allow k ≤lm not every pair of attributes need have a ranking, but we do insist that every individual attribute must enter into the ranking at least once. That is, every row or column must have at least one ranked cell. In particular k ≥max{l, m}. One can consider the remaining cells as left blank or filled in with zeros.
For a given l and m we refer to such an object as a k-conjoint ranking table. This is quite a general concept and it is easy to imagine many examples. Here is one other. A graduate student is signing up to take qualifying exams, and is required to fill out an l × m table indicating preferences. Each of the l rows refers to a subject area, such as algebra, combinatorics, analysis, etc.. Each of the m columns refers to a particular professor; we assume that they are all equally capable of asking about any of subjects. The student will be asked k questions, where k ≤lm, and may put the numbers 1 through k in any of the cells, indicating preferences for who asks what kinds of questions. However, the student is not allowed to avoid any professors, and must put at least one number in each column.
Similarly, the student is not allowed to avoid any subject areas, and must put at least one number in each row.
For more background on how conjoint analysis is used in marketing, see Green .
Theorem 2. The number of k-conjoint ranking tables of size l × m is l! m! c(k) l,m.
Proof of Theorem 2. Let d(k) l,m be the number of l × m conjoint tables containing k ones.
By definition of c(k) l,m, it suffices to show that (xy)k = Pk l,m=1 d(k) l,m(xl/l!)(ym/m!). It also suffices to prove this for x and y restricted to the positive integers, since polynomials coin-ciding on the positive integers are identical.
The number of ways to place the numbers 1, 2, . . . , k into distinct cells of a x×y rectangle is clearly (xy)k. However, this placement can be done in the following way: choose l rows (in xl/l! ways), choose m columns (in ym/m! ways), and then place the numbers in the chosen rows and columns so that every row or column is used (in d(k) l,m ways). Summing over l and m gives the result.
Theorem 2 has the following immediate corollary: Corollary 3. The c(k) lm satisfy c(k) l,m = 0 for lm < k and c(k) l,m > 0 for lm ≥k .
Proof. When lm < k, the conjoint ranking table is too small to fit all k numbers, so c(k) l,m = 0.
However, when lm ≥k, there are enough cells in the table to be ranked from 1 to k.
6 Furthermore, there must exist at least one way of placing the numbers that satisfies the row and column constraints since the conditions 1 ≤l, m ≤k imply that k ≥max(l, m). Thus c(k) l,m > 0.
One can see the pattern of zeros for k = 9 in (2). The hyperbolic shape of the boundary between the zero and nonzero coefficients becomes more pronounced as k increases.
Here we make contact with earlier work, for the core of the proof above is counting a set of binary matrices, and these have been counted in other ways; see Chapter 9 in . One approach is to use the inclusion-exclusion principle. We would like to give this argument to show how it leads to another expression for c(k) l,m (which also appears in older literature, but not as interpreted here).
Fix l and m, take k ≤lm, and let O be the set of binary matrices of size l × m with 1’s in exactly k positions. Then |O| = lm k .
Now let k ≥max{l, m} and let C be the subset of O which have at least one 1 in every row and in every column. We want to find |C|.
For the index i running from 1 to l let Ai be the subset of O whose i’th row has all 0’s, and for the index i running from 1 to m let Ai+l be the subset of O whose i’th column has all 0’s. Then C = A1 ∩A2 ∩· · · ∩Al+m , where Ai = O \ Ai .
By the inclusion-exclusion principle |C| = |O|− l+m X i=1 |Ai|+ X i1<i2 |Ai1∩Ai2|− X i1<i2<i3 |Ai1∩Ai2∩Ai3|+· · ·+(−1)l+m|A1∩A2∩· · ·∩Al+m| .
To compute the general sum X |Ai1 ∩· · · ∩Aih| we have to have to distinguish the matrices that have zeroed rows from those that have zeroed columns. Suppose among the h sets Ai1, . . . , Aih that p of them have zeroed rows.
Then h −p have zeroed columns. For a fixed p there are l p ways to select p rows to zero out and there are m h−p ways to select h −p columns to zero out. After these choices, there are (l −p)(m −h + p) cells remaining amongst which we can place k ones. Thus X |Ai1 ∩· · · ∩Aih| = h X p=0 l p m h −p (l −p)(m −h + p) k and |C| = l+m X h=0 h X p=0 (−1)h l p m h −p (l −p)(m −h + p) k .
7 Multiplying |C| by k! distinguishes the nonzero elements, whether they are k distinguished balls tossed into bins or the numbers from 1 to k. The end result is evidently the same as counting the number of k-conjoint ranking tables, and hence c(k) l,m = 1 l!m! · k!|C| = k!
l!m!
l+m X h=0 h X p=0 (−1)h l p m h −p (l −p)(m −h + p) k .
(8) There is one more approach and one more formula. The equation (1) can also be written in the form xy k = k X l,m=1 b(k) l,m x l y m , (9) where b(k) l,m = l!m!
k! c(k) l,m .
We understand the binomial coefficient to be defined for nonintegral x by x k = Γ(x + 1) k!Γ(x −k + 1) , and we use x(x −1) . . . (x −k + 1) = Γ(x + 1) Γ(x −k + 1) .
We see that the b(k) l,m give a count of the number of the l ×m binary matrices with exactly k ones such that each row and column has at least one 1. We have a further comment on the combinatorics of (9) when x and y are integers. The left-hand side, xy k is just the number of ways to select k elements from an x × y array. How does this jibe with the right-hand side?
For any given selection of k cells there is a unique minimal subarray (smallest number of rows and columns) such that every row and column of that subarray contains a selected cell. This is illustrated in Figure 1. On the left, the darkened cells correspond to a selection of k = 8 cells, and the arrows indicate the rows and columns of the subarray in which the selected cells are contained. On the right we see the subarray extracted, and note that every row and column contains a selected square. Counting all possible such subarrays thus counts the ways of choosing k elements from the big array. This is what the right-hand side in equation (9) does, for x l y m is the number of ways to choose an l × m subarray, and then for each such subarray we count the number of ways to populate it with k entries such that no row or column is void. That multiplier is precisely b(k) l,m.
From Theorem 1, the b’s satisfy the recurrence b(k+1) l+1,m+1 = (l + 1)(m + 1)(b(k) l,m + b(k) l,m+1 + b(k) l+1,m) + ((l + 1)(m + 1) −k)b(k) l+1,m+1 .
(10) But it was noted explicitly by Cameron that M¨ obius inversion applied to (9) implies b(k) l,m = k X s,t=1 (−1)l+s+m+t l s m t st k .
(11) 8 contains selections column corresponding subarray row indices selection of 8 entries from 9 x 6 array 3 4 5 7 8 2 5 6 3 2 in which every row and column indices 7 1 2 5 6 1 4 5 6 8 9 4 3 3 2 Figure 1: Combinatorial Interpretation of equation (9).
Furthermore, Maia and Mendez found a similar formula using combinatorial species.
M¨ obius inversion is kin to matrix inversion and it is interesting to see how the latter can be used to derive (11).
For k2 pairs (xi, yi), i = 1, . . . , k2, we treat (9) as a system of linear equations for the k2 unknowns b(k) l,m. In matrix form, Aβ = ξ where x1 1 y1 1 x1 1 y1 2 · · · x1 1 y1 k x1 2 y1 1 x1 2 y1 2 · · · x1 k y1 k x2 1 y2 1 x2 1 y2 2 · · · x2 1 y2 k x2 2 y2 1 x2 2 y2 2 · · · x2 k y2 k .
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· · · .
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xk2 1 yk2 1 xk2 1 yk2 2 · · · xk2 1 yk2 k xk2 2 yk2 1 xk2 2 yk2 2 · · · xk2 k yk2 k | {z } A b(k) 1,1 b(k) 1,2 .
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b(k) 1,k b(k) 2,1 b(k) 2,2 .
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b(k) k,k | {z } β = x1y1 k x2y2 k .
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xiyi k .
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xk2yk2 k | {z } ξ .
Thus if the sequences (xi) and (yi) are chosen such that A is invertible, we will have an expression for the coefficients in β. The matrix inversion can be done quite simply because 9 one can choose sequences that make A the Kronecker product of the Pascal matrix with itself. Recall that the lower-triangular Pascal matrix P is defined by Pij = i j , 1 ≤i, j ≤k .
The Kronecker square of P is the k2 × k2 matrix with entries (P ⊗P)ij = P⌈i k⌉⌈j k⌉· P(i−1)k+1,(j−1)k+1 , where ik denotes i mod k. It is easily confirmed that P ⊗P equals the matrix A when xi = i k and yi = (i −1)k + 1. Since the inverse of a Kronecker product is the Kronecker product of the inverses, A−1 = (P ⊗P)−1 = P −1 ⊗P −1 .
Applying the Pascal matrix inverse formula P −1 ij = (−1)i+j i j , (P ⊗P)−1 ij = (−1)⌈i k⌉+⌈j k⌉+(i−1)k+1+(j−1)k+1 i k j k (i −1)k + 1 (j −1)k + 1 .
From the formula for A−1 let us derive (11). We have β = A−1ξ , b(k) l,m = βi , where i = k(l −1) + m, or b(k) l,m = βi = k2 X j=1 A−1 i,j ξj = k2 X j=1 A−1 i,j j k ((j −1)k + 1) k = k2 X j=1 (−1)⌈i k⌉+⌈j k⌉+(i−1)k+1+(j−1)k+1 i k j k (i −1)k + 1 (j −1)k + 1 j k ((j −1)k + 1) k .
Both i and j run from 1 to k2. From the equation i = k(l−1)+m, it follows that i k = l and (i−1)k+1 = m. Similarly, writing j = ks+t where s ∈{0, 1, . . . , k−1} and t ∈{1, 2, . . . , k}, it follows that j k = s + 1 and (j −1)k + 1 = t. Thus, b(k) l,m = k−1 X s=0 k X t=1 (−1)l+s+1+m+t l s + 1 m t (s + 1)t k = k X s=1 k X t=1 (−1)l+s+m+t l s m t st k , which is (11).
We conclude this part of the discussion by noting that we have three different expressions for the c’s (or the b’s), (7), (8) and (11) – and we do not know algebraically how to derive one from another!
10 5 More Than Two Variables For an expansion (x1x2 · · · xn)k = X L c(k) L x1 l1x2 l2 · · · xn ln , L = (l1, l2, . . . , ln) , 1 ≤li ≤k , of a product of more than two variables all the results in the preceding sections have natural extensions. Modifications to the earlier arguments are straightforward and so we record the outcomes with little additional detail – the chief problem is notation. We follow the generally accepted conventions on multi-indexing. In particular L! = l1!l2! · · · ln! .
The formula for the c’s in terms of Stirling numbers is c(k) L = k X p=1 (−1)k−p k p n Y i=1 p li .
or simply c(k) L = k X p=1 (−1)k−p k p p L .
if we allow ourselves the analog to the the multi-indexed case of binomial coefficients and write p L = n Y i=1 p li .
There is a natural extension of the product rule for the forward difference operator and it can be applied just as before to obtain a recurrence relation. It will pay to invest in a little extra notation. We write L + 1 = (li + 1: i = 1, . . . , n) and for a subset S ⊆{1, . . . , n} we write LS = (li : i ∈S) , LS + 1 = (li + 1: i ∈S) , LS = (li : i ∈{1, . . . , n} \ S) .
The result is c(k+1) L+1 = c(k) L −kc(k) L+1 + n X m=1 X |S|=m Y i∈S li !
c(k) LS+1,LS .
The generalization of a 2-dimensional conjoint ranking table allows for n independent attributes (color, price, shape, . . . ) with li choices for the i’th attribute. Based on the recurrence one can then show that L!c(k) L is the number of ways to fill in an l1 × l2 × · · · × ln conjoint ranking table with the numbers 1 through k, insisting, as before, that each attribute 11 must enter into the ranking at least once. Again this implies the nonnegativity of the c’s, more precisely c(k) L = 0 if Y i li < k and c(k) L > 0 if Y i li ≥k.
Extensions of the alternate formulas (8) and (11) are more complicated to write. For the former, to keep the final result from being too cluttered we use the notation ∥L∥= X li∈L li and for a multi-index M = (m1, m2, . . . , mn) with 1 ≤mi ≤h and ∥M∥= h we let Φ(L, M) = n Y i=1 (li −mi).
Then c(k) L = k!
L!
∥L∥ X h=0 (−1)h X M,∥M∥=h Φ(L, M) k n Y i=1 li mi .
While the derivation of this formula is only an extension of the argument for two variables, some discussion will help make the form clearer.
Let “slices” refer to the objects of dimension n −1 in a multidimensional table which are given by fixing the entry in one coordinate position. (One could say this is the higher-dimensional generalization of rows and columns for matrices.) As before, we use inclusion-exclusion to count b(k) L , the number of distinct l1×. . .×ln (0, 1)-tables with exactly k ones and no zeroed slices; that is the formula without the factorials in front. In the outer summation, the index h is the number of zeroed slices. The inner summation then runs over possible ways to distribute the h zeroed slices across the n different dimensions.
The number of zeroed slices in dimension i is mi, and li mi is the number of ways to select a particular set of mi slices to zero out.
After removing the zeroed slices, the number of remaining cells is given by Φ(L, M), and Φ(L,M) k is the number of ways to distribute k ones amongst these remaining cells.
Lastly, the analog of formula (11) based on Kronecker products and matrix inversion is c(k) L = k!
L!
k X r1,r2,...,rn=1 (−1) P i(ri+li) Q i ri k Y i li ri .
6 Acknowledgements We would like to thank Jiehua Chen, John T. Gill III, Michael Godfrey, and Donald Knuth for their interest and suggestions, and particularly the referee for a very meticulous report and many insightful comments.
This research was supported by a scholarship from the Frank H. and Eva Buck Founda-tion.
12 References P. J. Cameron, T. Prellberg, and D. Stark, Asymptotics for incidence matrix classes, Electron. J. Combin. 13 (2006), #R85.
Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC Press, 2002.
P. E. Green and V. Srinivasan, Conjoint analysis in consumer research: issues and outlook, J. Consumer Research 5 (1978), 103–123.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics, Addison Wesley, second edition, 1994, pp. 257–267.
M. Maia and M. Mendez, On the arithmetic product of combinatorial species, Discrete Math. 308 (2008), 5407–5427.
R. P. Stanley, Log-Concave and Unimodal Sequences in Algebra, Combinatorics, and Geometry, Ann. New York Acad. Sci. 576 (1989), 500–535.
H. S. Wilf, Generatingfunctionology, AK Peters Ltd., 2006, pp. 136–139.
2000 Mathematics Subject Classification: Primary 05A10, 11B65, 05A15; Secondary 05A19, 11B73.
Keywords: Binomial coefficients, Stirling numbers, conjoint ranking tables, recurrence rela-tions, generating functions.
(Concerned with sequences A007318, A008275, A008277, and A068424. ) Received August 31 2009; revised version received November 6 2009. Published in Journal of Integer Sequences, November 16 2009.
Return to Journal of Integer Sequences home page.
13 |
190289 | https://mangolanguages.com/resources/learn/grammar/english/what-is-the-possessive-construction-in-english | English Articles
What is the possessive construction in English?
By: Erin Kuester Tue Sep 23 2025
EnglishNouns, Sentence Structure
English speakers use the possessive construction to say that one noun belongs to another noun (X has Y). We also use the possessive to express other close relationships between two nouns. There are two main ways to form a possessive construction:
possessor's
John's iPad
→ John has the iPad
noun
the branch of a tree
→ a tree has the branch
But you can also use (e.g. my, your, our...) or (e.g. mine, yours, ours).
In this post, we’ll explore what a possessive is and how we use it in English. Let’s go ahead and dive right in!
What is a possessive phrase?
Possessive phrases show that one noun belongs to another noun. Here are some of the situations where you can use a possessive phrase in English:
To show show that one noun owns another noun (possessions belong to an owner):
Maria’s purse
→ Maria owns the purse
+ the teacher’s house
→ the teacher owns the house
To show that one noun usually uses another noun (tools belong to the user):
the dog’s leash
→ the dog uses the leash
Often this is used when someone regularly uses an institution (institutions belong to the people who use them):
Andy’s school
→ Andy uses or goes to that school
To show that one noun is the creator of another noun (art belongs to an artist):
Jane Austen’s novel
→ Jane Austen wrote the novel
+ a goat's milk
→ the goat made the milk
To show that one noun is part of another noun (pieces belong to a whole):
the pages of the book
→ the book doesn’t own the pages, but the pages belong inside the book
+ the leaves of a tree
→ the tree doesn’t own the leaves, but the leaves are pieces of the tree
To describe the role that a noun plays in someone’s life (or the “life” of an object!):
Mary’s mother
→ describes someone who is a mother to Mary
+ the author of Crime and Punishment
→ describes someone who is an author to Crime and Punishment
To show that something happened at a particular time (an event belongs to a time):
yesterday’s meeting
→ the meeting happened yesterday
+ tomorrow’s weather
→ the weather that will happen tomorrow
To show that one noun is generally or always at a location (something belongs to a place):
the mayor of New York
→ the mayor for/in the city of New York
+ San Francisco’s bridges
→ the bridges that exist in San Francisco
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What are the types of possessive phrases in English?
There are four ways to form a possessive phrase in English: with the possessive -'s, with the preposition of, with possessive adjectives, and with possessive pronouns.
Possessive -'s:
John’s school
→ the school belongs to John
With of:
the sound of rain
→ the sound belongs to the rain
Possessive adjective:
my house
→ the house belongs to me
Possessive pronoun:
a friend of mine
→ the friend belongs to me
We use possessive -'s and possessives with ofwhen the possessor is a noun and we use the other two when the possessor is a pronoun. In this post, we’ll focus on the first two.
Check out our posts on English possessive adjectives and English possessive pronouns to learn more about these other two types.
Let’s look more closely now at the possessive -'s and the possessive with of!
Possessives with the possessive -’s
The possessive -’s (sometimes called the “Saxon genitive”) is the most common way to form a possessive phrase when the possessor is a noun for a person or an animal. Form a possessive phrase with the possessive -’s like this:
possessor noun
-'s
possessed noun
Marina's bicycle
a young boy's favorite yellow ball
the fish's food
It is important to remember that when you use the possessive -'s, the possessor noun comes first! This is not the same for all kinds of possessive phrases!
Tip
In English, we usually use an apostrophe (‘) to show that part of a word is missing (eg: is not → isn’t, he will → he’ll). A long time ago the possessive -'s was a short form of a longer ending, but in modern English there is no “long form” of the possessive -'s, it’s just -'s!
Important
When you form a possessive phrase with the possessive -'s, the possessed noun cannot have an article.
❌the dog's the ball
❌a driver's a car
This is because the possessor + -'s acts as a determiner and in English a noun can have only one determiner. Check out our mini-post to learn more about English determiners.
It sounds simple, but there are some tricky things about using the possessive -'s that even native English speakers sometimes get confused about. Let’s take a closer look at the possessive -'s.
How to add possessive -’s to singular nouns?
In writing, we add -'s to the end of a singular noun, no matter how it ends.
brother → brother's
Jess → Jess's
We pronounce singular words with the possessive -'s in the same way we would pronounce words with the plural -s. So remember, if possessive -'s comes after “hissy” sounds like s, z, ch, sh, or dge, add a syllable.
Jess [jess] → Jess's [jess·iz], ❌[jesss]
How to add possessive -’s to plural nouns?
If the plural possessor is an plural noun (meaning that it doesn’t end in plural -'s), we follow exactly the same rule that we follow for singular nouns. Just add the possessive -'s to the end of the plural noun:
children → children's
the children’s lunch
→ a lunch for more than one child
But unfortunately, things get tricky when the possessor is a regular plural noun!
In English, regular plural nouns, like brothers, swings, dogs, and parents, already have plural -s ending. So if you added both the plural -s and the possessive -'s, you’d have a lot of extra [s] sounds. Here is how we solve this problem to form possessives out of regular plural nouns.
In writing, we add the plural -s, but we only add the apostrophe (’) from the possessive -'s, not the whole ending.
dogs → dogs'
possessor nounplural -spossessive apostrophepossessed noun
the dogs' toys
→ the toys belong to a group of dogs
When speaking, we do not pronounce the -'s if it follows the plural -s. We pronounce these possessive forms of a regular plural noun exactly the same way that we pronounce the plural noun on its own:
possessor nounplural -spossessive apostrophepossessed noun
my parents' car
+ ✅[mai parents kar] (just like my parents)❌[mai parents·iz car]
Native English speakers think that this is confusing too, so you may hear some errors!
###### ImportantDid you notice this differenece?
When you add possessive -'s to a singular noun that ends with an -s (e.g. James), you do add a syllable and an -'s at the end.
James [jamez] → James’s house [jamez·iz]
But when you add the possessive -'s to a plural noun ending with the plural -s (e.g. the Millers), you do not pronounce the possessive -'s, and you only add the apostrophe:
the Millers [mil·erz] → the Millers’ house [mil·erz]
How to add possessive -’s to a list of nouns
If there are two nouns that are equal possessors together, we put the possessive -'s at the end of the list of owners:
possessor nounpossessivepossessed noun
I went to Beth and Alissa's house for dinner this week.
→ Beth and Alissa own the house together, it belongs to both of them.
If there are two nouns that own two separate things, we put the possessive -'s after each noun:
possessor nounpossessivepossessed noun
I went to Beth's and Alissa's houses for dinner this week.
→ Beth owns one house and Alissa owns a different house.
Which nouns can have a possessive -’s?
Technically, any noun can have the possessive -'s. However, native speakers usually only add the possessive -'s to nouns for living things.
Marina’s bicycle
→ Marina is a person.
The dog’s toy
→ The dog is an animal.
It is also okay to add possessive -'s to time expressions, words for natural events, locations, or some words for for people. These are non-living nouns that we think of as having some “life.”
this weekend’s forecast
→ A weekend is a period of time.
the storm’s strength
→ A storm is a natural event.
New York's mayor
→ New York is a location.
the government’s policies
→ The government is run by people.
Want to know a more common way to form a possessive when the possessor is not a human or an animal? Keep reading!
Possessives with ‘of’: How and when to use them in English?
We usually make possessive phrases with of when the possessor noun is not a person or an animal. Here is how to use of to make a possessive phrase:
possessed noun
of
possessor noun
the wall of the building
the rules of this organization
the grammar of English
Did you notice that in this type of possessive phrase, the first noun is the possessed noun comes first?
This is because when you form a possessive with of, you are forming a prepositional phrase (of + possessor noun) that describes the possessed noun. Just like when we use other prepositional phrases to describe nouns (e.g. the dog in the window, the pencil on the table), a prepositional phrase comes after the noun it describes, not before.
Important
In a possessive phrase with of, both the possessed noun and the possessor noun can have articles:
✅the window of a car
✅the sound of the rain
When to use ‘of’ to form a possessive phrase?
We most often use possessors with of to show that something is part of a whole.
the engine of the airplane
→ The engine is a part of the airplane.
the wheels of the bus
→ The wheels are a part of the bus.
the roofs of the houses
→ The roofs are part of the houses.
But we do use of to express other kinds of possession. For example:
the colors of my favorite Basketball team
→ the colors belong to the team
the story of my life
→ the story is about my life – we often use the story of + noun to name a story
And of is even used before people to sound a bit more formal or impressive. Often these nouns sound like titles:
the mother of the bride
→ a title for the bride’s mother during a wedding
the crown of Queen Elizabeth
→ this sounds like a museum sign. It is more impressive than Queen Elizabeth’s Crown
Important
Remember that not every noun that is followed by of is a possessive phrase! Some of them use of to do something else. Here are some examples:
the engine of an airplane
→ The engine is part of the airplane.
← possessive with of
a piece of paper
→ The piece is made out of paper.
← of + mass noun
a group of cats
→ The group is made up of cats.
← collective noun + of
You can only rephrase a possessive with of using possessive -'s!
✅an airplane’s engine
❌a paper’s piece
❌a cats' group
How to layer possessive phrases?
In English, we can layer possessive phrases. What does that mean? Let’s look at an example:
If Jane has a mother, and her mother has a father, you can describe him like this:
Jane’s mother’s father
Do you see how that works? First create one possessive phrase, Jane’s mother, then use the whole possessive phrase as the possessor of another possessive phrase, Jane’s mother’s father.
In English you can layer possessive phrases of all kinds:
the signature of the mayor of Cincinnati
→ The signature belongs to the mayor. The mayor belongs to Cincinnati.
Kaitlyn’s daughter’s birthday
→ The birthday belongs to the daughter. The daughter belongs to Kaitlyn.
their dog’s toy
→ The toy belongs to the dog. The dog belongs to them.
You can even make very long layered possessive phrases, like this:
the window of Leo’s best friend’s older brother’s car
Can you figure out what that means?
You are describing a window. That window belongs to a car. The car belongs to an older brother. The older brother belongs to a best friend. That best friend belongs to Leo! Just look at all that possession!
In this post, we’ve taken a look at possessives in English. We saw that…
A possessive construction in English shows that one thing belongs to another thing.
There are two main possessive constructions in English: possessive -'sand x of y.
To make a possessive with -'s, just add -'s to the owner and then say the thing that is owned: owner’s owned. If the owner has a plural -s, just add -’.
If the possessor is not a person or an animal, use x Phrase y: owned ofpossessor.
Possessive adjectives and possessive pronouns also show that one thing belongs to another thing.
If you’re feeling more confident about forming the possessive in English, take a look at our English possessives practice activities!
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190290 | https://www.vanityfair.com/news/2020/04/cnns-raw-unfolding-ronald-reagan-coverage-heralded-the-nonstop-news-cycle?srsltid=AfmBOoqX0o7RC24r6sRH2tPoMVnHmduB2tMeCL6P_Fe_5it4vEQG0A8R | “Shots Fired. Hilton Hotel”: How CNN’s Raw, Unfolding Reagan Coverage Heralded the Nonstop News Cycle | Vanity Fair
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TV News
“Shots Fired. Hilton Hotel”: How CNN’s Raw, Unfolding Reagan Coverage Heralded the Nonstop News Cycle
Ted Turner’s upstart cable network beat the Big Three in reporting on the 1981 assassination attempt, though it—along with its broadcast rivals—made a major mistake amid the studio chaos, an early sign of the perils of breaking news on TV.
By Lisa Napoli
April 27, 2020
Chaos surrounds shooting victims immediately after the assassination attempt on President Reagan in 1981 outside the Hilton Hotel in Washington, D.C.By Dirck Halstead/Liaison/Getty Images.
Save this story
Save this story
On a rainy spring Monday in March, Cissy Baker wound up sending her White House crew to a snoozer of a time-filler: the ballroom of the sprawling Washington Hilton on Connecticut Avenue, where President Ronald Reagan was about to address the national Conference of the Building and Construction Trades of the mighty trade labor union, AFL-CIO.
As was the custom in the carefully orchestrated universe of Washington politics, the text of the speech had been released to the press corps in advance. Most television viewers were unaccustomed to seeing routine events of the day in their entirety, but this was the kind of typical governmental affair that helped CNN burn through many an hour. There was always the chance that at some point the affable president might “commit news,” as the broadcasters cheekily referred to any unexpected development. Maybe there’d be boos from the audience; a bit of mileage could be had from that. As far as Atlanta was concerned, a speech by the president was far preferable to a five-minute, thumb-sucking analysis from Daniel Schorr. No wonder his nickname at CBS had been “Jukebox.”
The camera lingered on the president as he shook hands and beamed his movie-star grin. Anchor Bernie Shaw smoothly deployed his inside-the-Beltway knowledge in summarizing the remarks. Being able to offer this sort of live, postgame analysis was precisely what had lured him to this job. Who cared if there was no audience?
“President Reagan, in a speech that lasted about 19 minutes, drew applause four times from this group,” Shaw observed, with such authority that a viewer might actually believe there was a significance to the number of rounds of applause.
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His midday assignment complete, he tossed the baton back to Atlanta. And during the next commercial break, Baker’s wish for a more interesting day suddenly materialized.
The words rang out from the police scanner at 2:27 p.m. “Shots fired” followed by “Hilton Hotel.”
In that instant, Baker frantically connected the dots: The Hilton? That’s where the president was, with one of her crews wrapping up inside. Her mind raced strategically over the map of the city. The chess game of routing personnel, particularly at a time of crisis, was a crucial part of running an assignment desk. Her back-of-the-hand knowledge of the nation’s capital was precisely the reason she’d been offered this job. It didn’t hurt that she ranked as a Washington insider. Her father happened to be the Senate majority leader, Howard Baker.
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The next words that bleated out of the scanner offered a disturbing new clue: “Rainbow to GW.” Baker knew the code. “GW” meant the George Washington Hospital, and “Rainbow,” the first lady. If Nancy Reagan was heading for the hospital, that must be because the president was headed there too. But why?
Hearing the fracas among his anxious colleagues, Shaw demanded to know what was going on. A desk assistant said sarcastically, “I think they’re shooting at your president.”
“Don’t joke,” Shaw scolded.
For a veteran newsman, he was curiously unjaded—patriotic, and respectful of authority, even. (That didn’t equal passive. As a young member of the Marine Corps in Hawaii, he’d tracked down Walter Cronkite when he’d learned the anchorman was coming to town, urgently hoping for guidance on how to get into the business.)
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The assistant responded to Shaw: “I’m not joking.”
A split second later, Atlanta dumped out of a taped report on education in China to anchor Bob Cain on the set.
“We interrupt…there’s been a late development,” he said urgently. “Shots reported fired outside the hotel where President Reagan spoke a short while ago. Here’s Bernard Shaw in our Washington bureau.”
Shaw knew little more than what Cain had just said, but he began to speak, masking the shivers and chills he felt. The mere suggestion of an assassination attempt could plunge the world’s security and economy into a tailspin. His job—his responsibility—was to inform the public in a measured, sober, deliberate tone. It was crucial not to fuel hysteria.
“Bob, as you can understand, details are very sketchy. We don’t know precisely what happened, nor…Pardon me.”
His voice was hollow. In his rush to get into position behind the anchor desk, Shaw forgot to clip on his microphone. He calmly reached over to grab it and fastened it in place.
“Okay, my apology,” he said, looking down to consult the fragments of information being rushed before him by Sandy Kenyon, his producer. The young man had been so eager to work at the network that he’d bought a one-way ticket to D.C. from New York and talked his way into a job. At this grave moment, he sat at Shaw’s feet, out of the camera’s view, pecking away on a state-of-the-art IBM Selectric typewriter and synthesizing details as his colleagues collected them.
“Details are very sketchy at this moment,” Shaw repeated. “We don’t know precisely what happened. We don’t know the sequence…First of all, the president is safe. We are told that shots were fired at his party as he left the hotel…We can report that shots were fired as President Reagan left the Washington Hilton hotel following that address we carried live here on CNN. The president did not appear to be hurt, according to United Press International.” He continued to read the wire copy Kenyon had handed him.
For this vague report, the upstart CNN could now claim a triumph: It had beaten the other networks in announcing the shooting by four whole minutes. To its tiny audience, this didn’t matter. To their broadcast competitors, it was proof CNN meant business.
The only pandemonium greater than the scene of a shooting is the unfolding madness of a newsroom trying to sort out the aftermath. Before now, the mechanics of both were, except for Hollywood depictions and the assassination of President John F. Kennedy, shrouded from public view.
A few minutes later, over on ABC, Shaw’s former colleague, newsman Frank Reynolds, took a deep breath as he summarized the same bits of information Shaw had just delivered. He, however, had a visual aid: videotape shot by the White House press pool.
CNN had been angling to join the pool but had been denied admission for several reasons—because it employed non-union labor; because how could anyone trust this upstart, from second-string cable; because it had never been done any other way before.
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“This is the first time any of us has seen this tape,” Reynolds told viewers as the dramatic video, just rushed into the studio, began to roll. The eight minutes of footage, he cautioned, were not edited and thus not as neat as it could be. That it was raw made it all the more compelling. Television had changed radically since just six years earlier, when two women in the span of weeks attempted to shoot then President Gerald Ford. TV crews then, not yet equipped to use videotape in the field, captured those incidents on film. Then, the networks had interrupted programming to inform viewers what had occurred, but, with little else to report, waited till their regular nightly newscasts to add to the story. Now, with the emerging influence of CNN, broadcasters could not afford to fact-check and wait. Reynolds narrated the footage as he watched it for the first time himself.
This is not live, Reynolds remembered to mention as he improvised, but it is fresh tape of the shooting, which had occurred just 15 minutes earlier. It appeared, he observed, that Press Secretary James Brady had been struck in the head. This prompted a phone call by the anchor, on-set, to a reporter. Where is the president? The president had not yet returned to the White House. He was on the way to the hospital. Wait, the president is on the way to the hospital? Is he on his way, or is he being taken there?
After playing the tape on the air again, Reynolds signed off for the moment. There were so many unanswered questions.
“There really is nothing more that we can tell you at this point,” the ABC newsman told his audience, recapping what he knew so far. “So, that’s it. As soon as we get any more information on this we will come back on the air as quickly as we can.”
Back to the soap opera The Edge of Night.
At the first all-news channel, Shaw didn’t have the luxury to break away and wait for the facts to click into place. This was exactly the kind of developing story made for CNN, the kind CNN cofounder Reese Schonfeld had been waiting for—a golden opportunity to “capture the surfers.”
Besides, CNN had nothing to break away to. A producer tried to fight the edict to stay on the story. “We don’t have any information,” he argued. “We don’t know anything!”
“It doesn’t make any difference,” came the order. “Get Bernie back in the chair, and get ready to go.”
Nearly 20 years earlier, during the stone ages of television news, Schonfeld had had to watch as the networks grabbed the glory during the Kennedy assassination—with nary a frame of film of the actual shooting.
Here, now, was his chance to play alongside the networks, in their league—thanks to the invention of videotape, thanks to portable cameras and satellites, thanks to this crazed lunatic of a gunman whose name no one knew yet. Thanks, most of all, to Ted Turner.
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But Schonfeld would only be sure he’d arrived completely when he gained admission to that press pool. To him, the clubby collusion of the networks—how the existence of their troika silenced other competition—boiled his blood and embodied all that was wrong with television.
By bizarre coincidence, this was the day he planned to fire his greatest salvo in this fight to force his way into the inner sanctum. CNN had prepared a lawsuit against the White House and the networks, charging them with antitrust and violation of the First Amendment for blocking CNN from the pool. Now, because of these bullets, the suit would have to wait. But to prove his point, he rolled tape on that pooled video and ran a copy of it on his airwaves, anyway. “A pool for one was a pool for all.” Let the networks sue him if they weren’t happy.
Now it was Shaw’s turn to narrate that jarring video for CNN’s tiny audience, his voice competing with a cacophony of sounds—the clanking electric typewriter, bleating television monitors, agitated voices of his behind-the-scenes colleagues, working the phones, hawking the wires, in search of the latest. Details, meanwhile, dribbled in, some tiny, some large, some ultimately incorrect, all absolutely raw. That’s all we have. That’s all we know. We still don’t have that for you. Is that correct? I’m not sure what we’re getting right. Things are in a state of confusion. Chaos around the hotel. The president is okay. Press Secretary James Brady is on the ground and may not be. A Secret Service agent and a cop have been shot too.
Soon it would be discovered that, despite what they’d been confidently reporting, the president was not okay. This stark fact hastened the networks back on the air, though they were still scrambling for details. Over the next hours, confusion reigned at all four newsrooms on display for viewers, echo chambers all. The unfolding drama of the news was as riveting as the shooting itself. With a blank slate of airtime to fill, video of the heinous act replayed again and again, in real time, in slow motion, examined, dissected, frame by frame, as reporters stitched together their facts and the confusion morphed into a complete story.
At CNN, Daniel Schorr joined Shaw on set and bantered to fill the time, sharing that he just returned from medical leave in the same hospital where the president of the United States was now being treated by the same surgeon who’d recently treated him.
Senator Howard Baker, having announced to Congress the turn of events, called his daughter Cissy, the CNN assignment editor, to deliver a grim scoop: He’d been told that Press Secretary James Brady was dead. She, in turn, passed along the information to Sandy Kenyon, who quickly crafted a script for Shaw—who refused to read it. The other networks began to report the news: James Brady had succumbed to gunshot wounds. Dan Rather even called for a moment of silence in his honor.
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Around the CNN newsroom, watching competition report the news, the anxious staff confronted Kenyon. Why won’t Bernie say it? We had it first.
Since it wasn’t clear how Baker had received his information, and since it seemed that he hadn’t witnessed Brady’s demise himself, Shaw deemed the senator’s information unreliable. Despite his instincts, the anchor capitulated: CNN had learned from a “top-level congressional source,” he told viewers, that James Brady had died. Quickly, he hedged his bets: “We are not sure, we have no official confirmation. This is just one report. Brady, in fact, may be alive.”
Not long after came word that, in fact, he was—and the three mighty networks were in the awkward position of having to retract the story. But the mistake was less of a reminder about rushing to be first than it was one of the first post-CNN examples of how television journalism would be redefined. For “news” no longer meant reporting an event in its aftermath. Forevermore, news would mean following an endless shower of unfolding details, right before your very eyes. News, in other words, had become sports.
On this day, side by side, as newbie CNN resembled the Big Three networks, and the networks looked and sounded like CNN, media critics decried that the shooting of the president was evidence that television news had collectively sunk to a new low.
Who cared, syndicated columnist Nicholas von Hoffman wrote, if a reporter had been in the same hospital as the president and attended by the same doctor? Rumor, gossip, hearsay, and tongue-wagging: “While a worried nation sought information,” he lamented of the coverage, “it got incompetent, if ardent hysterics.”
From Up All Night: Ted Turner, CNN, and the Birth of 24-Hour News by Lisa Napoli, published by Abrams Press.
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190291 | https://www.gamelogic.co.za/downloads/HexMath2.pdf | Hex Grid Geometry for Game Developers (1.3) Herman Tulleken June 7, 2018 Contents Introduction 2 Credit and Further Reading 2 Changes v.1.3 2 Changes v.1.2 3 Hex Vectors 3 Coordinate System 3 Norms 6 Transformations 8 Trigonometry 9 Products 9 Shapes 10 Lines 10 Halfplanes 12 Polygons 13 Circles 14 Axis-aligned Triangles 14 Axis-aligned Quadrangles, Pentagons, Hexagons and Hexagrams 15 Grid Point Division 18 Definitions 18 Uniform Spatial Partitioning 19 Wrapped Grids 21 Recursive Spatial Partitioning 23 The Gosper Curve 24 Grid Colorings 26 Triangular grids 27 hex grid geometry for game developers (1.3) 2 Introduction This document is an introduction to the math for working on a hex grid, and should be useful to you if you want to implement geometric algorithms on such a grid.
I assume you know how vectors work, and how we use them in rectangular systems to do geometry.
This article is divided into three main sections. The first deals with vectors on a hex grid, and basic operations defined on them. It uses concepts that should already be familiar to you from “square” math.
The second part deals with shapes; their equations, and properties of special shapes. Like the the previous section, the ideas should be familiar, although the formulas are different.
The third deals with algorithms and techniques based grid point division. This tool is not commonly used for square grids, but the ideas follow from integer division for 1D math.
Credit and Further Reading The hex-rounding function is from Amit Patel’s page on hexagons: Hexagonal Grids. This page has many interactive examples, and also deals with some other hex coordinate systems. It is also from Patel’s Implementation of Hex Grids that I first got the idea to use floats instead of only integers for coordinates, which drastically simplifies a lot of the math, and makes the system a lot closer to “ordinary” rectangular coordinate math. Patel wrote quite a bit on various implementation issues, and his work is worth a read: • Hex grids: simplifying the variants • Hex grids: procedurally generating code • Hex grids: choosing data structures and generating algorithms • Hex grids: code generator • Hex grids: more on coordinate names • Hex grids: finishing up the code generator project Changes v.1.3 Fixed a typo1 on page 18: “where p, q and r are integers” now reads 1 Thanks to Daniel Toffetti who alerted me to this.
“where d, q and r are integers”.
hex grid geometry for game developers (1.3) 3 Changes v.1.2 This document was first published in June 2015. Version 1.2 is only a minor revision: it fixes a few typos2, and the layout is slightly 2 Some of which were pointed out by “user69513” on Math Stackexchange.
different. Because I lost the original source, I had to recreate it, and it is possible a few errors crept in in the process. If you spot any, please let me know3.
3 herman@gamelogic.co.za Hex Vectors Coordinate System By choosing a simple coordinate system, we can get many of the conveniences of square grids to apply to hex grids, so that we can deduce geometric information by performing calculations on coordi-nates. For a rectangular coordinate system: • we put down two identical real lines on the plane at a 90 degrees angle from each other, • we call the one the x-axis and the other y-axis, • and for any point, we can draw parallel lines towards the axes, • and jot down the two numbers where these parallel lines cross the axes, • and call them the x and y coordinates of the point.
There is a bit of mathematical machinery before we get to a vector (which is really an offset from one point to another), but since we won’t distinguish between points and vectors, I will skip over this.
Once we have vectors, we can define a few operations that allow us to do geometric computations using their coordinates. These include vector addition, multiplication with a real number, multi-plication with matrices, and a handful of other multiplication-like operators (such as the dot product).
To get to vectors for a hex system, we follow the same procedure as above, with a minor difference in the first step: • we put down two identical real lines on the plane at a 60 degree angle from each other.
This difference is minor, but it gives this system the following property: integer coordinates fall on the faces of a hex grid, and we call these points grid points.
As with the square system, we can define some operations, and for the most part things work exactly the same: hex grid geometry for game developers (1.3) 4 Figure 1: A hex Grid • We can add and subtract vectors to denote displacements in this hex world.
• We can multiply vectors with scalars to denote scaling.
• We can write equations for lines, circles, and other shapes.
• We can use matrices for transformations such as scaling, reflection, rotation, shearing.
But there some differences: • Distances are given by a different formula.
• Rotation and reflection matrices have different formulas.
• Angles between vectors are calculated using different formu-las.
• We have slightly different identities that relate coordinates with magnitude and rotation with respect to the x-axis.
• We have a different definition of the dot product.
We will from time to time find it useful to use an auxiliary coordi-nate, which can be calculated from the other two: z = −x −y.
(1) This is mostly to make formulas more concise and symmetrical.
This also makes it easier to see certain patterns when comparing formulas with the square system. The three lines x = 0, y = 0, and z = 0 are the major axes; the three lines x = y, y = z and z = x are the minor axes.
We sometimes use special notation for the unit vectors: hex grid geometry for game developers (1.3) 5 Figure 2: Hex coordinate system. The major axes are shown in red (x), green (y) and blue (z). The minor axes are shown in grey. If we draw lines parallel to the x- and y-axes, they intersect these axes at the points marked 3 and 2, so the point has coordinates (3, 2).
ex = (1, 0) (2) ey = (0, 1) (3) ez = (−1, −1).
(4) Figure 3: The major axes shown on a hex grid.
Grid points and hex rounding.
We now introduce a special operation, which we call “hex rounding”; it gives for any point, the closest grid point. This operation divides the plane into hexagons–except for at the edges of these hexagons, the operation gives us the coordinate at the center of the hexagon the point “belongs” to.
If v is a vector, then the closest grid point is denoted by ⟨v⟩. It is this operator that makes hex-coordinates useful; without it, there is little benefit of using hex-coordinates over rectangular coordinates.
hex grid geometry for game developers (1.3) 6 Let ⟨a⟩be the integer closest to a, and let dx = |x −⟨x⟩|, dy = |y −⟨y⟩|, and dz = |z −⟨z⟩|. Then we can calculate ⟨v⟩as ⟨v⟩= (−⟨y⟩−⟨z⟩, ⟨y⟩) if dx = max dx, dy, dz (⟨x⟩, −⟨x⟩−⟨z⟩) if dy = max dx, dy, dz (⟨x⟩, ⟨y⟩) if dz = max dx, dy, dz .
(5) Grid points have a few special properties: • The sum (and difference) between two grid vectors is another grid vector.
• Scaling a grid-vector by an integer is always a grid vector.
• Any grid point can be written as a linear combination of (1, 0) and (0, 1), with integer coefficients.
• Rotating a grid vector by 60 degrees (and any integer multiple thereof) gives another grid vector.
• Reflecting a grid vector about a line which contains two neigh-boring grid points gives another grid vector.
Vector addition and scalar multiplication.
For hex grids, vector addition and scalar multiplication is the same as for square systems. As an example of these vectors in action, let’s find the coordinates of the center of an edge, and of a vertex.
The midpoint of an edge between two hexagons is given by (u + v)/2, where u and v are the coordinates of the faces of the hexagons.
These points are always in the form (m/2, n/2) where m and n are integers.
The vertex between three neighboring hexagons is given by (u + v + w)/3. These are always in the form (m/3, n/3) where m and n are integers.
Norms Norms are functions that assign real numbers to vectors that coincide with our notion of length. They are useful not only for determining distances between points, but also to write equations for common shapes.
The Euclidean length is the norm we know from square geometry, and for hex coordinates is given by: |v| = r x2 + y2 + z2 2 .
(6) hex grid geometry for game developers (1.3) 7 As with square systems, we can find the distance between two points by taking the magnitude of their difference: d(u, v) = |v −u|.
(7) In calculations on a hex grid, the hex norm is occasionally useful4: 4 It is a good exercise to prove that the two quantities in the equation are in fact equal.
|v|h = |x| + |y| + |z| 2 = max {|x|, |y|, |z|} .
(8) We can define a related distance metric: dh(u, v) = |v −u|h.
(9) For grid points, this gives the minimum number of hexes we have to enter moving from one point to the next. It is analogous to two different norms in the square world: the taxicab norm and the chess norm. The taxicab norm (the sum of the absolute values of the coordinates) gives the minimum number of squares we have to enter to move from one square cell to another if we are only allowed orthogonal moves. The chess distance (the maximum of te absolute values of the coordinates) is the minimum number of squares we have to enter to move from one cell to another if we are also allowed diagonal movements.
We can either view the hex distance as a taxicab norm, where the roads are at 60 degree angles instead of 90 and there are three directions instead of two; or as the chess norm where there are not any diagonals because of the structure of the hex grid.
The following are not norms; we will call them pseudonorms.
The down-triangle pseudo-norm and up-triangle pseudo-norm are defined by |v|▽= max {x, y, z} (10) |v|△= max {−x, −y, −z} .
(11) These pseudo-norms are used to write compact equations for equilateral triangles on a hex grid. We will get back to this later, but to give you an idea, |v|▽≤r is an equilateral triangle pointing down, with sides of length r.
The triangle psuedo-norms are related by the identity | −v|▽= |v|△.
(12) These can be combined to define the star norm: |v|s = min |v|▽, |v|△ .
(13) The star norm is used to write compact equations for regular hexagrams (or stars).
hex grid geometry for game developers (1.3) 8 The triangle pseudonyms are also related to the hex norm: |v|h = max |v|▽, |v|△ .
(14) Transformations In our normal geometry on a square coordinate system, we can represent a large class of useful transformations (of points and sets of points) as multiplication with matrices. The same trick works in a hex coordinate system, although the matrices we use look slightly different.
Rotation.
This is the general formula for the rotation matrix: R(θ) = sin(θ + 2π 3 ) sin θ −sin θ sin(θ + π 3 ) !
.
(15) To rotate a vector, we simply multiply it with the matrix above.
And for multiples of 60 degrees, the rotation matrix only has in-teger components, and so rotating a grid point by multiples of 60 degrees will give another grid point. Notice the occurrences of the sine shifted by various multiples of 60 degrees; we will see this of-ten. In square systems, we have occurrences of the sine shifted by 90 degrees (remember that cosine is just a convenient abbreviation for sin(x + π 2 ).
Scaling.
Uniform scaling works the same as in square systems S(s) = s 0 0 s !
.
(16) The following matrix “scales” the coordinates independently, and is really a shear: S(s1, s2) = s1 0 0 s2 !
.
(17) Combining this shearing with rotations, we can get more general shears. For this purpose, it is more useful to only scale one compo-nent. We can get all possible shears along one shearing-axis by using a matrix of the form: R(−θ)S(s, 1)R(θ).
(18) We can get all two-axes shears by combining two of these: R(−θ1)S(s1, 1)R(θ1)R(−θ2)S(s2, 1)R(θ2).
(19) hex grid geometry for game developers (1.3) 9 Reflection about x-axis.
Reflection about the x-axis can be done using the following transformation matrix.
Rx = 1 1 0 −1 !
.
(20) We can get reflections about other lines through the origin using rotations R(−θ)RxR(θ).
(21) Trigonometry When using vectors for doing geometry, it is not very common to use trigonometry, but it is useful to know how hex coordinates are related to lengths and angles of the vectors. This is, for example, helpful in finding equations for certain shapes. Here are a few basic identities x = 2r √ 3 sin θ + 2π 3 (22a) y = 2r √ 3 sin θ (22b) z = 2r √ 3 sin θ −2π 3 (22c) And x2 + y2 + z2 = 2r2 (23) cot θ = x −z √ 3y (24) sin θ + 2π 3 + sin θ + sin θ −2π 3 = 0 (25) sin2 θ + 2π 3 + sin2 θ + sin2 θ −2π 3 = 3 2 (26) Products The reason why we don’t use trigonometry bso often is that we can use special products (the dot and perp dot product) to by-step trigonometric calculations.
For a hex system, we define the dot product as: u · v = uxvx + uyvy + uzvz.
(27) The cosine of the angle between two vectors is given by: cos α = u · v √ 3 |u| |v| .
(28) hex grid geometry for game developers (1.3) 10 We can also define a cross-like product. Although it has the same definition as the perp dot product, it really involves a 60 degree rotation on the vector rather than a 90 degree one, so it would be a bit of a misnomer.
u × v = uxvy −uyvx.
(29) If it bothers you that z-coordinates are not involved in this defini-tion, you would be glad to know that the following holds: u × v = uxvy −uyvx (30a) = uyvz −uzvy (30b) = uzvx −uxvz (30c) (30d) We can get the (signed) angle from u to v using this: sin α = 2u × v √ 3 |u| |v| .
(31) Shapes We will be mostly interested in grid shapes, shapes that contain only grid points. We will also talk about grid lines, grid triangles, grid circles, and so on to mean in each case these shapes only contain grid points. Non-grid shapes are still useful in interim calculations, and we will use the terms real shapes, real lines, and so on (real as in real number, not real as in not fake) Lines Axis-aligned grid lines are described by one of these equations: x = k (32a) y = k (32b) z = k, (32c) with x, y and k integers.
We can also write equations in parametric form: v = p + nex (33a) v = p + ney (33b) v = p + nez.
(33c) hex grid geometry for game developers (1.3) 11 Figure 4: The line z = −2.
where p is any point in the line and n goes through all integers.
It is trivial to determine whether these lines are parallel (they must have the same equation form), and if not, where they intersect. For example, the lines y = 3 and x = 5 intersect in the point (5, 3). The lines y = 3 and z = −x −y = −5 intersect in the point (2, 3).
What about “lines” that look the one in Figure 5?
Figure 5: An impure line.
For these types of lines, it is necessary to recognize it as a grid approximation of a real line. Now there are many lines that have the same approximation, and we can choose any one to work with. The above grid line can be described by this equation: v = ⟨(1/4, 0) + n(−1/2, 1)⟩.
(34) With these type of equations we usually try to avoid choices that give points on the edges of hexagons, so that the behavior of our rounding function on edges does not affect the line. If, in the hex grid geometry for game developers (1.3) 12 example above, we chose v = ⟨n(−1/2, 1)⟩, we could get either of these lines, depending on how the rounding function works.
More generally, then, we have this general description for grid lines: v = ⟨p + ntq⟩ (35) where p and q are not necessarily grid points, and n goes through all integers, and t is a number with which we control the density of lines. If it is too big, we get holes in the lines and we call them sparse.
If it is too small, we get repeated points, and we call the lines dense.
There is a unique value for t that ensures that the grid line has no holes and no repeated points, and we call such lines pure.
Unlike real lines, non-parallel grid lines do not always intersect in a single point, and parallel lines that don’t coincide can intersect.
Finding all the points of intersection can be somewhat tricky. For non-parallel lines one point of intersection can be found easily: the grid point closest to the intersection of the associated real lines, and we know that other intersection points must lie close to this.
• Pure lines that are not parallel always intersect in at least one point. All points of intersection are distinct.
• Sparse lines may not always intersect.
• Dense lines that are not parallel always intersect in at least one point. Points of intersection may not be distinct.
• Two lines are parallel if their q-values are the same.
• Two parallel grid lines are the same if ⟨p⟩= ⟨p′⟩and ⟨p + tq⟩= ⟨p′ + tq⟩.
Halfplanes Halfplanes are shapes with all points to the side of the borderline and including the borderline. They are described with inequalities with this general form (for real half-planes): (v −p) × q ≤0, (36) where p is a point on the borderline of the half-plane, and q is in the direction of the borderline (so that the defined halfplane lies on the right of q).
For real halfplanes, the edges are the line (v −p) × q = 0. For grid lines, it is often more useful to exclude the border, and use (v −p) × q < 0, (37) instead so that potential rounding errors are avoided.
hex grid geometry for game developers (1.3) 13 Figure 6: The grid halfplane 2x + y − 3 < 0.
Polygons Figure 7: The polygon P((−3, 0), (3, −3), (3, 0), (0, 3); v) ≤0.
A convex polygon T with n vertices ti (in anti-clockwise order) is given by the intersection of halfplanes (v −ti)(ti+1 −ti) ≤0.
We define the polygon function P(T; v) = P(t0, · · · , tn−1; v) = max i=0..n−1 {(v −ti) × (ti+1 −ti)} , (38) where t0 = tn. A polygon is then defined by P(T; v)0.
(39) Arbitrary polygons can be represented as the union of convex polygons. The union of polygons Ti is given by min i {P(Ti; v)} ≤0.
(40) The area of any parallelogram is given by the cross product of the hex grid geometry for game developers (1.3) 14 vectors from one vertex to the two adjacent ones: A = √ 3 2 |u × v| .
(41) The area of a triangle between vectors u and v is given by A = √ 3 4 |u × v| .
(42) The area of a polygon with vertices (in clockwise order) t0, · · · , tn−1 is given by A = √ 3 4 n−2 ∑ i=1 (ti −t0) × (ti+1 −t0) .
(43) The formula holds even if the polygon is not convex.
Circles Figure 8: The grid circle |v −p| < 9 4.
The equation for a circle of radius r at a point p is given by: |v −p| ≤ √ 2r.
(44) The parametrical form is given by v = √ 2r 3 sin t + 2π 3 , sin t + p.
(45) Axis-aligned Triangles The triangle psuedo-norms we defined earlier can be used to write equations for axis-aligned triangles. For a down-triangle centered at p with side length 3r, we have |v −p|▽≤r (46a) hex grid geometry for game developers (1.3) 15 Figure 9: The triangles |v −(1, 1)|▽≤1 and |v + (1, 1)|Delta ≤1.
and for an up-triangle, we have |v −p|△≤r (46b) Up triangles are bounded by the lines x = px + r, y = py + r, and z = pz + r, and down triangles by the lines x = px −r, y = py −r, and z = pz −r. These triangles are always equilateral, and the length of their sides is given by 3r. By taking the equations of the lines of the triangle sides two at a time, we can easily calculate the three vertices of the triangle. For up triangles: t1 = (px + r, py + r) (47a) t2 = (px + r, py −2r) (47b) t3 = (px −2r, py + r) (47c) and for down triangles: t1 = (px −r, py −r) (48a) t2 = (px −r, py + 2r) (48b) t3 = (px + 2r, py −r).
(48c) The center p can be found as the average of the three vertices: p = t1 + t2 + t3 3 .
(49) The radius is a third of any of the three sides: r = t2 −t1 3 = t3 −t2 3 = t1 −t3 3 .
(50) Axis-aligned Quadrangles, Pentagons, Hexagons and Hexagrams In hex space, convex polygons can have up to six sides.
hex grid geometry for game developers (1.3) 16 All quadrangles in hex-space are either parallelograms or isosce-les trapezoids. Pentagons always have two pairs of sides parallel.
Hexagons have three pairs of sides parallel.
Figure 10: Parallelogram Figure 11: Trapezoid Axes-aligned rhombi have one of the following equations: max |x −px|, |y −py| ≤r (51a) max {|x −px|, |z −pz|} ≤r (51b) max |y −py|, |z −pz| ≤r.
(51c) Regular hexagons are given by the equation |v −p|h ≤r.
(52) Regular hexagrams (star shapes) are given by the following equa-tion: |v −p|s ≤r.
(53) hex grid geometry for game developers (1.3) 17 Figure 12: Hexagon Figure 13: Regular Hexagram hex grid geometry for game developers (1.3) 18 Grid Point Division Definitions This section extends the ideas of division in 1D to 2D For integers, recall the division algorithm: each integer n can be written in the form n = qd + r, where d, q and r are integers, and 0 ≤r < d. It will be useful for us to extend this to real numbers. The above statement hold if we let n, d and r be any real numbers.
For real numbers, we define v div d = jv d k (54a) v mod d = v −(v div d)d.
(54b) We then have, for the division algorithm n = qd + r (55) q and r given by q = n div d (56a) r = n mod d.
(56b) For real numbers, let’s look at what these operations really mean, so that the 2D version is more intuitive. Let us partition the real line into half-open segments of length d, with each segment given by Pq = (qd, (q + 1)d) with q some integer.
Now for any real number n, if n ∈Pq, then q is given by n div d.
So this operation tells us in which partition the number lies. The number r = n mod d tells us how “far” into the partition the number lies. When we view the partitions as being the same, then r is a useful identifier that we can use to determine if two numbers in different partitions are the same.
For example, let d = 2π, and let’s view the real number as angles.
Then the partitions are really the same (as each partitions has the same angles as any other partition). In this case, r = n mod 2π gives us the canonical angle, and we can use it to determine whether two angles are really the same. For example, −π mod 2π = π, and 3π mod 2π = π, so we know that −π and 3π are really the same angle, and moreover, that they are both the same as the angle π.
When it comes to 2D, we want to partition the plane into parallel-ograms, and we want the mod and div operators to tell us, for any point in the plane, which partition that point is in, and its position within that partition. In the following sections we will see how this is useful for a wide variety of algorithms.
hex grid geometry for game developers (1.3) 19 Let D = g h !
be a 2 × 2 matrix, with g and h two sides of a parallelogram. Then we make the following definitions: v div D = j vD−1k (57a) v mod D = v −(v div D)D.
(57b) where the floor of a matrix can be obtained by flooring all the compo-nents of the matrix. We can write every vector v in the form v = mg + nh + r = qD + r, (58) where q = (m, n) is a grid point, and r is a vector contained in the parallelogram bounded by g and h, and given by q = v div D (59a) and r = v mod D.
(59b) As in the 1D case, the div operator tells us in which partition a point lies, and the mod operator tells us where in that parallelogram that points lies. Let us formalise this a bit. Let P0,0 be the paralel-logram bounded by vectors g and h. Then if the plane is partitioned into paralellograms Pi,j = {(x, y) | (x, y) −ig −jh ∈P0,0}, then the div operator tells us in which parallelogram any points lies, and the mod operator tells us where inside the parallelogram the point lies.
The following identities are useful for computing these operations: v div D = (v × h, g × v) div (g × h) (60a) v mod D = (v × h, g × v) mod (g × h) (60b) Uniform Spatial Partitioning We saw how we can easily partition the plane into parallelograms.
But what about other shapes, such as in Figure ]reffig:HexagonPartitions?
Suppose P is a parallelogram with matrix D, and R is any set of points, and suppose that the function fD : R →P defined by fD(r) = r mod D is a bijection.
Then we can partition space now into regions with the same shape as R.
For a point v, we calculate which point in R has the same remain-der when divided by D as v. This is the location within the region of the point, and is given by f −1 D (v mod D). We subtract this from v; hex grid geometry for game developers (1.3) 20 Figure 14: Partition with g = (4, −1) and h = (0, 2). In each cell, the top coordinate is the coordinate of the point v, and the bottom coordinate is the partition index v div D.
Figure 15: Partitioning a plane in hexagons.
hex grid geometry for game developers (1.3) 21 this gives as the corner of the paralellogram. We can now divide this by D to find the partition index.
The partition index is given by v div (R, D) = (v −f −1 D (v mod D)) div D (61a) and the location in the partition is v mod (R, D) = f −1 D (v mod D) div D (61b) Wrapped Grids Wrapped grids are often used in games. They have some properties that makes them attractive for game worlds: • They are finite.
• They allow infinite movement along any direction.
• They have a different topology which makes different things possible. For example, they allow the player to catch an enemy from behind or the front.
These combine to allow us to simulate the feeling of being on a sphere, as is used in Civilization type games (even though it is really a topological torus). In square grid worlds, we easily implement wrapping using modular arithmetic on coordinates. And in hex worlds, we can use the same approach if our grid is in the shape of a parallelogram. But hex grids allows for another type of wrapping if our grid is in the shape of a hexagon.5 5 There is a nice interactive example here: grids/hexagons/#wraparound.
While parallelogram wrapping resembles the surface on a torus, hexagon wrapping resembles the surface of a twisted torus. Figure 14 shows the same grid as Figure 15.
This wrapping has the strange effect that if you continue in a straight line, you will visit each cell before returning to the original cell. You can always hit an enemy by shooting in any (axis-aligned) direction!
Without the concept of point division we introduced earlier, this type of wrapping is difficult to implement: if you go out of the one end, it is not clear how to calculate the coordinate of where you will come in.
But with point division it is really trivial.
What we usually want is to calculate the following: where we are when we start at point v when travelling along u. In a non-wrap grid, our location would be given by u + v. In a wrapped grid, the calculation is the same, except that we apply a wrapping function W to the result.
hex grid geometry for game developers (1.3) 22 Figure 16: Wrapped Hex Grid on Torus.
Notice the line formed by the grid points (0, −2), (0, −1), (0, 0), (0, 1), (0, 2), and (−2, 0).
The wrapping function is given by W(v) = v mod (R, D) (62) where R is the points of our grid (the points for which W(v) = v), and D is the matrix of parallelogram that describes the repitition of our wrapping. There are two basic wrappings for hexagons that corresponds to whether the topological torus is twisted left or right.
When R is a hexagon with sides of length n, two possible values for each type D are given by DL = 2n −1 1 −n n −1 n !
(63a) and DR = 2n −1 −n n n −1 !
.
(63b) Figure 17: Hexagon wrapped with DL(3). Notice the line formed by the grid points (0, −2), (0, −1), (0, 0),(0, 1), (0, 2), and (−2, 0).
hex grid geometry for game developers (1.3) 23 Figure 18: Hexagon wrapped with DR(3).
Recursive Spatial Partitioning Squares have the nice property that we can dissect them and get squares again. Or put differently, we can stack n × n squares and get a square. We make use of this property in many algorithms and data-structures: quad trees, sampling, noise generation. Moreover, using modular and integer division, we can work on multi-res grids with ease (for example, if we impose a grid of half the resolution on an existing grid, we can get low res coordinates by integer-diving high-res by two). These manipulations are so trivial we hardly notice that we are doing something special.
When we try to apply those ideas to hex grids, it seems they break down. No number of hexes can be stacked to get a pure hexagon shape, or put differently, we cannot dissect a hexagon into smaller hexagons.
But hexagons can be stacked to form approximate hexagons, as shown in the figures below. And these can be stacked together again, and again. So some form of recursive partitioning is possible. The trick is now to get it into a form that makes it possible for us to write useful algorithms.
We already saw that we can partition space into hexagons. By applying this recursively, we can get partitions of the shapes in the images shown.
Let P0(v) = v.
(64) Then the k-level partition is given by: Pk(v) = Pk−1(v) div (R, D) (65) where R is the 7-point hexagon, and D = DL(1).
hex grid geometry for game developers (1.3) 24 While we used the ideas here to find somewhat regular subdivi-sions, they still work when our stacks are not hexagon(ish). In fact, they work for any region R if translated copies of R tile the plane and each copy has 6 neighboring copies. And because we don’t ever use any property of hex grids (we only calculate directly on the co-ordinates), we can use the exact same scheme on square grids. For example, we can use the 5-cell cross as a primitive element, that gives us recursive space divisions that are not squares. And we can use the same scheme of noise generation on these cross shapes, and get results with far less artifacts (admittedly, at a higher computational cost).
And remember now that we can also represent other grids as hex or square grids, and the same ideas can be applied directly to those grids. So we have found an algorithmic tool that can be applied to any grid that is represented as either a square or hex grid. It is a bit tricky to find the coloring to use, but it can be made easier by offsetting the parallelogram slightly so that it does not fall on any edges or vertices. For example, if we want to recursively subdivide a tri-grid into hexes, we have to use a 9-coloring.
Figure 19: Stack hexagons together.
(Some extra tools are necessary for dealing with tilings which involve rotations or reflections, such as if we wanted to recursively subdivide a grid into triangle or rhomb shapes. I have not investi-gated these, but it seems that not much work should be required to handle these as well.) The Gosper Curve Suppose we have a grid partitioned at multiple resolutions, and we want to iterate through all points such that: 1. each point is one unit vector away from the previous point, 2. we don’t leave a partition at any resolution before we visited all points in the partition.
For hex grids, one type of curve that has this property is called the hex grid geometry for game developers (1.3) 25 Gosper curve. (Strictly speaking, the Gosper curve is a fractal, and we are really interested in discrete approximations of it. We will call these approximations Gosper curves too.) These curves are useful for a variety of applications; essentially, they give us a way to fake 2D algorithms using 1D. For example, the following images were generated using 1D spatially correlated noise.
Figure 20: Noise.
These types of curves are neatly described by L-systems, and is easy to implement. If you don’t know what an L-system is, here is a brief introduction: Figure 21: A Gosper traversal of a hex grid.
1. An L-system is a set of rules that tells you how to replace symbols in a string with new strings.
2. The system has an axiom–a string of symbols as the starting string.
You can now take the axiom, and rewrite it using the rules, re-placing each symbol with the appropriate string. You can do this repeatedly, giving you a new string each time.
These strings can then be interpreted as a sequence of commands to drive some other system to generate an object. To generate a Gosper curve, we will interpret the string as turtle graphics com-mands; commands that tell a “turtle” with a pen how to move with simple commands such as “go forward”, “turn 60 degrees left” and so on. (This terminology comes from LOGO, an educational program-ming language famous for its use of turtle graphics.) Here is the L-system for the Gosper curve, shown below: Axiom: X Rewrite Rules: X →X + YF + +YF −FX −−FXFX −YF+ Y →−FX + YFYF + +YF + FX −−FX −Y.
hex grid geometry for game developers (1.3) 26 Here, we interpret F to mean move one unit forward, −to mean turn 60 degrees left, and + to mean turn 60 degrees right. After k iterations of rewriting, we get a curve that spans yk hexes.
Unfortunately, this does not describe cell-to-cell movements that we really want. However, each segment falls in a unique hexagon, in particular, their midpoints fall into the hexagons, and we can therefore easily get the movement that we want by calculating the midpoints of the segments and rounding those values to grid points.
The resulting sequence of points describes a path in a hex grid that performs a depth first traversal.
Figure 22: Procedural generation using a Gosper curve.
Grid Colorings For the remainder of this section, we will only be concerned with grid points, so we restrict v and D to have only integer components, which means r will only have integer components too.
In many cases, we will only need r as a label; we don’t actually care about the values of r. For these cases, we make the following definition.
Let c(v) be a function that assigns for each grid point within the parallelogram of D a unique integer between 0 and N −1, where N is the number of grid points in the parallelogram, given by N = |g × h|.
(66) We extend this function to arbitrary grid points: c(v) = c(v mod D) (67) We call this function a coloring of the grid by D. A coloring c by D has the property that for any two integers m and n, with q = (m, n), c(v + mg + nh) = c(v + qD) = c(v).
(68) Figure 23: Regular 3-coloring, with D = 0 3 1 1 .
It is a usual visual aid to color grid points by different colors depending on the value of c(v), below we show a few such colorings.
I don’t know of an algorithm to implement colorings for arbitrary g and h. However, there is a relatively straightforward implementa-tion if g is restricted to have y = 0. We can always choose suitable g and h with gy = 0 for any coloring.
When gy = 0, our coloring parallelogram always have n rows of equal length m. We can now use simple modular arithmetic and hex grid geometry for game developers (1.3) 27 offsets to calculate the colors as follows: yc = y mod n (69a) xc = x mod m (69b) We combine these to get the final index: c = ycm + xc.
(70) Figure 24: Regular 4-coloring, with D = 0 2 2 2 .
In this scheme, the numbers in each row is typically not in order.
However, the order is not important, since we use these color indices mostly for labeling. The fact that one point maps to 0 and another to 1 is not important. What is important is that they are different.
Three colorings are of particular importance; they are useful in a variety of ways.
Regular 3-coloring.
This coloring is the only way to color a hex grid with the smallest number of colors so that no neighbors have the same color. This coloring is used for hexagonal chess, because it gives diagonally opposite cells the same colors, and therefor is a visual aid to movement of bishops, queens and kings. This coloring is used for representing triangular grids using hex grids (as explained in the next section).
Regular 4-coloring.
In this coloring, opposite neighbors have the same color. We can use this coloring is we want to use a hex grid to represent faces and edges or faces and vertices simultaneously. The former makes this coloring useful in maze generation algorithms (such as Prim’s algorithm). This coloring is used for representing rhombille grids using hex grids.
Figure 25: Regular 7-coloring, with D = 0 7 4 1 .
Regular 7-coloring.
This coloring gives unique colors to each hex in partitioning into hexes. This coloring is used for representing floret pentagonal grids using hex grids.
Triangular grids Triangular grids, despite their regular nature, are hard to work with algorithmically.
The usual coordination hides many of the symmetries, and make algorithms ugly and error prone, similar to the way in which differ-ent choices of coordinates for hex grids make algorithms ugly and error prone.
But there is a neat trick we can use to work with these grids, and get all the nice geometric advantages of hex grids.
hex grid geometry for game developers (1.3) 28 We simply treat points on the hex grid as points on a tri grid grid.
Figure 26 should explain what I mean: Figure 26: I hex grid superimposed oin a tri grid can be used to give a tri grid coordinates.
As you can see, not all hex points correspond to tri points. And at first glance, it looks like the coordinates do not capture all the information–when is a triangle up or down? How do we know a hex is not part of the system? But this information is actually nicely recovered using the 3-coloring you can see in the image.
Using this scheme, we can do all the geometry using the simple tools of hex grids.
In triangular grids, the grid lines we think of as important are parallel to the minor axes. They are either parallel, or intersect in two grid points. The grid point closest to the intersection of the associated real lines give you one of these points. The other can be found by checking all this point’s neighbors.
Figure 27: Lines corresponding to the equations v = ( 1 2, 1 2 ) + nt(−1, 2), v = (1, 1 2 ) + nt(1, 1), and v = ( 1 2, 1 2 ) + nt(−2, 1).
We can use the same trick for other grids as well: for example, using appropriate colorings we can also represent rhombille grids and floret pentagon grids.
hex grid geometry for game developers (1.3) 29 Figure 28: The intersection of two lines.
Figure 29: Rhombille Grid Figure 30: Floret Pentagon Grid |
190292 | https://www.bbc.co.uk/bitesize/articles/zm8rcmn | KS3
Rounding whole numbers
Part of MathsRounding and estimating
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Key points
When it’s not necessary to give an exact value, numbers can be rounded. Rounding means making a number simpler but keeping its value close to what it is.
When rounding numbers the result is less accurate, but easier to use. For example, if there were 39,828 spectators at a music concert the headline in a newspaper report might say 40,000 fans attended the event.
Depending on the size of the numbers, numbers can be rounded up or down to the nearest:
whole number
ten
hundred
thousand
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Rounding Numbers
There are steps which can be followed to decide whether a number should be rounded up or rounded down.
Identify the digitOne of the symbols of a number system, most commonly the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 in the column that you are rounding to. For example, if rounding to the nearest hundred then identify the digit in the hundreds place value column.
Leave the digit the same if the next digit is less than 5 (this is called rounding down)but increase it by 1 if the next digit is 5 or more (this is called rounding up)
Using a number lineA straight line on which numbers are marked at equal intervals. as a representation is a useful way of visualising whether a number is closer to one value or another.
When using values on a number line, marking the halfway point can help to identify whether to round up or down.
Examples
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Slide 1 of 9, A number line from zero to ten., Use a number line to identify if you need to round up or round down. The whole numbers are equally spaced on this number line. The values increase by each step to the right.
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A range of values
Sometimes you are given a rounded version of a number and need to explain what the original value was. In effect you need to work backwards.
Drawing a number line is a helpful representation to identify the minimum and maximum values that could have been used.
The number line should be increasing in the same sized step that the rounded value has been given to. (For example, if the rounded value is 70 to the nearest 10, then the number line should increase in steps of 10)
Examples
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Slide 1 of 7, Written: Forty pounds to the nearest ten pounds., The price of a meal is rounded to the nearest £10 to cost £40. What different amounts could the original bill have been?
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Question
Which of these numbers when rounded to the nearest hundred give 11,600?
When rounded to the nearest hundred, 11,550 and 11,601 would give 11,600
11,669 is rounded to 11,700
11,701 is rounded to 11,700
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Practise rounding whole numbers
Practise rounding whole numbers in this quiz.
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Real-world maths
Rounding whole numbers is used regularly in everyday life.The exact figure or value is not always needed initially, particularly if people are deciding whether they can afford a product or not.
When shopping people often round the value of items they are buying to the nearest £1 and add them together to make sure they don’t over-spend.
Newspapers and magazines may round the attendance at sporting events or festivals to the nearest thousand or ten thousand in their headlines, rather than printing the actual attendance. This makes it easier for the audience to read and remember.
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Game - Divided Islands
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Using your maths skills, help to build bridges and bring light back to the islands in this free game from BBC Bitesize.
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190293 | https://personal.math.vt.edu/jwells13/teaching/mat2534_spring2024/mat2534notes_written.pdf | MAT 2534 Discrete Math Joe Wells Virginia Tech Spring 20241 Last Updated: May 4, 2024 1This courses is using Susanna Epp’s Discrete Mathematics with Applications, 4th Edition.
The chapter/section titles have been retained, but otherwise internal numbering of theorems, examples, etc. will likely disagree with the course text.
2 Contents 2 The Logic of Compound Statements 5 2.1 Logical Form and Logical Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . .
5 2.1.1 Truth Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 2.1.2 Table of Logical Equivalences . . . . . . . . . . . . . . . . . . . . . . . . . .
9 2.2 Conditional Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 2.2.1 Related Conditionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 2.2.2 Biconditional Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 2.3 Valid and Invalid Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 2.3.1 Rules of Inference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 2.3.2 Logical Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24 2.3.3 Fallacies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 2.3.4 Sound Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29 3 The Logic of Quantified Statements 31 3.1 Predicates and Quantified Statements I . . . . . . . . . . . . . . . . . . . . . . . . .
31 3.1.1 The Universal Quantifier: ∀. . . . . . . . . . . . . . . . . . . . . . . . . . .
32 3.1.2 The Existential Quantifier: ∃ . . . . . . . . . . . . . . . . . . . . . . . . . .
33 3.1.3 Universal Conditional Statement . . . . . . . . . . . . . . . . . . . . . . . .
34 3.1.4 Implicit Quantification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35 3.1.5 Relationship between ∀and ∧; Relationship between ∃and ∨ . . . . . . . .
37 3.2 Predicates and Quantified Statements II . . . . . . . . . . . . . . . . . . . . . . . .
38 3.2.1 Negating Universal and Existential Statements . . . . . . . . . . . . . . . . .
38 3.2.2 Conditionals - Related Conditionals and Negations . . . . . . . . . . . . . .
38 3.3 Statements with Multiple Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . .
41 3.3.1 ∀∀Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41 3.3.2 ∃∃Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42 3.3.3 ∀∃and ∃∀Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44 3.4 Arguments with Quantified Statements . . . . . . . . . . . . . . . . . . . . . . . . .
49 3.4.1 Rules of inference with quantifiers . . . . . . . . . . . . . . . . . . . . . . . .
49 3.4.2 Logical Proofs with Nested Quantifiers . . . . . . . . . . . . . . . . . . . . .
52 3.4.3 An Actual Math Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53 4 Elementary Number Theory and Methods of Proof 55 4.1 Direct Proof and Counterexample I: Introduction . . . . . . . . . . . . . . . . . . .
55 4.1.1 Context/Relation to Formal Proofs . . . . . . . . . . . . . . . . . . . . . . .
55 4.1.2 Important Hypotheses and Definitions . . . . . . . . . . . . . . . . . . . . .
56 4.1.3 Proofs – Scaffolding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58 4.1.4 Proving an Existential Statement . . . . . . . . . . . . . . . . . . . . . . . .
59 4.1.5 Disproving Universal Statements . . . . . . . . . . . . . . . . . . . . . . . .
60 4.1.6 Proving a Universal Statement . . . . . . . . . . . . . . . . . . . . . . . . . .
61 4.1.7 Disproving an Existential Statement . . . . . . . . . . . . . . . . . . . . . .
62 4.1.8 Proof by Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63 4.1.9 Direct Proof with Nested Quantifiers . . . . . . . . . . . . . . . . . . . . . .
65 4.2 Direct Proof and Counterexample II: Writing Advice . . . . . . . . . . . . . . . . .
69 3 4 CONTENTS 4.2.1 Common Mistakes in Proof-Writing . . . . . . . . . . . . . . . . . . . . . . .
69 4.3 Direct Proof and Counterexample III: Rational Numbers . . . . . . . . . . . . . . .
71 4.4 Direct Proof and Counterexample IV: Divisibility . . . . . . . . . . . . . . . . . . .
74 4.6 Proof By Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78 4.7 Indirect Argument: Contradiction and Contraposition . . . . . . . . . . . . . . . . .
80 4.7.1 Contradiction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80 4.7.2 Contraposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81 4.8 Indirect Argument: Two Three Famous Theorems . . . . . . . . . . . . . . . . . . .
85 4.8.1 √ 2 is Irrational . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85 4.8.2 The Infinitude of Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86 4.8.3 Area of a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87 5 Sequences, Mathematical Induction, and Recursion 89 5.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89 5.2 Mathematical Induction I: Proving Formulas . . . . . . . . . . . . . . . . . . . . . .
94 5.3 Mathematical Induction II: Application . . . . . . . . . . . . . . . . . . . . . . . . .
99 5.4 Strong Mathematical Induction and the Well-Ordering Principle for the Integers . . 103 5.4.1 Well-Ordering Principle for the Integers . . . . . . . . . . . . . . . . . . . . . 106 5.6 Defining Sequences Recursively . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 6 Set Theory 111 6.1 Set Theory: Definitions and the Element of Proof . . . . . . . . . . . . . . . . . . . 111 6.1.1 Set Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 6.1.2 Arithmetic Operations and Set Theory - Some Interesting History . . . . . . 119 6.2 Properties of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 6.2.1 Properties of Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 6.3 Disproofs and Algebraic Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 6.4 Boolean Algebras, Russell’s Paradox, and the Halting Problem . . . . . . . . . . . . 134 7 Properties of Functions 139 7.1 Functions Defined on General Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 7.1.1 Arrow Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 7.1.2 Range, Preimage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 7.2 One-to-One, Onto, and Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . 146 7.3 Composition of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 7.4 Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 7.4.1 Infinity Infinities: “To Infinity and Beyond” . . . . . . . . . . . . . . . . . . 159 7.4.2 New Bijections from Old . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 8 Properties of Relations 169 8.1 Relations on Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 8.1.1 Arrow Diagrams/Directed Graphs . . . . . . . . . . . . . . . . . . . . . . . . 170 8.1.2 Inverse Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 8.2 Reflexivity, Symmetry, and Transitivity . . . . . . . . . . . . . . . . . . . . . . . . . 174 8.2.1 Proving and disproving properties of binary relations . . . . . . . . . . . . . 176 8.3 Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 .1 Proofs Skipped In Class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 CONTENTS 5 Index 185 6 CONTENTS Chapter 2 The Logic of Compound Statements 2.1 Logical Form and Logical Equivalence Definition A statement (or a proposition) is a sentence which is either true or false, but not both.
The truth value of a statement either “true” or “false.” Example 2.1.1 • “1 + 2 = 3” is a true statement.
• “1 + 2 = 4” is a false statement.
• “x + 2 = 5” is neither true nor false; not a statement. Since x is unspecified. Usually when we are solving for x, we are trying to find an x-value that makes the statement true.
To make life simpler when breaking down compound statements, we introduce some logical notation: Logical Connectives and Order of Operations symbol English translation ∨ “or” ∧ “and” ¬ or ∼ “not” Order of Operations 1. Parentheses () 2. ¬ 3. ∧ 4. ∨ Remark. “not” should be interpreted generally as negating a statement, which is more commonly how one would use it in English.
Example 2.1.2 Let p and q be the following statements.: p : Trey drinks water.
q : Sandy eats cookies.
Interpret the following statements in plain English.
• p ∨q • p ∧q • ¬p ∨q • ¬p ∧¬q 7 8 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS • p ∨q means “Trey drinks water or Sandy eats cookies.” • p ∧q means “Trey drinks water and Sandy eats cookies.” • ¬p ∨q means “Trey does not drink water or Sandy eats cookies.” • ¬p ∧¬q means “Trey does not drink water and Sandy does not eats cookies.” Logical form of common expressions “Neither a nor b” means not a and not b.
“a but not b” means a and not b.
“a ≥2” means a > 2 or a = 2.
“1 ≤b < 5” means 1 ≤b and b < 5.
2.1.1 Truth Tables When considering a sentence comprised of several component statements, we want to know if the entire compound sentence is actually a statement (i.e. has a well-defined truth value). In order to do this, we’ll need to analyze how the logical symbols (i.e. logical connectives) relate to the validity of the compound statement.
Example 2.1.3 Let x be a fixed real number and consider the sentence “2 < x < 5,” which we know is “x > 2 and x < 5.” Fix a couple of different x-values and record the truth values of the three statements x > 2, x < 5, and 2 < x < 5 in a truth table.
x-value x > 2 x < 5 2 < x < 5 −1 F T F 3 T T T 7 T F F The sentence “2 < x < 5” is only true for x-values in which both of “2 < x” and “x < 5” are true.
Definition If p and q are both statements, then the conjunction of p and q is the statement p ∧q. The truth table for conjunctions is below.
p q p ∧q T T T T F F F T F F F F 2.1. LOGICAL FORM AND LOGICAL EQUIVALENCE 9 Example 2.1.4 Let x be a fixed real number and consider the sentence “x ≤2,” which we know is “x < 2” or “x = 2.” Let’s fix a couple of different x-values and record the truthfulness of x < 2, x = 2, and x ≤2 in a truth table: x-value x < 2 x = 2 x ≤2 −1 T F T 2 F T T 7 F F F The sentence “x ≤2” is only true for x-values in which at least one of “x < 2” and “x = 2” are true.
Definition: Disjunction If p and q are both statements, then the disjunction of p and q is the statement p ∨q. The truth table for conjunctions is below.
p q p ∨q T T T T F T F T T F F F Remark. Common English tends to use an “exclusive or.” At a restaurant, when asked “Soup or salad?” we implicitly understand it to mean that you can have either soup or salad, but not both.
In logic, disjunction represents an “inclusive or”, which would allow for “soup, salad, or both” as valid answers. While this is a bit of a conventional choice, by comparing the final column of the conjunctive and disjunctive truth tables, we see that the inclusivity keeps them similar. Sometimes the symbol ⊕will be used to denote an exclusive or (although we will not use that in these notes).
Definition If p is any statement, then the negation of p is the statement ¬p. The truth table for negation is below.
p ¬p T F F T Now that we know about the basic logical connectives, let’s fill in their truth tables.
Definition Let p and q be statements. Then exclusive or, denoted p ⊕q is true when precisely one of p and q is true. The truth table for ⊕is below.
10 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS p q p ∨q T T F T F T F T T F F F In plain English, this corresponds to the phrase “either... or...” Definition A compound statement is a logical statement involving multiple logical connectives.
Example 2.1.5: 3-variable truth table Complete the following truth table for the statement: p ∧¬(q ∨r).
p q r q ∨r ¬(q ∨r) p ∧¬(q ∨r) T T T T F F T T F T F F T F T T F F T F F F T T F T T T F F F T F T F F F F T T F F F F F F T F Definition Two (compound) statements, P and Q, with all of the same truth values are called logically equivalent.
Symbolically we write P ≡Q.
Example 2.1.6: Exclusive Or Let p and q be statements. Show that p ⊕q ≡ (p ∨q) ∧¬(p ∧q) by filling out a truth table and verifying both statements have the same truth values.
p q p ⊕q p ∨q p ∧q ¬(p ∧q) (p ∨q) ∧¬(p ∧q) T T F T T F F T F T T F T T F T T T F T T F F F F F T F 2.1. LOGICAL FORM AND LOGICAL EQUIVALENCE 11 This table justifies the interpretation of the exclusive or: “p or q is true, but not both.” Example 2.1.7: Negation is Not Distributive Show that ¬(p ∨q) ̸≡(¬p) ∨(¬q).
p q p ∨q ¬(p ∨q) ¬p ¬q (¬p) ∨(¬q) T T T F F F F T F T F F T T F T T F T F T F F F T T T T Definition: Tautology and Contradiction A tautology is a statement (call it t) that is always true and a contradiction is a statement (call it c) that is always false.
2.1.2 Table of Logical Equivalences Theorem 2.1.8: Table of Logical Equivalances Let p, q, and r be statements, let t be a tautology, and let c be a contradiction. We then have the following table of equivalences: 12 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS Commutative Laws p ∧q ≡q ∧p p ∨q ≡q ∨p Associative Laws (p ∧q) ∧r ≡p ∧(q ∧r) (p ∨q) ∨r ≡p ∨(q ∨r) Distributive Laws p ∧(q ∨r) ≡(p ∧q) ∨(p ∧r) p ∨(q ∧r) ≡(p ∨q) ∧(p ∨r) Identity Laws p ∧t ≡p p ∨c ≡p Negation Laws p ∨¬p ≡t p ∧¬p ≡c Double Negative Laws ¬(¬p) ≡p Idempotent Laws p ∧p ≡p p ∨p ≡p Universal Bound Laws p ∨t ≡t p ∧c ≡c De Morgan’s Laws ¬(p ∧q) ≡¬p ∨¬q ¬(p ∨q) ≡¬p ∧¬q Absorption Laws p ∨(p ∧q) ≡p p ∧(p ∨q) ≡p Negation of t and c ¬t ≡c ¬c ≡t Example 2.1.9: Proof of Commutative Laws Use a truth table to show the Commutative Laws in Theorem 2.1.8.
p q p ∧q q ∧p T T T T T F F F F T F F F F F F p q p ∨q q ∨p T T T T T F T T F T T T F F F F 2.1. LOGICAL FORM AND LOGICAL EQUIVALENCE 13 Example 2.1.10: Proof of Associative Laws Use a truth table to show the Associative Laws in Theorem 2.1.8.
p q r p ∧q (p ∧q) ∧r q ∧r p ∧(q ∧r) T T T T T T T T T F T F F F T F T F F F F T F F F F F F F T T F F T F F T F F F F F F F T F F F F F F F F F F F p q r p ∨q (p ∨q) ∨r q ∨r p ∨(q ∨r) T T T T T T T T T F T T T T T F T T T T T T F F T T F T F T T T T T T F T F T T T T F F T F T T T F F F F F F F Example 2.1.11: Proof of Distributive Laws.
Use a truth table to show the Distributive Laws in Theorem 2.1.8.
p q r q ∨r p ∧(q ∨r) p ∧q p ∧r (p ∧q) ∨(p ∧r) T T T T T T T T T T F T T T F T T F T T T F T T T F F F F F F F F T T T F F F F F T F T F F F F F F T T F F F F F F F F F F F F 14 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS p q r q ∧r p ∨(q ∧r) p ∨q p ∨r (p ∨q) ∧(p ∨r) T T T T T T T T T T F F T T T T T F T F T T T T T F F F T T T T F T T T T T T T F T F F F T F F F F T F F F T F F F F F F F F F Example 2.1.12: Proof of Identity Laws Use a truth table to show the Identity Laws in Theorem 2.1.8.
p t p ∧t T T T F T F p c p ∨c T F T F F F Example 2.1.13: Proof of Negation Laws.
Use a truth table to show the Negation Laws in Theorem 2.1.8.
p ¬p p ∨¬p t T F T T F T T T p ¬p p ∧¬p c T F F F F T F F Example 2.1.14: Proof of Double Negative Law Use a truth table to show the Double Negation Law in Theorem 2.1.8.
p ¬p ¬(¬p) T F T F T F Example 2.1.15: Proof of Idempotent Laws.
Use a truth table to show the Idempotent Laws in Theorem 2.1.8.
p p ∧p T T F F p p ∨p T T F F 2.1. LOGICAL FORM AND LOGICAL EQUIVALENCE 15 Example 2.1.16: Proof of Universal Bound Laws Use a truth table to show the Universal Bound Laws in Theorem 2.1.8.
p t p ∨t T T T F T T p c p ∧c T F F F F F Example 2.1.17: Proof of De Morgan’s Laws Use a truth table to show the De Morgan’s Laws in Theorem 2.1.8.
p q p ∧q ¬(p ∧q) ¬p ¬q ¬p ∨¬q T T T F F F F T F F T F T T F T F T T F T F F F T T T T p q p ∨q ¬(p ∨q) ¬p ¬q ¬p ∧¬q T T T F F F F T F T F F T F F T T F T F F F F F T T T T Example 2.1.18: Proof of Absorption Laws Use a truth table to show the Absorption Laws in Theorem 2.1.8.
p q p ∧q p ∨(p ∨q) T T T T T F F T F T F F F F F F p q p ∨q p ∧(p ∨q) T T T T T F T T F T T F F F F F Example 2.1.19: Proof of Negation of t and c Use a truth table to show the Negation of t and c in Theorem 2.1.8.
t c ¬t T F F t c ¬c T F F 16 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS 2.2 Conditional Statements Consider the following promise made by your instructor to his broccoli-averse child.
If you eat your dinner, then I will give you cookies for dessert.
If the child eats dinner and your instructor gives the child cookies for dessert, then the promise is upheld.
If the child eats dinner and your instructor does not gives the child cookies for dessert, then the promise is not upheld.
If the child does not eat dinner, then cookies or not, it would be unfair to claim that the instructor did not uphold the promise.
Definition: If p then q If p and q are statements, then the statement “if p, then q” is called the conditional of q by p and is denoted “p = ⇒q.” p is called the hypothesis and q is called the conclusion. The truth table for the conditional is below.
p q p = ⇒q T T T T F F F T T F F T When the hypothesis is false, the conditional is called vacuously true.
Remark. One could also write q ⇐ = p, but since English is read left-to-right, we will typically avoid using the left-pointing arrow.
Remark. Your book uses two different arrows, →and = ⇒, which have different meanings in the realm of formal logic. In this class, we will be a little bit sloppy and always use = ⇒.
Since p →q is only false when p is true and q is false, then the following observation is immediate.
Example 2.2.1: conditional identity Let p, q be statements. Use a truth table to show that (p ⇒q) ≡(¬p ∨q).
p q p = ⇒q ¬p ¬p ∨q T T T F T T F F F F F T T T T F F T T T Proposition 2.2.2: Negation of a conditional statement If p, q are statements, then ¬(p ⇒q) ≡(p ∧¬q).
2.2. CONDITIONAL STATEMENTS 17 Proof. Using the results of Example 2.2.1, this follows immediately from DeMorgan’s Law.
¬(p = ⇒q) ≡¬(¬p ∨q) ≡¬¬p ∧¬q (DeMorgan’s Law) ≡p ∧¬q (double negation law) Now that we have a new symbol, we revisit the order of operations.
Order of Operations 1. Parentheses () 2. ¬ or ∼ 3. ∧ 4. ∨ 5. = ⇒or → Remark. Although the order of operations suggests that everything to the left of “ = ⇒” implies everything to the right of “ = ⇒,” it is perfectly reasonable to use parentheses to clarify.
Example 2.2.3: Division Into Cases Let p, q, r be statements. Show the following logical equivalence p ∨q = ⇒r ≡ (p = ⇒r) ∧(q = ⇒r) .
p q r p ∨q p ∨q ⇒r p ⇒r q ⇒r (p ⇒r) ∧(q ⇒r) T T T T T T T T T T F T F F F F T F T T T T T T T F F T F F T F F T T T T T T T F T F T F T F F F F T F T T T T F F F F T T T T 2.2.1 Related Conditionals Definition: Converse, inverse, contrapositive Given statements p, q and the conditional statement, p = ⇒q, there are three closely-related conditionals: • The converse is p ⇐ = q.
• The inverse is ¬p = ⇒¬q.
18 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS • The contrapositive is ¬p ⇐ = ¬q.
Example 2.2.4 Consider the following conditional statement: If my car is in the repair shop, then I cannot get to class.
Write the converse, inverse, and contrapositive statements.
• [Converse] If I cannot get to class, then my car is in the repair shop.
• [Inverse] If my car is not in the repair shop, then I can get to class.
• [Contrapositive] If I can get to class, then my car is not in the repair shop.
Notice that some of these sound like they could be logically equivalent to one another. The contrapositive, for example, may be logically equivalent to the original statement. The converse, however, doesn’t seem like it should be - there may be other reasons that one cannot attend class that are independent of the car needing repairs.
Proposition 2.2.5: Conditional Equivalences Let S be the conditional statement p = ⇒q. Then 1. S is logically equivalent to the contrapositive of S, and 2. the converse of S is logically equivalent to the inverse of S.
Proof of 1.
p = ⇒q ≡¬p ∨q (Conditional Identity) ≡¬p ∨¬(¬q) (Double Negative Law) ≡¬(¬q) ∨¬p (Commutative Law) ≡¬q = ⇒¬p (Conditional Identity).
Proof of 2.
q = ⇒p ≡¬q ∨p (Conditional Identity) ≡¬q ∨¬(¬p) (Double Negative Law) ≡¬(¬p) ∨¬q (Commutative Law) ≡¬p = ⇒¬q (Conditional Identity).
One may use the phrase that p happens “only if” q happens. In other words, if q doesn’t occur, then p doesn’t occur. This is now phrased like the contrapositive, so it must be equivalent to p ⇒q.
We record this and some other typical phrases below 2.2. CONDITIONAL STATEMENTS 19 Logical form of common expressions “p implies q” means “p = ⇒q” “p only if q” means “p = ⇒q” “p if q” means “p ⇐ = q” “p is a sufficient condition for q” means “p = ⇒q” “p is a necessary condition for q” means “p ⇐ = q” 2.2.2 Biconditional Statements Definition: biconditional If p, q are logical statements, then the biconditional of p and q, denoted p ⇐ ⇒q, is true when p and q have the same truth values, and false when p and q have opposite truth values.
p q p ⇐ ⇒q T T T T F F F T F F F T Example 2.2.6 Show that that p ⇐ ⇒q ≡ (p = ⇒q) ∧(p = ⇒q) .
p q p ⇔q p ⇒q p ⇐q (p ⇒q) ∧(p ⇐q) T T T T T T T F F F T F F T F T F F F F T T T T Because of this connection, we often use the following English phrases to mean “p ⇐ ⇒q”: “p if and only if q.” “p iff q.” “p is both necessary and sufficient for q.” Example 2.2.7 For each set of statements below, fill in the blank to correctly identify whether the statement is an “if”, “only if”, or “if and only if” statement.
Note: Any mathematical statements below were chosen only to provide context for results you may have encountered before; none of them is actually mandatory prerequisite knowledge for this class.
20 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS It is cloudy if it is raining.
A polynomial has odd degree only if it has at least one real root.
The glass is half empty if and only if the glass is half full.
A matrix is invertible if and only if its determinant is nonzero.
Lassie is a dog only if Lassie is a mammal.
A function f has a critical point at x = a if f ′(a) = 0.
Exercise 2.2.8 Let p, q be statements. What is the relationship between p ⇐ ⇒q and p ⊕q?
2.3. VALID AND INVALID ARGUMENTS 21 2.3 Valid and Invalid Arguments Definition: argument, validity An argument is a sequence of statements. All statements in an argument, except for the final one are called premises. The final statement is called the conclusion. An argument is said to be valid when it satisfies the following criterion: if the premises are all true, then the conclusion is also true.
In logical symbols, an argument is typically written in the following way: p1 (hypothesis 1) p2 (hypothesis 1) .
.
.
pn (hypothesis 1) ∴ c (conclusion) The symbol ∴is read as “therefore.” Remark. An argument is valid whenever the proposition (p1 ∧p2 ∧· · · ∧pn) = ⇒c is a tautology. In this way, it is clear that the order of the premises do not actually matter.
We can check validity of an argument using a truth table by looking for any rows with all true premises (called critical rows) and verifying that the conclusion table entry is also true in every critical row.
If one or more critical rows has a false conclusion, the argument is invalid.
Example 2.3.1 Determine whether the following argument is valid or invalid.
p p = ⇒q ∴ q Premise 1 Premise 2 Conclusion p q p p ⇒q q T T T T T T F T F F T F F F F There is exactly one row where all hypotheses are true (highlighted), and that row has a true conclusions, so it must be valid.
22 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS Example 2.3.2 Determine whether the following argument is valid.
If we meet the God of Death, then we say “Not today.”a The phrase “not today” was not spoken.
Therefore, we did not meet death.
aGame of Thrones Season 1, Episode 6.
The argument above can be simplified with symbols and variables.
p = ⇒q ¬q ∴ ¬p and so we set up a truth table.
Premise 1 Premise 2 Conclusion p q p ⇒q ¬q ¬p T T T F F T F F T F F T T F T F F T T T There is only one row in the truth table where all premises are true, and that row also has a true conclusion, so it must be valid.
Example 2.3.3 Determine whether the following argument is valid.
p = ⇒q ¬p ∴ ¬q Premise 1 Premise 2 Conclusion p q p ⇒q ¬p ¬q T T T F T F F F T T T F F F T T T There are two critical rows (highlighted), but one of them has a false conclusion. The argument is invalid.
2.3. VALID AND INVALID ARGUMENTS 23 Example 2.3.4 Determine whether the following argument is valid.
p ∨q ¬q ∴ p Premise 1 Premise 2 Conclusion p q p ∨q ¬q p T T T F T F T T T F T T F F F F There is only one row in the truth table where all hypotheses are true (highlighted), and that row also has a true conclusion, so it must be a valid argument.
2.3.1 Rules of Inference Definition: rule of inference A rule of inference is the form of argument that is valid.
Theorem 2.3.5: Common Rules of Inference Let p, q, r be propositions. The following are all valid logical argument forms.
(a) modus ponens a p = ⇒q p ∴ q (b) modus tollens b p = ⇒q ¬q ∴ ¬p (c) contradiction rule p = ⇒c (a contradiction) ∴ ¬p (d) division of cases 24 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS p ∨q p ⇒r q ⇒r ∴ r (e) addition p ∴ p ∨q (f) specialization (or simplification) p ∧q ∴ p p ∧q ∴ q (g) conjunctionc p q ∴ p ∧q (h) transitivity (or hypothetical syllogism) p = ⇒q q = ⇒r ∴ p = ⇒r (i) elimination (or disjunctive syllogism) d p ∨q ¬p ∴ q p ∨q ¬q ∴ p (j) resolution p ∨q ¬p ∨r ∴ q ∨r aThis is latin for “method of affirming” and was proven in Example 2.3.1.
bThis is latin for “method of denying” and was proven in Example 2.3.2.
cThe proof of this trivially follows from the definition of conjunction dThis was proven in Example 2.3.4.
2.3. VALID AND INVALID ARGUMENTS 25 Example 2.3.6: Proof of contradiction rule Use a truth table to prove the contradiction rule in Theorem 2.3.5.
Premise 1 Conclusion p p ⇒c ¬p T F F F T T Example 2.3.7: Proof of division of cases Use a truth table to prove the division of cases rule in Theorem 2.3.5.
Premise 1 Premise 2 Premise 3 Conclusion p q r p ∨q p ⇒r q ⇒r r T T T T T T T T T F T F T F T T T T T T F F T F F T T T T T T F T F T T F F F T F F F F F Example 2.3.8: Proof of addition rule Use a truth table to prove the addition rule in Theorem 2.3.5.
Premise 1 Conclusion p q p p ∨q T T T T T F T T F T F T F F F F Example 2.3.9: Proof of simplification rule Use a truth table to prove the simplification rule in Theorem 2.3.5.
Premise 1 Conclusion p q p ∧q p T T T T T F F T F T F F F F 26 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS Example 2.3.10: Proof of transitivity Use a truth table to prove the transitivity rule in Theorem 2.3.5.
Premise 1 Premise 2 Conclusion p q r p ⇒q q ⇒r p ⇒r T T T T T T T T F T F T F T F T F F F F T T T T T F T F T F F F T T T T F F F T T T Example 2.3.11: Proof of resolution Use a truth table to prove the resolution rule in Theorem 2.3.5.
Premise 1 Premise 2 Conclusion p q r p ∨q ¬p ¬p ∨r q ∨r T T T T F T T T T F T F F T F T T F T T T F F T F F F T T T T T T F T F T T T T F F T F T F F F F T 2.3.2 Logical Proofs Definition: logical proof A logical proof is a method of validating an argument by applying a sequence of rules of inference to deduce the stated conclusion from the hypotheses.
Remark. Logical proofs allow us to avoid truth tables. We note that, although the final proof is recorded in an enumerated list, you will likely think about things nonlinearly.
Remark. When writing down a logical proof, the only rule is that earlier steps cannot follow from later steps, otherwise you have freedom of choice as to when to introduce various statements. The analog in programming are the competing conventions of when to introduce variables – all at the beginning, or only before they are needed.
2.3. VALID AND INVALID ARGUMENTS 27 Example 2.3.12 Write a logical proof demonstrating the validity of the following argument.
¬A = ⇒(C ∧D) A = ⇒B ¬B ∴ C ¬A = ⇒(C ∧D) (premise 1) A = ⇒B (premise 2) ¬B (premise 3) Begin by noticing that the last two lines of premises can be combined using modus tollens.
A = ⇒B ¬B ∴ ¬A So we can rewrite the original collection of premises ¬A = ⇒(C ∧D) A = ⇒B ¬B − → ¬A = ⇒(C ∧D) ¬A Now notice that we can apply modus ponens ¬A = ⇒(C ∧D) ¬A ∴ C ∧D and our premises become ¬A = ⇒(C ∧D) ¬A − → C ∧D Finally, we can apply simplification to get C ∧D ∴ C which is precisely the conclusion we wanted to reach.
Now let’s collect this all in a single procedural list.
1.
A = ⇒B (Hypothesis) 2.
¬B (Hypothesis) 3.
¬A (Modus Tollens, Steps 1 and 2) 4.
¬A = ⇒(C ∧D) (Hypothesis) 5.
C ∧D (Modus Ponens, Steps 3 and 4) 6.
C (Simplification in Step 6) 28 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS Example 2.3.13: Proof by Contradiction Write down a logical proof demonstrating the validity of the following argument.
P = ⇒R P = ⇒¬R ∴ ¬P First we apply the conditional identity to both of our premises. Begin by noticing that we can apply conjunction to our first two hypotheses P = ⇒R ∴ (¬P ∨R) P = ⇒¬R ∴ (¬P ∨¬R) so we rewrite the original collection of premises P = ⇒R P = ⇒¬R − → ¬P ∨R ¬P ∨¬R We can now apply conjunction to get ¬P ∨R ¬P ∨¬R ∴ (¬P ∨R) ∧(¬P ∨¬R) so we rewrite the premises ¬P ∨R ¬P ∨¬R − → (¬P ∨R) ∧(¬P ∨¬R) Using the distributive property, we rewrite the premises again (¬P ∨R) ∧(¬P ∨¬R) − → ¬P ∨(R ∧¬R) Applying the conditional identity ¬P ∨(R ∧¬R) − → P = ⇒(R ∧¬R) The complement law yields P = ⇒(R ∧¬R) − → P = ⇒c and finally the contradiction rule gives P = ⇒c − → ¬P Now let’s collect this all in a single procedural list.
1.
P = ⇒R (hypothesis) 2.
¬P ∨R (conditional identity, 1) 3.
P = ⇒¬R (hypothesis) 4.
¬P ∨¬R (conditional identity,3) 5.
(¬P ∨R) ∧(¬P ∨¬R) (conjunction,2 and 4) 6.
¬P ∨(R ∧¬R) (distributive law, 5) 7.
P = ⇒(R ∧¬R) (conditional identity, 6) 8.
P = ⇒c (complement law, 7) 9.
¬P (contradiction rule) 2.3. VALID AND INVALID ARGUMENTS 29 2.3.3 Fallacies Definition: fallacy A fallacy is an error in reasoning that results in an invalid argument.
Remark. An argument is invalid precisely when all premises are true, but the conclusion is false.
Example 2.3.14: Ambiguous Premises Ambiguous premises can arise in many ways. For example, using words with multiple meanings and equivocating them.
6 is an odd number of legs for a horse.
Odd numbers cannot be divided by 2.
Therefore 6 cannot be divided by 2.
The word “odd” in line 1 is a synonym for “unusual” and in line 2 it is being used to describe a number not divisible by 2.
Example 2.3.15: Ambiguous Premises Ambiguous premises can arise in many ways. For example, using words that cannot be quantified: If you have a good understanding of Discrete Math, then you will do well on the exam.
You have a good understanding of Discrete Math.
Therefore you get an “A” on the exam.
“Doing well” on a test is ambiguous – arguably a grade of “B” or “C+” would be considered “doing well” to most.
Example 2.3.16: Circular Reasoning Circular reasoning occurs when you use the conclusion as a premise.
You can’t give me a “C” – I’m an “A” student!
You cannot claim to be an “A” student until you receive an “A” grade.
Example 2.3.17: Jumping to the Conclusion Jumping to the conclusion happens when some premises are missing.
Drake and Rihanna have been seen together in public.
Therefore Drake and Rihanna are dating.
30 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS Example 2.3.18: Converse Error Show that the following argument is invalid.
If Zeke is a cheater, then Zeke sits in the back of the classroom.
Zeke is sitting in the back of the classroom.
Therefore Zeke is a cheater.
Let’s assign some variables b, c to the above statements c : “Zeke is a cheater” b : “Zeke sits in the back” Then we have the following truth table Premise 1 Premise 2 Conclusion c b c ⇒b b c T T T T T T F F F T T T F F F T F This is called the “converse error” beause it implicitly assumes that q ⇒p is logically equivalent to p ⇒q, which is not the case.
Example 2.3.19: Inverse Error Show that the following argument is invalid.
If this polygon P is a square, then it has four sides.
P is not a square.
Therefore P does not have four sides.
There are plenty of 4-sided polygons that are not squares, so already this argument seems problematic. Let’s assign some variables s, f to the above statements s : “P is a square” f : “P has four sides” Then we have the following truth table Premise 1 Premise 2 Conclusion s f s ⇒f ¬s ¬f T T T F T F F F T T F F F T T F This is called the “inverse error” because it implicitly assumes that ¬p ⇒¬q is logically equivalent to p ⇒q, which is not the case.
2.3. VALID AND INVALID ARGUMENTS 31 2.3.4 Sound Arguments Definition An argument is called sound if is it valid and the premises are all actually true. An argument is unsound otherwise.
The above is more of a philosophical distinction than one detectable in a truth table. For example, consider the two following arguments If an animal is a fluffy god, then it has fur.
An animal does not have fur.
Therefore it is not a dog.
If a potato is green, then it is from Mars.
A potato is not from Mars.
Therefore that is not green.
Both of the above arguments are examples of modus tollens and are thusly valid. However, the one on the left is sound (ignore the pedantry of “hair vs. fur”, but the one on the right is not – any discussion of Martian potatoes is pretty outlandish and absurd at present.
Remark. Don’t eat green potatoes. Not only are they almost certainly not from Mars, they carry a high risk of solanine poisoning.
32 CHAPTER 2. THE LOGIC OF COMPOUND STATEMENTS Chapter 3 The Logic of Quantified Statements 3.1 Predicates and Quantified Statements I Definition: predicate, domain A predicate is a sentence that contains a finite number of variables and becomes a statement when specific values are substituted for the variables. The domain of a predicate is the collection of values that may be substituted in place of the variable(s).
Example 3.1.1 Let P(x) be the predicate x2 > x and let D be the domain R - the set of real numbers. Assess the truth values of the following statements: P(−1), P(1), P(10).
P(−1): (−1)2 = 1 > −1 True P(1): (1)2 = 1 ̸> 1 False P(10): (10)2 = 100 > 10 True Example 3.1.2 Let P(x, y) be the predicate y ≥x and let D be the domain R × R (that is, x and y are both real numbers). Assess the truth values of the following statements: P(0, 1), P(1, 1), P(1, 0.9).
P(0, 1): 1 ≥0 True P(1, 1): 1 ≥1 True P(1, 0.9): 0.9 ≥1 False Definition: truth set If P(x) is a predicate and x has domain D, then the truth set of P(x) is the set of all elements of D that make P(x) true when they are substituted for x. The truth set of P(x) is denoted {x ∈D | P(x)} .
The symbol ∈is short for “in” in English.
Definition The following short-hand notation is used for some commonly-occuring sets.
33 34 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS N The natural numbers: 0, 1, 2, 3, . . .
Z The integers: . . . , −2, −1, 0, 1, 2, . . .
Z+ The positive integers: 1, 2, 3, 4, . . .
Q The rational numbers (i.e. all possible fractions) R The real numbers Remark. Some people take the convention that 0 is not a natural number, and some take the convention that that 0 is a positive integer. These competing conventions are commonplace, and it’s not often a big deal in practice if an author doesn’t make their particular conventions explicit at the onset. Nevertheless, mathematicians are human and need something to argue about, so your instructor will staunchy insist that anyone who doesn’t adhere to his conventions is patently wrong.
Example 3.1.3 Let P(x) be the predicate x2 < 10. Find the truth set for P when the domain D is ...
a. ... N.
b. ... Z.
c. ... Z+.
d. ... R.
a. The natural numbers x which make P(x) true are {0, 1, 2, 3} b. The integers x which make P(x) true are {−3, −2, −1, 0, 1, 2, 3} c. The positive integers Z+ which make P(x) true are {1, 2, 3} d. The real numbers R which make P(x) true are {x ∈R | −3 < x < 3} 3.1.1 The Universal Quantifier: ∀ Definition: universal statement, counterexample Let P(x) be a predicate and let D be the domain of x. A universal statement is a statement of the form For every x in D, P(x) is true. (In symbols, ∀x ∈D, P(x)) The universal statement ∀x ∈D, P(x) is true if and only if it is true for every x in D. The universal statement is false if there is at least one x′ in D where P(x′) is false. Such an x′ is called a counterexample.
Example 3.1.4 Let P(x) be the predicate x2 ≥x. Determine whether or not the universal statements are true or false.
1. ∀x ∈Z, P(x) 2. ∀x ∈R, P(x) 3.1. PREDICATES AND QUANTIFIED STATEMENTS I 35 1. This is true.
2. This is false. When x = 1 2, then x2 = 1 4 ̸≥x, so x = 1 2 is a counterexample. In fact, any x in the interval 0 < x < 1 will be a counter-example.
Example 3.1.5: Method of Exhaustion Let D be the following set of prime numbers D = {29, 41, 47, 53, 59} and let P(x) be the predicate “x divided by 6 has a remainder of 5.” Is the universal statement ∀x ∈D, P(x) true?
Statements about prime numbers are often nontrivial, but since we only have a few, we can check them all explicitly: 29 = 4(6) + 5 41 = 6(6) + 5 47 = 7(6) + 5 53 = 8(6) + 5 59 = 9(6) + 5 so the universal statement is true.
3.1.2 The Existential Quantifier: ∃ Definition: existential statement Let P(x) be a predicate and let D be the domain of x. An existential statement is a statement of the form There exists some x in D for which P(x) is true. (In symbols, ∃x ∈D, P(x)) The existential statement ∃x ∈D, P(x) is true if and only if it is true for at least one x in D.
It is false if and only if it is false for every x in D.
Example 3.1.6 Let P(x) be the predicate x2 < x. Determine whether or not the existential statements are true or false.
1. ∃x ∈R, P(x) 2. ∃x ∈Z, P(x) 1. This is true. When x = 1 2, then x2 = 1 4 < 1 2 = x.
2. This is false. We don’t yet have a means of proving it, but you can be convinced by 36 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS comparing the graphs of y = x and y = x2 (where x, y are integers).
−2 −1 1 2 −4 −2 2 4 In English, of course, there are numerous ways in which we can communicate these quantifiers.
Below is an incomplete list of such things.
Quantifiers and common expressions ∀ ∃ “for each” “for some” “for all” “there is” “for arbitrary” “there exists” “for any” “at least one” “for every” “can find a” Remark. It’s also worth noting that, in symbolic logic, the quantifiers are always written first, but in English the quantifier may come at the end. For example “x2 ≥0 for every x ∈R.” 3.1.3 Universal Conditional Statement Definition: universal conditional statement Let P(x), Q(x) be predicates with the same domain D. A universal conditional statement is a statement of the form For every x in D, if P(x) is true, then Q(x) is true.
Symbolically, this is ∀x ∈D, P(x) = ⇒Q(x).
and it is true if and only if P(x) = ⇒Q(x) is true for every x in the domain D. It is false if and only if it is false for at least one x in D (which occurs when P(x) is true and Q(x) is false).
3.1. PREDICATES AND QUANTIFIED STATEMENTS I 37 Example 3.1.7 Which of the following conditionals are true for the domain R?
1. x2 > 4 = ⇒x > 2 2. x2 > 4 ⇐ ⇒|x| > 2 For simplicity we use the following notation: P(x) “x2 > 4” Q(x) “x > 2” R(x) “|x| > 2” 1. x2 > 4 = ⇒x > 2 is false. This universal conditional has the form ∀x ∈R, P(x) = ⇒Q(x).
But P(−3) is true and Q(−3) is false, so x = −3 is a counterexample.
2. x2 > 4 ⇐ ⇒|x| > 2 is true. This universal conditional has the form ∀x ∈R, P(x) ⇐ ⇒R(x).
3.1.4 Implicit Quantification As is often the case, many times the quantifier is not ever stated explicitly, which we refer to as implicit quantification. It is up to you, the reader, to correctly determine which quantifier applies here.
Example 3.1.8: Implicit Quantification “The sum of even integers is even.” “For all integers x and for all integers y, if x and y are even, then x + y is even.” ∀x ∈Z, ∀y ∈Z, x even ∧y even = ⇒(x + y) even Example 3.1.9: Sentences: Informal to Formal For each of the following statements, identify the predicate(s), domain(s), and variable(s).
Then rewrite the sentence formally using logical symbols and quantifiers.
1. Whenever an integer is non-zero, its square is positive.
2. Every integer is even if its square is even.
3. a2 + 2 = 6 for some integer a.
4. There’s at least one ghost in this classroom right now.
1. Whenever an integer is non-zero, its square is positive.
38 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS P(x): x ̸= 0 Q(x): x2 > 0 D: Z With the above notation, this sentence is written ∀x ∈Z, P(x) = ⇒Q(x).
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Alternatively, one could take P(x): x2 > 0 D: nonzero integers in which case the sentence is written ∀x ∈D, P(x).
2. Every integer is even if its square is even.
P(x): x is even Q(x): x2 is even D: Z With the above notation, this sentence can be written ∀x ∈Z, Q(x) = ⇒P(x).
3. a2 + 2 = 6 for some integer a.
P(x): x2 + 2 = 6 D: Z With the above notation, this sentence can be written ∃a ∈Z, P(x).
4. There’s at least one ghost in this classroom right now.
P(x): x is in this classroom right now D: Ghosts With the above notation, this sentence can be written ∃a ∈D P(x).
3.1. PREDICATES AND QUANTIFIED STATEMENTS I 39 Exercise 3.1.10: Sentences: Informal to Formal For each of the following statements, identify the predicate(s), domain(s), and variable(s).
Then rewrite the sentence formally using logical symbols and quantifiers.
1. Some people have tattoos.
2. Among all basketball players, some are tall.
3. Somebody in your group likes the orange Starburst.
4. There is an even prime number.
5. No Tech student has class on Sundays.
6. Integers are also real numbers.
7. All dogs go to heaven.
8. If a real number is rational, then so is its multiplicative inverse.
9. The sum of even integers is even.
10. Any factor of 4 is also a factor of 8.
11. John likes the taste of every Starburst.
12. For any Starburst flavor, there’s someone out there who likes the taste it.
3.1.5 Relationship between ∀and ∧; Relationship between ∃and ∨ Suppose D = {x1, x2, x3, . . . , xn} is a finite domain and P(x) is some predicate. Using the method of exhaustion, one can verify the claim ∀x ∈D, P(x) by checking that P(x1), P(x2), ..., and P(xn) are all true. In other words ∀x ∈D, P(x) ≡P(x1) ∧P(x2) ∧· · · ∧P(xn) Similarly, by exhaustion, one can verify the claim ∃x ∈D, P(x) by checking that at least one of P(x1), P(x2), ..., or P(xn) is true. In other words ∃x ∈D, P(x) ≡P(x1) ∨P(x2) ∨· · · ∨P(xn) Example 3.1.11 Rewrite the following statements in terms of ∧and ∨.
1. Every integer x with −1 ≤x ≤1 satisfies x3 = x.
2. There is a natural number x < 4 for which x3 = x.
In both of these, let P(x) be the statement x3 = x.
1. ∀x ∈{−1, 0, 1}, P(x) ≡P(−1) ∧P(0) ∧(P1) 2. ∃x ∈{0, 1, 2, 3}, P(x) ≡P(0) ∨P(1) ∨P(2) ∨P(3) 40 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS 3.2 Predicates and Quantified Statements II 3.2.1 Negating Universal and Existential Statements Recall that the universal statement statement ∀x ∈D, P(x) is false when there is some x in D (that is, one or more) where P(x) is false.
Thus, its negation ¬(∀x ∈D, P(x)) is true when there is some x in D where P(x) is false.
This observation (and the equivalent one when negating an existential statement) yields the following: Theorem 3.2.1: Negation of Universal/Existential Statements Let P(x) be a predicate with domain D. Then we have the following: ¬ (∀x ∈D, P(x)) ≡∃x ∈D, ¬P(x) ¬ (∃x ∈D, P(x)) ≡∀x ∈D, ¬P(x) Example 3.2.2 Negate the following statements.
1. All cats have wings.
2. y2 = −7 for some integer y.
3. x2 > 1 for all real numbers x.
4. No mathematicians are interesting.
1. The universally quantified object is “cats” and the predicate regards the number of wings, so the negation is Some cats have no wings.
2. The existentially quantified object is “y” and the predicate is y2 = −7, so the negation is y2 ̸= −7 for all integers y.
3. The universally quantified object is “x” and the predicate is x2 > 1, so the negation is x2 ≤1 for some real number x.
3.2.2 Conditionals - Related Conditionals and Negations There are, of course, universal conditionals that are related to the usual universal conditional ∀x ∈D, P(x) = ⇒Q(x).
3.2. PREDICATES AND QUANTIFIED STATEMENTS II 41 Definition: Related Universal Conditionals Let P(x), Q(x) be predicates with domain D and consider the universal conditional statement ∀x ∈D, P(x) = ⇒Q(x). The contrapositive is ∀x ∈D, ¬Q(x) = ⇒¬P(x).
The converse is ∀x ∈D, Q(x) = ⇒P(x).
The inverse is ∀x ∈D, ¬P(x) = ⇒¬Q(x).
Example 3.2.3 Write down the contrapositive, converse, and inverse of the following statement: For all pairs of integers, x and y, if x is even and y is even then x + y is even.
Let Z2 denote pairs of integers. Define the following predicates P(x, y): “x is even and y is even.” Q(x, y): “x + y is even.” The statement is then written symbolically as ∀(x, y) ∈Z2, P(x, y) = ⇒Q(x, y).
Now we have (Contrapositive). Symbolically the contrapositive is ∀(x, y) ∈Z2, ¬Q(x, y) = ⇒ ¬P(x, y), which can be written in plain English as For all pairs of integers, x and y, if x + y is not even then x and y are not both even.
(Converse). Symbolically the converse is ∀(x, y) ∈Z2, Q(x, y) = ⇒P(x, y), which can be written in plain English as For all pairs of integers, x and y, if x + y is even then x and y are both even.
(Inverse). Symbolically the converse is ∀(x, y) ∈Z2, Q(x, y) = ⇒P(x, y), which can be written in plain English as For all pairs of integers, x and y, if x and y are not both even then x + y is not even.
42 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS Proposition 3.2.4: Negation of Universal Conditional Statement Let P(x), Q(x) be predicates with domain D. Then we have the following: ¬ (∀x ∈D, P(x) = ⇒Q(x)) ≡∃x ∈D, P(x) ∧¬Q(x).
Proof.
¬ (∀x ∈D, P(x) = ⇒Q(x)) ≡∃x ∈D, ¬ (P(x) = ⇒Q(x)) (??) ≡∃x ∈D, ¬ (¬P(x) ∨Q(x)) (??) ≡∃x ∈D, ¬¬P(x) ∧¬Q(x) (DeMorgan’s Law) ≡∃x ∈D, P(x) ∨¬Q(x) (Double Negative Law) Example 3.2.5 Negate the following conditional statements: 1. ∀x, if x < −1, then x2 > 1 2. Whenever VT students attend a football game, they sing “Enter Sandman.” 1. We first begin by recalling that the negation of a > b is a ≤b. Now, ¬ ∀x, if x < −1, then x2 > 1 ≡∃x, ¬ if x < −1, then x2 > 1 ≡∃x, x < −1 and ¬ x2 > 1 ≡∃x, x < −1 and x2 ≤1 2. First we have to interpret this symbolically to understand the (universal) conditional.
This statement is about all VT students (by Hokie Law, you are required to know the lyrics of Enter Sandman...). If any student attends a football game, then they must sing “Enter Sandman”. More symbolically, one might write ∀student∈VT, student attends football game = ⇒student sings “Enter Sandman”.
So the negation is ∃student∈VT, student attends football game land student does not sing “Enter Sandman”.
Written in plain English, one would probably write something like There is a VT student who attends a foodball game and doesn not sing “Enter Sandman.” 3.3. STATEMENTS WITH MULTIPLE QUANTIFIERS 43 3.3 Statements with Multiple Quantifiers As is often the case in math, quantified statements may involve multiple variables with multiple quantifiers. Anyone who has already taken a calculus class and has seen the formal definition of a limit has experienced this.
Definition of Limit of a Function A function f has a limit L at x = a if it has the following property: ∀ε ∈R+, ∃δ ∈R+, ∀x ∈R, 0 < |x −a| < δ = ⇒|f(x) −L| < ε Definition: nested quantifiers A logical statement with multiple variables and multiple quantifiers is said to have nested quantifiers.
3.3.1 ∀∀Statements Recall that, for a finite domain D = {d1, . . . , dn}, we have ∀x ∈D, P(x) ≡P(d1) ∧P(d2) ∧· · · ∧P(dn) Example 3.3.1 Let Dx = {x1, x2} and Dy = {y1, y2}. Rewrite the following doubly-quantified statements ∀x ∈Dx, ∀y ∈Dy, P(x, y) ∀y ∈Dy, ∀x ∈Dx, P(x, y) using only logical connectives ∧, ∨, ¬.
44 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS ∀x(∀yP(x, y)) ≡∀x(P(x, y1) ∧P(x, y2)) ≡ P(x1, y1) ∧P(x1, y2) ∧ P(x2, y1) ∧P(x2, y2) which by commutativity and associativity can be rewritten as ≡ P(x1, y1) ∧P(x2, y1) ∧ P(x1, y2) ∧P(x2, y2) ≡∀x(P(x, y1) ∧P(x, y2)) ≡∀y(∀xP(x, y)) Example 3.3.2 Write the following sentence formally and rearrange the order of the quantifiers. Then translate this back into plain English.
Every student must do all homework problems.
Symbolically, the above sentence is ∀x ∈{students}, ∀y ∈{homework problems}, x must do y.
Changing the order ∀y ∈{homework problems}, ∀y ∈{students}, y must do x which translates to All homework problems must be done by everyone and this has the same meaning as the original statement.
Nested Universal Quantifiers Let P(x, y) be a predicate involving two variables from domains Dx and Dy, respectively.
∀x ∈Dx, ∀y ∈Dy, P(x, y) ≡∀y ∈Dy, ∀x ∈Dx, P(x, y) 3.3.2 ∃∃Statements Recall that, for a finite domain D = {d1, . . . , dn}, we have ∃x ∈D, P(x) ≡P(d1) ∨P(d2) ∨· · · ∨P(dn) 3.3. STATEMENTS WITH MULTIPLE QUANTIFIERS 45 Example 3.3.3 Let Dx = {x1, x2} and Dy = {y1, y2}. Rewrite the following doubly-quantified statements ∃x ∈Dx, ∃y ∈Dy, P(x, y) ∃y ∈Dy, ∃x ∈Dx, P(x, y) using only the logical connectives.
∃x(∃yP(x, y)) ≡∃x(P(x, y1) ∨P(x, y2)) ≡ P(x1, y1) ∨P(x1, y2) ∨ P(x2, y1) ∨P(x2, y2) which by commutativity and associativity can be rewritten as ≡ P(x1, y1) ∨P(x2, y1) ∨ P(x1, y2) ∨P(x2, y2) ≡∃x(P(x, y1) ∨P(x, y2)) ≡∃y(∃xP(x, y)) Example 3.3.4 Write the following sentence formally and rearrange the order of the quantifiers. Then write the resulting sentence back in plain English.
Some kid is stealing cookies from one of the boxes.
Symbolically, the above sentence is ∃x ∈{kids}, ∃y ∈{boxes of cookies}, x stole from y.
Changing the order ∃y ∈{boxes of cookies}, ∃y ∈{kids}, y stole from x.
which translates to There is a box from which some kid stole cookies.
and this has the same meaning as the original statement.
Nested Existential Quantifiers Let P(x, y) be a predicate involving two variables from domains Dx and Dy, respectively.
∃x ∈Dx, ∃y ∈Dy, P(x, y) ≡∃y ∈Dy, ∃x ∈Dx, P(x, y) 46 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS 3.3.3 ∀∃and ∃∀Statements Example 3.3.5 Let Dx = {x1, x2} and Dy = {y1, y2}. Rewrite the following doubly-quantified statements ∀x ∈Dx, ∃y ∈Dy, P(x, y) ∃y ∈Dy, ∀x ∈Dx, P(x, y) using only the logical connectives. Are these logically equivalent?
∀x(∃yP(x, y)) ≡ P(x1, y1) ∨P(x1, y2) ∧ P(x2, y1) ∨P(x2, y2) ∃y(∀xP(x, y)) ≡ P(x1, y1) ∧P(x2, y1) ∨ P(x1, y2) ∧P(x2, y2) They do not appear to be logically equivalent. This can be seen looking at a truth table.
P(x1, y1) P(x1, y2) P(x2, y1) P(x2, y2) ∀x∃yP(x, y) ∃y∀xP(x, y) .
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T F F T T F .
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Example 3.3.6 Write the following symbolically, rearrange the quantifiers, and translate it back into plain English. How do the quantifiers change the meaning?
Everybody is good at something.
In symbols, this would read ∀x ∈{people}, ∃y ∈{things}, x is good at y.
Switching the order of the quantifiers, we get ∃y ∈{things}, ∀x ∈{people}, x is good at y.
which translates to There is one thing that everyone is good at (i.e., everyone is good at the same thing).
and this sentence is not equivalent to the original sentence.
3.3. STATEMENTS WITH MULTIPLE QUANTIFIERS 47 Nested Quantifiers of Mixed Type Let P(x, y) be a predicate involving two variables from domains Dx and Dy, respectively.
∀x ∈Dx, ∃y ∈Dy, P(x, y) ̸≡∃y ∈Dy, ∀x ∈Dx, P(x, y) ∃x ∈Dx, ∀y ∈Dy, P(x, y) ̸≡∀y ∈Dy, ∃x ∈Dx, P(x, y) Remark. To be clear, even switching the placement of the variables results in inequivalent statements. In other words, none of ∀x∃yP(x, y), ∃y∀xP(x, y), ∀y∃xP(x, y), or ∃x∀yP(x, y) are equivalent. Below is the full truth table using the same domains above.
∀x∃y ∃y∀x ∀y∃x ∃x∀y P(x1, y1) P(x1, y2) P(x2, y1) P(x2, y2) P(x, y) P(x, y) P(x, y) P(x, y) T T T T T T T T (1) T T T F T T T T (2) T T F T T T T T (3) T T F F F F T T (4) T F T T T T T T (5) T F T F T T F F (6) T F F T T F T F (7) T F F F F F F F (8) F T T T T T T T (9) F T T F T F T F (10) F T F T T T F F (11) F T F F F F F F (12) F F T T F F T T (13) F F T F F F F F (14) F F F T F F F F (15) F F F F F F F F (16) One then sees that ∀x∃yP(x, y) ̸≡∃y∀xP(x, y) by row (7) ∀x∃yP(x, y) ̸≡∀y∃xP(x, y) by row (4) ∀x∃yP(x, y) ̸≡∃x∀yP(x, y) by row (4) ∃y∀xP(x, y) ̸≡∀y∃xP(x, y) by row (4) ∃y∀xP(x, y) ̸≡∃x∀yP(x, y) by row (4) ∀y∃xP(x, y) ̸≡∃x∀yP(x, y) by row (7) Example 3.3.7 The following table encodes the truth values of P(x, y) for all ordered pairs of x = 1, . . . , 5 and y = 1, . . . , 5.
48 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS P(x, y) x = 1 x = 2 x = 3 x = 4 x = 5 y = 1 T F T F F y = 2 T F T T T y = 3 F F T F F y = 4 T T T T T y = 5 F T T F T Determine which of the following are true: (a) ∀x, ∃y, P(x, y) (b) ∃x, ∀y, P(x, y) (c) ∀y, ∃x, P(x, y) (d) ∃y, ∀x, P(x, y) (a) ∀x, ∃y, P(x, y).
One should think about this statement as a game: If I hand you any x-value, can you find a y-value making P(x, y) true? Yes, for example: P(1, 1), P(2, 4), P(3, 3), P(4, 4), P(5, 2) (we remark that, for each x, the above choices of y are not unique possible choices).
Thus the statement is true.
(b) ∃x, ∀y, P(x, y) One should think about this statement as a game: Can you find a single x-value, so that P(x, y) is always true no matter what y is? Yes, for example: P(3, 1), P(3, 2), P(3, 3), P(3, 4), P(3, 5) Thus the statment is true.
(c) ∀y, ∃x, P(x, y) One should think about this statement as a game: If I hand you any y-value, can you find a x-value making P(x, y) true? Yes, for example: P(3, 1), P(1, 2), P(3, 3), P(2, 4), P(5, 5).
(we remark that, for each y, the above choices of x are not unique possible choices).
Thus the statement is true.
(d) ∃y, ∀x, P(x, y) One should think about this statement as a game: Can you find a single y-value, so that P(x, y) is always true no matter what x is? Yes, for example: P(1, 4), P(2, 4), P(3, 4), P(4, 4), P(5, 4).
Thus the statement is true.
3.3. STATEMENTS WITH MULTIPLE QUANTIFIERS 49 Example 3.3.8 The following table encodes the truth values of P(x, y) and Q(x, y) for all ordered pairs of x = 1, 2, 3 and y = 1, 2, 3.
P(x, y) x = 1 x = 2 x = 3 y = 1 T F T y = 2 F T F y = 3 T F T Q(x, y) x = 1 x = 2 x = 3 y = 1 T F T y = 2 T T F y = 3 F T T Determine which of the following are true: (a) ∀x, ∃y, P(x, y) = ⇒Q(x, y) (b) ∃x, ∀y, P(x, y) = ⇒Q(x, y) (c) ∀y, ∃x, P(x, y) = ⇒Q(x, y) (d) ∃y, ∀x, P(x, y) = ⇒Q(x, y) Recall that a conditional statement p = ⇒q is only false when p is true and q is false. So for simplicity, we begin by making a single table corresponding to the conditional P(x, y) = ⇒ Q(x, y). That is, we look at the conditional P(x, y) = ⇒Q(x, y) for every possible pair of (x, y).
P(x, y) = ⇒Q(x, y) x = 1 x = 2 x = 3 y = 1 T T T y = 2 T T T y = 3 F T T And now we’ve reduced this problem to something like Example 3.3.7 INCOMPLETE Exercise 3.3.9 Three people, Carey, Yuen, and Tim, are going to eat at a buffet. The buffet has four main areas: salads, main courses, desserts, and beverages. The figure below shows which person (P) selected which item (I) from each area (A).
50 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS Carey Yuen Tim Desserts pie cake Salads Main Courses Beverages green salad fruit salad soda coffee milk fish spaghetti Rewrite each of the following statements informally and find its truth value.
1. ∃I, ∀P, P chose I.
2. ∃P, ∀I, P chose I.
3. ∃P, ∀A, ∃I in A, P chose I.
4. ∀P, ∀A, ∃I in A, P chose I.
Theorem 3.3.10: Negations of Multiple Quantifiers Let P(x, y) be a predicate and Dx, Dy the domains for x, y, respectively. Then ¬ (∀x ∈Dx, ∃y ∈Dy, P(x, y)) ≡∃x ∈Dx, ∀y ∈Dy, ¬P(x, y) ¬ (∃x ∈Dx, ∀y ∈Dy, P(x, y)) ≡∀x ∈Dx, ∃y ∈Dy, ¬P(x, y) Example 3.3.11 Negate each of the following statements.
1. ∀x ∈R, ∃y ∈R such that xy < 0.
2. ∀ε ∈R+, ∃δ ∈R+, ∀x ∈R, 0 < |x −a| < δ = ⇒|f(x) −L| < ε .
1. ∃x ∈R such that ∀y ∈R, xy ≥0.
2. ∃ε ∈{positive reals} such that ∀δ ∈{positive reals}, 0 < |x−a| < δ and |f(x)−L| ≥ε.
3.4. ARGUMENTS WITH QUANTIFIED STATEMENTS 51 3.4 Arguments with Quantified Statements 3.4.1 Rules of inference with quantifiers Definition: arbitrary, particular An element x of a domain D is called arbitrary if it is indistinguishable from all other elements in the domain. x is called particular if it has some additional properties not shared by the other elements in the domain.
Remark. In a proof, you will never see the word “particular,” but they are usually pretty easy to identify as there are usually other conditions that are satisfied. For example, if the domain is Z, then you might see something like “take x = 1” or “Let x be an integer greater than 2.” Arbitrary elements are sometimes labeled as arbitrary, usually with a phrase like “arbitrary” or “fixed, but arbitrary.” However, it is often the case that the arbitrary-ness is only communicated implicitly.
Theorem 3.4.1: Quantified Rules of Inference Let P(x) be a predicate. The following are all valid logical argument forms.
(a) universal instantiation c is an element (arbitrary or particular) ∀x ∈D, P(x) ∴ P(c) (b) universal generalization c is an arbitrary element P(c) ∴ ∀x ∈D, P(x) (c) existential instantiation ∃x ∈D, P(x) ∴ (c is a particular element) ∧P(c) (d) existential generalization c is an element (arbitrary or particular) P(c) ∴ ∃x ∈D, P(x) Remark. Instantiation allows us remove the quantifiers and focus on a single statement P(c). We can then apply rules of inference to this single statement.
Generalization allows us to add quantifiers to a single statement P(c).
We should be a bit more verbose and explanatory about why each of the above works.
52 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS Universal instantiation. If P(x) is true for every x in the domain, then we can pick a single element c from the domain and it follows that P(c) must be true. We may choose c arbitrarily or not – it doesn’t matter because c is in the domain and P(x) is true for every x in the domain.
Universal generalization. Since an arbitrary element c is indistinguishable from all other elements, proving P(c) represents the exact same strategy that we would use to prove P(x) for every x ∈D. Thus we can conclude that our singular proof is enough to conclude ∀x ∈D, P(x). Note that c being arbitrary is necessary: if c is particular and satisfies some additional conditions not shared by the rest of the domain, we can’t actually conclude that P(x) holds for every x in the domain.
Existential instantiation. If P(x) is true for some x in the domain, then there must be some element c for which P(c) is true. However, such a c is probably pretty special (P(x) may be true for only a handful of domain elements), so c is necessarily particular.
Existential generalization. If we can show that there is an element c (either arbitrary or particular – it doesn’t matter) for which P(c) is true, then we can claim there is an element in the domain for which the predicate evaluates to true. Hence ∃x ∈D, P(x).
Example 3.4.2: Universal Modus Ponens Let P and Q be predicates with domain D. Use the Rules of Inference to see that the following argument is valid.
∀x ∈D, P(x) = ⇒Q(x) ∃x ∈D, P(x) ∴ ∃x ∈D, Q(x) Proof.
1.
∀x ∈D, P(x) = ⇒Q(x) (Hypothesis) 2.
∃x ∈D, P(x) (Hypothesis) 3.
c ∈D is particular (Existential Instantiation, 2) 4.
P(c) (Existential Instantiation, 2) 5.
P(c) = ⇒Q(c) (Universal Instantiation, 1 & 3) 6.
Q(c) (Modus Ponens, 4 & 5) 7.
∃x ∈D, Q(x) (Existential Generalization, 3 & 6) Example 3.4.3: Universal Modus Tollens Let P and Q be predicates with domain D. Use the Rules of Inference to see that the following argument is valid.
∀x ∈D, P(x) = ⇒Q(x) ∃x ∈D, ¬Q(x) ∴ ∃x ∈D, ¬P(x) 3.4. ARGUMENTS WITH QUANTIFIED STATEMENTS 53 Proof.
1.
∀x ∈D, P(x) = ⇒Q(x) (Hypothesis) 2.
∃x ∈D, ¬Q(x) (Hypothesis) 3.
c ∈D is particular (Existential Instantiation, 2) 4.
¬Q(c) (Existential Instantiation, 2) 5.
P(c) = ⇒Q(c) (Universal Instantiation, 1 & 3) 6.
¬P(c) (Modus Tollens, 4 & 5) 7.
∃x ∈D, ¬P(x) (Existential Generalization, 3 & 6) Example 3.4.4: Proof by Cases Let P, Q, R be predicates with domain D. Use the Rules of Inference to see that the following argument is valid.
∀x ∈D, (P(x) ∨Q(x)) ∀x ∈D, (P(x) = ⇒R(x)) ∀x ∈D, (Q(x) = ⇒R(x)) ∴ ∀x ∈D, R(x) We begin by applying universal instantiation to each of the premises.
c is arbitrary P(c) ∨Q(c) P(c) = ⇒R(c) Q(c) = ⇒R(c) Now, each of P(c), Q(c), R(c) is a statement and is definitively true or false. We can apply the conditional identity of Table of Logical Equivalances and rewrite some of the hypotheses.
c is arbitrary P(c) ∨Q(c) ¬P(c) ∨R(c) ¬Q(c) ∨R(c) We combine two of the hypotheses using the resolution rule of inference c is arbitrary Q(c) ∨R(c) ¬Q(c) ∨R(c) We combine two of the hypotheses using the resolution rule of inference c is arbitrary R(c) ∨R(c) Now we use the idempotent law to rewrite our premises as 54 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS c is arbitrary R(c) And finally some universal generalization to write ∀x ∈D, R(x) Now we’ll write down this sequence of steps in a logical proof.
Proof.
1.
∀x ∈D, (P(x) ∨Q(x)) (Hypothesis) 2.
∀x ∈D, (P(x) = ⇒R(x)) (Hypothesis) 3.
∀x ∈D, (Q(x) = ⇒R(x)) (Hypothesis) 4.
c ∈D is arbitrary (defining an arbitrary variable) 5.
P(c) ∨Q(c) (Universal Instantiation, 1 & 4) 6.
P(c) = ⇒R(c) (Universal Instantiation, 2 & 4) 7.
Q(c) = ⇒R(c) (Universal Instantiation, 3 & 4) 8.
¬P(c) ∨R(c) (Conditional Identity, 6) 9.
̸= Q(c) ∨R(c) (Conditional Identity, 7) 10.
Q(c) ∨R(c) (Resolution, 5 & 8) 11.
R(c) ∨R(c) (Resolution, 9 & 10) 12.
R(c) (Idempotent Law, 11) 13.
∀x ∈D, R(x) (Universal Generalization, 4 & 12) 3.4.2 Logical Proofs with Nested Quantifiers The same rules of inference work for nested quantifiers, applied successively left-to-right. But since order matters, we need to think very carefully about our instantiated variables.
Example 3.4.5: Mixed Type Quantifiers and Instantiation Explain the subtle (but very important) difference in the proofs below.
1.
∃x∀y, P(x, y) (Hypothesis) 1.
∀y∃x, P(x, y) (Hypothesis) 2.
c is particular (∃Inst., 1) 2.
d is arbitrary (def. of variable) 3.
∀y, P(c, y) (∃Inst., 1) 3.
∃x, P(x, d) (∀Inst., 1 & 2) 4.
d is arbitrary (def. of variable) 4.
c is particular (∃Inst., 3) 5.
P(c, d) (∀Inst., 3 & 4) 5.
P(c, d) (∃Inst., 3) .
.
.
.
.
.
3.4. ARGUMENTS WITH QUANTIFIED STATEMENTS 55 The difference between the two proofs is in the variable c.
• In the proof on the left, c is chosen so that P(c, y) is true, regardless of y. This means that c’s definition is independendent of d.
• In the proof on the right, c is chosen so that P(c, d) is true, which means that c’s definition is very much dependent upon d.a aAlthough it would make notation more cumbersome, it would be appropriate to write c(d) to stress the fact that c is a particular function of d. This is actually not uncommon in mathematics literature.
3.4.3 An Actual Math Proof Let’s see a logical proof in context.
Example 3.4.6: “If x is even, then x2 is even.” For simplicity, all variables below have domain Z.
Let E(x) be the predicate “∃k, x = 2k.” Use the rules of inference to see that the following argument is valid.
∀x∀y, [xy ∈Z] (closure of multiplication) ∀x∀y∀z∀w, [w = x(yz) = ⇒w = (xy)z] (associativity of multiplication) ∀x∀y∀z, [(x = y) = ⇒(xz = yz)] (multiplicative property of equality) ∴ ∀x, E(x) = ⇒E(x2) Proof. In what follows, all variables have domain Z.
1.
∀x∀y, [xy ∈Z] (hypothesis) 2.
∀x∀y, ∀z∀w[w = x(yz) = ⇒w = (xy)z] (hypothesis) 3.
∀x∀y∀z, [(x = y) = ⇒(xz = yz)] (hypothesis) 4.
n is arbitrary integer (element definition) 5.
k is arbitrary integer (element definition) 6.
(n = 2k) = ⇒ n2 = (2k)n (universal instantiation, 3) 7.
n2 = (2k)n = ⇒ n2 = 2(kn) (universal instantiation, 2) 8.
(n = 2k) = ⇒ n2 = 2(kn) (transitivity, 7 & 8) 9.
kn ∈Z (universal instantiation, 1) 10.
(n = 2k) = ⇒∃z n2 = 2z (existential generalization, 9) 11.
∃y (n = 2y) = ⇒∃z n2 = 2z (existential generalization, 10) 12.
∀x ∃y (x = 2y) = ⇒∃z x2 = 2z (universal generalization, 11) and this final line is precisely the statement ∀x, E(x) = ⇒E(x2).
56 CHAPTER 3. THE LOGIC OF QUANTIFIED STATEMENTS Chapter 4 Elementary Number Theory and Methods of Proof 4.1 Direct Proof and Counterexample I: Introduction 4.1.1 Context/Relation to Formal Proofs Definition: theorems,propositions,lemmas,corollaries A theorem or proposition or lemma or corollary is a statement that requires proof.
Remark. In math, the different naming is really just to convey nome heirarchy as to importance of results. For example • Theorems – these are the big, important results.
• Propositions – these are big results, but not as important as theorems.
• Lemmas – these are results (usually fairly technical) proven to aid in the proof of Theorems/Propositions.
• Corollaries – these are interesting results that follow immediately from one of the above and usually require almost no proof.
Remark. To avoid confusion with “proposition” as used synonymously with “statement,” we will not use that term in this class. In fact, we’ll default to “Theorem” for everything we prove.
Most theorems are simply stated as quantified statements of one of the following forms ∀x, P(x) ∀x, P(x) = ⇒Q(x) ∃x, P(x) but are really arguments of the following forms (hypotheses) ∴ ∀x, P(x) (hypotheses) ∴ ∀x, P(x) = ⇒Q(x) (hypotheses) ∴ ∃x, P(x) where the hypotheses are things like definitions, rules of arithmetic, etc.
Look again at Example 3.4.6 from the end of last class: Theorem 4.1.1 For all integers x, if x is even, then x2 is even.
Letting E(x) be the predicate “x is even,” the argument of the proof was more formally ∀x∀y, [xy ∈Z] (closure of multiplication) ∀x∀y∀z∀w, [w = x(yz) = ⇒w = (xy)z] (associativity of multiplication) ∀x∀y∀z, [(x = y) = ⇒(xz = yz)] (multiplicative property of equality) ∴ ∀x, E(x) = ⇒E(x2) 57 58 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF and the proof was Proof. In what follows, all variables have domain Z.
1.
∀x∀y, [xy ∈Z] (hypothesis) 2.
∀x∀y, ∀z∀w[w = x(yz) = ⇒w = (xy)z] (hypothesis) 3.
∀x∀y∀z, [(x = y) = ⇒(xz = yz)] (hypothesis) 4.
n is arbitrary integer (element definition) 5.
k is arbitrary integer (element definition) 6.
(n = 2k) = ⇒ n2 = (2k)n (universal instantiation, 3) 7.
n2 = (2k)n = ⇒ n2 = 2(kn) (universal instantiation, 2) 8.
(n = 2k) = ⇒ n2 = 2(kn) (transitivity, 7 & 8) 9.
kn ∈Z (universal instantiation, 1) 10.
(n = 2k) = ⇒∃z n2 = 2z (existential generalization, 9) 11.
∃y (n = 2y) = ⇒∃z n2 = 2z (existential generalization, 10) 12.
∀x ∃y (x = 2y) = ⇒∃z x2 = 2z (universal generalization, 11) 12.
∀x ∃y (x = 2y) = ⇒∃z x2 = 2z (universal generalization, 11) 12.
∀x, (x = 2y) = ⇒∃z x2 = 2z (universal generalization, 11) 13.
∀x, E(x) = ⇒E(x2) (rewriting, 12) Our goal is going to be to write proofs like this in a human-readable way.
4.1.2 Important Hypotheses and Definitions Generally we’ll assume all of the usual rules of arithmetic; see Appendix A of the course textbook.
Some specific ones that we’ll give names to are the following: Important Arithmetic Hypotheses 1. The integers are closed under addition. (The sum of two integers is again an integer.) Stated formally, ∀x ∈Z, ∀y ∈Z, [x + y ∈Z] 2. The integers are closed under multiplication. (The product of two integers is again an integer.) ∀x ∈Z, ∀y ∈Z, [xy ∈Z] Definition: even and odd integers An integer is said to be even precisely when it is twice another integer. An integer is said to be odd if and only if it is 1 more than twice another integer. Symbolically, 4.1. DIRECT PROOF AND COUNTEREXAMPLE I: INTRODUCTION 59 n is even ⇐ ⇒ ∃k ∈Z such that n = 2k.
n is odd ⇐ ⇒ ∃k ∈Z such that n = 2k + 1.
Remark. You may freely regard “odd” as meaning “not even.” Since even/oddness is an existential statement, any proof requires only finding a single integer k.
Example 4.1.2 Use the definitions of even and odd to justify the following statements.
1. 0 is even.
2. −401 is odd.
3. If a, b are integers, 6a2b is even.
1. Proof. Choosing k = 0, we have that 0 = 2(0) = 2k. Therefore 0 is even.
2. Proof. Choosing k = −201, we have that −401 = 2(−201) + 1 = 2k + 1 =. Therefore −401 is odd.
3. Notice that this statement nas multiple quantifiers ∀a ∈Z, ∀b ∈Z, ∃k ∈Z such that 6a2b = k.
Proof. Let a, b ∈Z and choose k = 3a2b. As the integers are closed under multiplication, k is an integer. We have that 6a2b = 2(3a2b) = 2k, therefore 6a2b is even.
Definition: prime and composite numbers An positive integer p is prime if and only if p > 1 and for any integers m, n with p = mn, then one of m or n is p. An integer is composite if and only if p > 1 and there are integers 1 < m, n < p for which p = mn. Symbolically, p is prime ⇐ ⇒ p > 1 and ∀m, n ∈Z+, if p = mn then m = p or n = p.
p is composite ⇐ ⇒ p > 1 and ∃m, n ∈Z+ such that 1 < m, n < p and p = mn.
Remark. “Composite” and “not prime” are not actually negations of each other. “Not prime” allows for the possibility that p = 1.
Remark. You may assume that both the m and n in the definition of prime satisfy 1 ≤m, n ≤p.
We’ll prove in a later section that this has to happen.
Example 4.1.3 Use the definitions of prime and composite to justify each of the following statements 1. 1 is not prime.
2. 2468 is composite.
3. 5 is prime.
60 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF 1.
Proof. By definition, a prime number is strictly greater than 1. Therefore 1 is not prime.
2.
Proof. Choose m = 2 and n = 1234, which satisfies 1 < m < n < 2468. Since 2468 > 1 and 2468 = 2(1234) = mn, then 2468 is a composite number.
3. Using the remark above, we have to show that the only possible integers m and n for which p = mn are m = 1, n = 5 or m = 5, n = 1. We can do this by checking all possible pairs with 1 ≤m, n ≤p.
Proof. Look at the following table.
mn m = 1 m = 2 m = 3 m = 4 m = 5 n = 1 1 2 3 4 5 n = 2 2 4 6 8 10 n = 3 3 6 9 12 15 n = 4 4 8 12 16 20 n = 5 5 10 15 20 25 4.1.3 Proofs – Scaffolding You should think of writing a proof as something akin to writing an essay. You need to organize your thoughts and scaffold it into a proof. Here is a simple list.
1. Restate the theorem symbolically You want to make sure you appropriately parse the theorem to understand the quantifiers, predicates, any conditionals, etc.
2. State relevant definitions 3. State your assumptions and list any variables (are they arbitrary or particular?) 4. State your goal (what are you trying to deduce from your assumptions?) 5. Scratch work/algebra 6. Write the proof 4.1. DIRECT PROOF AND COUNTEREXAMPLE I: INTRODUCTION 61 4.1.4 Proving an Existential Statement Proving a Existential Statement 1. Express the statement to be proved in the form ∃x ∈D, P(x).
2. Let x0 be a fixed particular element of the domain. This is for you to choose.
3. Show that P(x0) is true by using definitions and previously established rules.
4. Existential generalization allows us to conclude that the original existential proposition is true.
Example 4.1.4 Prove the following theorem.
Theorem. There is an odd composite number.
Our scratch work 1. Restate the theorem symbolically ∃x ∈Z+, x is odd ∧x is composite.
2. State relevant definitions x0 is “odd” iff that we can find an integer k so that x0 = 2k + 1.
x0 is “composite” iff we can find two integers m, n with x0 = mn and 1 < m, n < x0.
3. State your assumptions and list any variables (are they arbitrary or particu-lar?) Particular integers: x0 = 9, k = 4, m = 3, n = 3.
4. State your goal (what are you trying to deduce from your assumptions?) Want to show that x0 is odd using k, and x0 is composite using m, n above.
5. Scratch work/algebra x0 = 2(4) + 1 x0 = (3)(3) Proof. Choose x0 = 9, an integer. Since we can write x0 = 2(4) + 1 and 4 ∈Z, then 9 is odd. Since we can write x0 = 9 = (3)(3), and 3 ∈Z satisfies 1 < 3 < 9, then x0 is also composite.
Definition: Constructive and Nonconstructive Proofs of Existence A constructive proof of existence involves 1. finding an x in our domain for which Q(x) is true or 2. giving an set of directions for finding such an x in the domain.
A nonconstructive proof of existence involves showing either: 1. the existence of a value x that makes Q(x) true is guaranteed by an axiom or a previously 62 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF proved theorem 2. the assumption that there is no such x leads to a contradiction.
You’ve probably seen a non-constructive proof before; below is one such proof.
Example 4.1.5: non-constructive proof Prove the claim: The polynomial p(x) = x5 −x + 1 has a real root.
Proof. Recall from calculus that polynomials are continuous functions. Since p(−2) = −29 and p(1) = 1 and −29 < 0 < 1, then by the Intermediate Value Theorem, there is a real number x ∈(−2, 1) for which p(x) = 0.
4.1.5 Disproving Universal Statements Recall the negation ¬(∀x, P(x)) is logically equivalent to ∃x, ¬P(x).
Disproof by Counterexample: ∀x ∈D, P(x) 1. Pick a particular c in the domain.
2. Verify P(c) is false.(c is a counterexample) 3. By existential generalization, conclude ∃x, ¬P(x).
4. By logical equivalence, we conclude that ∀x, P(x) is false.
Recall the negation ¬(∀x, P(x) = ⇒Q(x)) is logically equivalent to ∃x, P(x) ∧¬Q(x).
Disproof by Counterexample: ∀x ∈D, P(x) = ⇒Q(x) 1. Pick a particular c in the domain.
2. Verify P(c) is true.
3. Verify Q(c) is false.(c is a counterexample) 4. By existential generalization, conclude ∃x, P(x) ∧¬Q(x).
5. By logical equivalence, we conclude that ∀x, P(x) is false.
4.1. DIRECT PROOF AND COUNTEREXAMPLE I: INTRODUCTION 63 Example 4.1.6 Prove or disprove the following claim: For all real numbers a, b, if a < b then a2 < b2.
Our scratch work 1. Restate the theorem symbolically INCOMPLETE 2. State relevant definitions INCOMPLETE 3. State your assumptions and list any variables (are they arbitrary or particu-lar?) INCOMPLETE 4. State your goal (what are you trying to deduce from your assumptions?) INCOMPLETE 5. Scratch work/algebra INCOMPLETE Disproof. Let a = −10 and b = 1. Then a = −10 < 1 = b, but a2 = 100 ≥1 = b2.
4.1.6 Proving a Universal Statement We’ve already seen the Method of Exhaustion before, which works well for small finite domains. In practice Direct Proof of Universal Statement ∀x, P(x) 1. Let c be a fixed, but arbitrary, element of the domain.
2. Show that P(c) is true by using definitions and previously established rules.
3. By universal generalization, conclude that ∀x ∈D, P(x).
Remark. Item 2 is doing a lot of heavy lifting here - that’s absolutely the hard part and there’s no one-size-fits-all strategy for doing it – this is where you, the human, have to think.
Direct Proof of Universal Statement ∀x, P(x) = ⇒Q(x) 1. Let c be a fixed, but arbitrary, element of the domain.
2. Suppose that P(c) is true.
3. Show that the conclusion Q(c) is true by using definitions and previously established rules.
4. By universal generalization, conclude that ∀x ∈D, P(x) = ⇒Q(x).
Remark. If c is arbitrary, it’s entirely possible that P(c) is false. This is okay – P(c) = ⇒Q(c) is vacuously true. As such, we lose nothing in the way of generality by assuming in Item 2 that P(c) is true.
64 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Example 4.1.7 Prove the statement: The sum of two even integers is even.
Our scratch work 1. Restate the theorem symbolically INCOMPLETE 2. State relevant definitions INCOMPLETE 3. State your assumptions and list any variables (are they arbitrary or particu-lar?) INCOMPLETE 4. State your goal (what are you trying to deduce from your assumptions?) INCOMPLETE 5. Scratch work/algebra INCOMPLETE We first acknowledge that, symbolically, this statement is ∀x ∈Z, ∀y ∈Z if x is even and y is even, then x + y is even. For scratch work, recalling the definition of even and odd ??, we know that, if x, y are even, there are integers k, ℓfor which x = 2k and y = 2ℓ= ⇒x + y = 2k + 2ℓ= 2(k + ℓ) and these rearrangement rules give us the clue into the proof.
Proof. Suppose x and y are arbitrary even integers. Then (by definition) there exist integers k and ℓsuch that x = 2k and y = 2ℓ. It follows then that x + y = 2k + 2ℓ= 2(k + ℓ). Since the integers are closed under addition, k + ℓis an integer, and therefore x + y is an even integer.
4.1.7 Disproving an Existential Statement Recall the negation ¬(∃x, P(x)) is logically equivalent to ∀x, ¬P(x) Disproving an Existential: ∃x ∈D, P(x) 1. Let c be a fixed, but arbitrary, element of D.
2. Prove ¬P(c) 3. By universal generalization, conclude that ∀x, ¬P(x).
4. By logical equivalence, conclude ∃x, P(x) is false.
4.1. DIRECT PROOF AND COUNTEREXAMPLE I: INTRODUCTION 65 Example 4.1.8 Show that the following statement is false: There is a positive integer n such that n2 + 3n + 2 is prime.
Our scratch work 1. Restate the theorem symbolically INCOMPLETE 2. State relevant definitions INCOMPLETE 3. State your assumptions and list any variables (are they arbitrary or particu-lar?) INCOMPLETE 4. State your goal (what are you trying to deduce from your assumptions?) INCOMPLETE 5. Scratch work/algebra INCOMPLETE Proving that this statement is false is equivalent to proving that its negation is true. The negation of this statement is For every positive integer n, n2 + 3n + 2 is composite (i.e. not prime).
We begin with some scratch work. Notice that we can factor this quadratic n2 + 3n + 2 = (n + 1)(n + 2) and neither of the factors are 1.
Proof. Let n be an arbitrary positive integer. Then n2 + 3n + 2 = (n + 1)(n + 2) and since n > 0 (by definition), then neither n + 1 = 1 nor n + 2 = 1. Therefore n2 + 3n + 2 cannot be a prime number.
4.1.8 Proof by Cases NOTE TO INSTRUCTOR: Proof By Cases is technically handled in Section 4.6, so this entire subsection has been copied there. In future iterations of these notes, delete this subsection.
Recall that (P(x) ∨Q(x)) = ⇒R(x) is logically equivalent to P(x) = ⇒R(x) ∧ P(x) = ⇒R(x) .
66 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Proof by Cases: ∀x ∈D, [P(x) ∨Q(x) = ⇒R(x)] 1. Let c be a fixed, but arbitrary element of D.
2. Case 1: (a) Assume P(c) is true.
(b) Prove R(c) is true.
3. Case 2: (a) Assume Q(c) is true.
(b) Prove R(c) is true.
4. By logical equivalence, conclude (P(c) ∨Q(c)) = ⇒R(c).
5. By universal generalization, conclude that ∀x, (P(x) ∨Q(x)) = ⇒R(x).
Remark. P(x) and Q(x) do not have to be distinct/disjoint!
If P(x) is “x is prime” and Q(x) is “x is even”, then there will be overlap when x = 2.
Remark. You may have to come up with P(x) and Q(x) on your own.
Remark. You may have to come up with far more than 2 cases. Famously, Kenneth Appel and Wolfgang Haken proved the Four Color Theorem using more than 1500 cases. (Neil Robertson has since proven that 663 cases is sufficient.) Example 4.1.9 Prove that for all integers n, n2 −3n is even.
Our scratch work 1. Restate the theorem symbolically INCOMPLETE 2. State relevant definitions INCOMPLETE 3. State your assumptions and list any variables (are they arbitrary or particu-lar?) INCOMPLETE 4. State your goal (what are you trying to deduce from your assumptions?) INCOMPLETE 5. Scratch work/algebra INCOMPLETE It’s not presently clear that n2 −3n becomes twice some integer, but maybe if we know more about n we can say more. Let’s do some scratch work.
n = 2k = ⇒ n2 −3n = (2k)2 −3(2k) = 2(2k2 −3k) n = 2k + 1 = ⇒ n2 −3n = (2k + 1)2 −3(2k + 1) = 4k2 + 4k + 1 −6k −3 = 4k2 −2k −2 = 2(2k2 −k −2) Proof. Suppose that n is an even integer. Then there exists another integer k for which n = 2k.
4.1. DIRECT PROOF AND COUNTEREXAMPLE I: INTRODUCTION 67 It follows that n2 −3n = 2(2k2 −3k).
Since Z is closed under addition and multiplication, 2k2 −3k is an integer. Therefore n2 −3n is even.
Suppose now that n is an odd integer. Then there exists another integer ℓfor which n = 2ℓ.
It follows that n2 −3n = 2(2k2 −k −2).
Since Z is closed under addition and multiplication, 2k2 −k −2 is an integer. Therefore n2 −3n is even.
We conclude that, regardless of the parity of n, n2 −3n is an even integer.
4.1.9 Direct Proof with Nested Quantifiers Proving a Statement with Nested Quantifiers: ∀x, ∃y, P(x, y) 1. Let x0 be a fixed arbitrary element of the domain Dx.
2. Pick y0 in the domain Dy. Your choice of y0 will likely be a function of x0.
3. Show that the conclusion P(x0, y0) is true by using definitions and previously established rules.
4. Existential/Universal generalization allows us to conclude that the original universal statement is true.
Example 4.1.10 Prove the following theorem.
Theorem. For every real number x ̸= 1, there is some real number y for which xy = x + y.
Scratch work.
Symbolically we can write ∀x ∈R, ∃y ∈R [(x ̸= 1) = ⇒(xy = x + y)] although it may be beneficial for the purposes of writing the proof to write ∀x ∈R [(x ̸= 1) = ⇒∃y ∈R(xy = x + y)] We try out some scratch work and manipulate our target equation to see if we can figure out how to choose y.
xy = x + y = ⇒xy −y = x = ⇒(x −1)y = x = ⇒y = x x −1 From here we probably have enough to write our proof.
68 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Proof. Let x be an arbitrary real number and suppose that x ̸= 1. Since x ̸= 1, then x−1 ̸= 0, so we choose y = x x −1. Now we have that xy = x2 x −1 and x + y = x + x x −1 = x(x −1) x −1 + x x −1 = x(x −1) + x x −1 = x2 −x + x x −1 = x2 x −1 therefore xy = x + y.
Exercise 4.1.11 Prove the following theorem.
Theorem. For all positive real numbers a, b, c, if a + b > c, a + c > b, and b + c > a there is a triangle △ABC in the Cartesian plane with side lengths a, b, c.
I don’t expect you to prove this, but it’s interesting to think about what this problem looks like symbolically and figure out how we might approach this.
∀a ∈R+, ∀b ∈R+, ∀c ∈R+ (a + b > c) ∧(a + c > b) ∧(b + c > a) = ⇒∃△ABC So we’ll let a, b, c be arbitrary, but fixed, and we’ll try to construct △ABC. Now, we can always think about rotating and translating the triangle in the plane, so we might as well assume that one vertex is at the origin and another vertex is along the x-axis.
C B A θ b a c Figure 4.1: The general setup for the triangle we construct After this setup, it comes down to finding the particular θ so that the distance between A and B has length c. That such a θ exists require some more thought, but this seems like a decent strategy to start with.
4.1. DIRECT PROOF AND COUNTEREXAMPLE I: INTRODUCTION 69 Proof. Let a, b, c be arbitrary positive integers, and suppose these numbers satisfy the following inequalities: a + b > c, a + c > b, and b + c > a Without loss of generality, let’s assume that a ≥b. We place the points C(0, 0), B(a, 0), and A(b cos θ, b sin θ) in the plane, where θ is some real number between 0 and π. In this way, the length of BC is a, the length of AC is b. All that’s left is to find the particular θ-value for which AB has length c. Notice that the distance between points A and B is given by the distance function d(θ) = q (b −a cos θ)2 + b2 sin2 θ.
This function is continuous on the interval [0, π]. In this interval, d satisfies a −b ≤d(θ) ≤a + b The first assumed inequality shows that c < a + b, and the third assumed inequality rearranges to show that a −b < c. Combining these two, we see that a −b < c < a + b so by the Intermediate Value Theorem, there is a value of θ in (0, π) for which d(θ) = c, as desired.
The proof above is kind of interesting because, although we tried to explicitly construct the triangle, the existence of the triangle was actually shown non-explicitly (the IVT does not tell you what the value is; just that it exists.
You could also have used the Law of Cosines to get the value of θ specifically: θ = cos−1 c2 −a2 −b2 2ab Proving a Statement with Nested Quantifiers: ∃x∀y, P(x, y) 1. Pick x0 as a particular element from the domain Dx.
2. Let y0 be a fixed, but arbitrary, element in the domain Dy. Your choice of y0 cannot depend on x0.
3. Show that the conclusion P(x0, y0) is true by using definitions and previously established rules.
4. Existential/Universal generalization allows us to conclude that the original universal statement is true.
Example 4.1.12 Prove the following claim.
70 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Theorem. f(x) = 1 x2 + 1 is a bounded function. That is, there exists some positive real number M so that |f(x)| ≤M for all x ∈R.
Scratch work.
Symbolically this statement says ∃M ∈R, ∀x ∈R, 1 x2 + 1 ≤M Proof. Take M = 1 and let x be an arbitrary real number. We know that x2 ≥0, so we have x2 ≥0 x2 + 1 ≥1 1 x2 + 1 ≤1 1 = 1 We also note that 1 x2+1 is always positive, hence is equal to its own absolute value. Therefore, for all x, 1 x2 + 1 ≤M.
4.2. DIRECT PROOF AND COUNTEREXAMPLE II: WRITING ADVICE 71 4.2 Direct Proof and Counterexample II: Writing Advice Guidelines for Writing Proofs 1. Begin by writing the statement which you wish to prove.
2. Identify the domain, hypotheses and conclusion.
3. Clearly mark the beginning of your proof with the word Proof.
4. Make your proof self-contained.
5. Write your proof in complete, grammatically correct sentences. In particular, do not use symbols to replace simple words in sentences (“Todd is happy and Tammi is sad” vs. “Todd is happy ∧Tammi is sad.”) 6. Keep your reader informed about the status of each statement in your proof.
7. Give a reason for each assertion in your proof.
8. Include connecting words and phrases to make your argument clear.
9. Display equations and inequalities.
10. Make sure that the conclusion has been explicitly shown.
11. Clearly mark the end of your proof, usually with a symbol like □or ■.
(Note that if you use LaTeX, then typing \begin{proof}...\end{proof} will handle items 3 and 11 for you.
Note that the above should not be taken as a rigid set of instructions, but more as a guide line.
Professional mathematicians will often skip item 10, for example, when the statement of the theorem immediately precedes the proof.
4.2.1 Common Mistakes in Proof-Writing Example 4.2.1: Arguing from examples Prove: The sum of any two even integers is even.
Proof. This is true because if m = 14 and n = 6, which are both even, then m + n = 20, which is also even.
Example 4.2.2: Using the same letter to mean two different things Prove: the sum of any two even integers is even.
Proof. Let m and n be fixed, but arbitrary, even integers. Then m = 2k and n = 2k for some integer k.
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72 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Example 4.2.3: Jumping to a conclusion Prove: The sum of any two even integers is even.
Proof. Let m and n be fixed, but arbitrary, even integers. Then by the definition of even, m = 2k1 and n = 2k2 for some integers k1 and k2. Thus, m + n = 2k1 + 2k2. So m + n is even.
Example 4.2.4: Circular reasoning Prove: The product of odd integers is odd.
Proof. Suppose m and n are odd integers. When any odd integers are multiplied, their product is odd. Hence mn is odd.
Example 4.2.5: Confusion between what is known and what is still to be shown Prove: The product of two odd integers is odd.
Proof. Suppose m and n are any odd integers. We must show that mn is odd. This means that there exists an integer s such that mn = 2s + 1.
Also by the definition of odd, there exist integers a and b such that m = 2a + 1 and n = 2b + 1.
Then mn = (2a + 1)(2b + 1) = 2s + 1.
So, since s is an integer, mn is odd by definition of odd.
This following issues are not serious on their own, but each reflects imprecise thinking that sometimes leads to problems later in a proof.
Example 4.2.6: Use of any rather than some Prove: The square of any odd integer is odd.
Proof. Suppose m is a fixed, but arbitrary, odd integer. By definition of odd, m = 2a + 1 for any integer a.
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Example 4.2.7: Use of if rather than because or since Prove: The square of any odd integer is odd.
Proof. Suppose m is a fixed, but arbitrary, chosen odd integer. If m is odd, m = 2a + 1 for any integer a. .
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4.3. DIRECT PROOF AND COUNTEREXAMPLE III: RATIONAL NUMBERS 73 4.3 Direct Proof and Counterexample III: Rational Num-bers Definition: rational numbers A real number r is a rational number if and only if, there are integer a, b with b ̸= 0 such that r = a b.
r is rational ⇐ ⇒ ∃p ∈Z and ∃q ∈Z such that r = p/q and q ̸= 0.
A real number that is not rational is called irrational.
Remark. The set of rational numbers is typically denoted Q, and the irrational numbers are denoted R \ Q or R −Q.
Remark. The name rational number comes from the fact that we can write a rational number as a ratio of integers.
Although this is contained in the properties of real numbers per Appendix A, we’ll state it in an attempt to be self-contained: Zero Product Property Z, Q, R have the following property: If x and y are both nonzero numbers, then xy is nonzero.
The contrapositive of this statement is: If the product of numbers xy = 0, then at least one of x and y is zero.
Example 4.3.1: Properties of Q Prove the following theorem.
Theorem. Q has the following properties: Q is closed under addition. The sum of two rational numbers is a ra-tional number..
Q is closed under multiplication. The product of two rational numbers is a rational number.
Proof. Let x, y be arbitrary rational numbers. By definition, we must have integers a1, a2 and nonzero integers b1, b2 such that x = a1 b1 and y = a2 b2 .
Q is closed under addition. Notice that x + y = a1 b1 + a2 b2 = a1b2 b1b2 + a2b1 b1b2 = a1b2 + a2b1 b1b2 .
74 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Since the integers are closed under addition and multiplication, then a1b2 + a2b1 and b1b2 are integers. Since b1, b2 are nonzero, then b1b2 is also nonzero by the zero product property for Z. Thus x + y is rational.
Q is closed under multiplication. Left as an exercise for the reader.
Example 4.3.2 Prove the following theorem.
Theorem. Every integer is a rational number.
In order to prove this, we begin by noting that this statement can be written in logical symbols as ∀z ∈Z, z is rational.
which again becomes ∀z ∈Z, ∃a ∈Z, ∃b ∈Z, such thatz = a b and b ̸= 0 Now we know that this is a universal statement, so we need to let z be fixed but arbitrary and then show that we can choose a and b appropriately.
Proof. Let x be an arbitrary integer, and choose integers a = x and b = 1. Then we have that x = x 1 = a b so x is a rational number, by definition.
Example 4.3.3 Prove or disprove the following claim.
Claim. Given any rational number x, there is an integer y for which xy is an integer.
Symbolically, this statement is ∀x ∈Q, ∃y ∈Q, such that xy ∈Z which means that x is arbitrary and we get to choose y (in a way that probably relies on x).
Proof. Let x be an arbitrary rational number. By definition, we can write x = a b where a, b ∈Z and b ̸= 0. Choose y = b. With this choice, we have that xy = a b (b) = a 1 = a and a is an integer, whence xy is an integer.
4.3. DIRECT PROOF AND COUNTEREXAMPLE III: RATIONAL NUMBERS 75 Alternate proof idea.
Proof. Let x be an arbitrary rational number and choose y = 0. Then we have that xy = x(0) = 0 and since 0 is an integer, xy is an integer.
Example 4.3.4 Prove or disprove the following claim.
Claim. The irrational numbers are closed under addition.
Symbolically, this statement is ∀x ∈R −Q, ∀y ∈R −Q, x −y ∈R −Q Disproof. Choose x = π and y = −π, two irrational numbers. Then x + y = π −π = 0, an integer, hence we’ve found a counterexample to the claim.
76 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF 4.4 Direct Proof and Counterexample IV: Divisibility Definition: divides Let n, d be integers. We say that d divides n if d ̸= 0 and there exists some integer k for which n = dk. Notationally, we write this as d | n.
Other phrases: • n is divisible by d.
• n is a multiple of d.
• d is a divisor of n.
• d is a factor of n.
Remark. Do not confuse the “divides” sign with a fraction. d | n is equivalent to saying that n d is an integer.
Example 4.4.1 Prove or disprove the following claim.
Claim. For all integers a, b, c, if a|b and a|c, then a|(b + c).
Proof. Let a, b, c be arbitrary integers and suppose both that a|b and a|c. By definition of divisibility, we must have that a is nonzero and that there are integers k, ℓfor which b = ak and c = aℓ.
By routine algebraic manipulation, we see that b + c = ak + aℓ= a(k + ℓ) and we note that k + ℓis an integer since Z is closed under addition. Since a ̸= 0, then we must have that a|(b + c).
Example 4.4.2 Prove or disprove the following claim.
Claim. For all integers a, b, c, if a|(b + c), then a|b and a|c.
Disproof. Take a = 4, b = 2 and c = 6.
Example 4.4.3 Prove or disprove the following claim.
Claim. For all integers a, b, c, if a|b and a|(b + c), then a|c.
4.4. DIRECT PROOF AND COUNTEREXAMPLE IV: DIVISIBILITY 77 Proof. Suppose a|b and a|(b + c). Then there are integers k and ℓfor which b = ak and b + c = aℓ. Combining this information and utilizing our algebra skills b + c = aℓ= ⇒c = aℓ−b = aℓ−(ak) = a(ℓ−k) since the integers are closed under addition and multiplication, ℓ−k is an integer. Therefore c is an integer multiple of a, whence a|c as desired.
Example 4.4.4: Transitivity of Division Prove or disprove the following claim.
Claim. For all integers a, b, c, if a|b and b|c, then a|c.
Proof. Let a, b, c ∈Z be arbitrary and suppose that a|b and b|c. By definition of division, there exist integers k, ℓfor which b = ak and c = bℓ. Combined this gives c = b(ak) = a(bk) (since integer multiplication is commutative). Moreover, by closure integer multiplication, bk is an integer, and therefore a|c.
Example 4.4.5 Prove or disprove the following claim.
Claim. For all integers a, b, c, if a|bc, then a|b and a|c.
Disproof, sketch. Take a = 4, b = 2, c = 2.
Example 4.4.6 Prove or disprove the following claim.
Claim. For all integers a, b, c, if a|b and a|bc, then a|c.
Disproof, sketch. Take a = 2, b = 4, c = 3.
Example 4.4.7 Prove or disprove the following claim.
Claim. For all integers a, b, c, d, if a|b and c|d, then ac|(bc + ad).
Proof sketch. b = ak and d = cℓ, so bc + ad = (ak)c + a(cℓ) = ac(k + ℓ) 78 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Lemma 4.4.8 For any positive integer k, 10k −1 is divisible by 9.
Proof. Notice that 10k −1 = (10 −1)(10k−1 + 10k−2 + · · · + 102 + 10 + 1) and 10 −1 = 9.
Theorem 4.4.9: Easy Test for Division by 9 For all positive integers n, n is divisible by 9 if and only if the sum of n’s digits is a multiple of 9.
Proof. Let n be an arbitrary (k + 1)-digit number “akak−1 · · · a2a1a0”, by which we mean that n = ak10k + ak−110k−1 + · · · + a2102 + a110 + a0.
(4.1) Through routine algebraic manipulation (which we’ve tried to color-code for simplicity for the reader) Equation (4.1) can be rewritten as n = ak10k + ak−110k−1 + · · · + a110 + a0 = ak + ak(10k −1) + ak−1 + ak−1(10k−1 −1) + · · · + a1 + a1(10 −1) + a0 = ak(10k −1) + ak−1(10k−1 −1) + · · · + a1(10 −1) + (ak + ak−1 + · · · + a1 + a0) (4.2) Lemma 4.4.8 guarantees the existence of integers m1, . . . , mk for which Equation (4.2) becomes = ak(9mk) + ak−1(9mk−1) + · · · + a1(9m1) + (ak + ak−1 + · · · + a1 + a0) = 9 (akmk + ak−1mk−1 + · · · + a1m1) + (ak + ak−1 + · · · + a1 + a0) (4.3) Case (⇒). Suppose that n is divisible by 9. By Equation (4.3), since n = 9 (akmk + ak−1mk−1 + · · · + a1m1) + (ak + ak−1 + · · · + a1 + a0) and 9 clearly divides 9 (akmk + ak−1mk−1 + · · · + a1m1), then by Example 4.4.3 9 divides (ak + ak−1 + · · · + a1 + a0).
Case (⇐). Suppose that ak + ak−1 + · · · + a1 + a0 is divisible by 9. It follows from the definition that there is some integer m0 for which ak + ak−1 + · · · + a1 + a0 = 9m0.
By substituting this into Equation (4.3), we obtain n = 9(akmk) + 9(ak−1mk−1) + · · · + 9(a1m1) + 9m0 = 9(akmk + ak−1mk−1 + · · · + a1m1 + m0) As Z is closed under addition and multiplication, this complicated expression is just 9 times an integer, whence n must be divisible by 9.
4.4. DIRECT PROOF AND COUNTEREXAMPLE IV: DIVISIBILITY 79 Corollary 4.4.10 A number n is divisible by 3 if and only if the sum of its digits are divisible by 3.
Exercise 4.4.11 Does the same test hold for multiples of 6?
80 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF 4.6 Proof By Cases Recall that (P(x) ∨Q(x)) = ⇒R(x) is logically equivalent to P(x) = ⇒R(x) ∧ P(x) = ⇒R(x) .
Proof by Cases: ∀x ∈D, [P(x) ∨Q(x) = ⇒R(x)] 1. Let c be a fixed, but arbitrary element of D.
2. Case 1: (a) Assume P(c) is true.
(b) Prove R(c) is true.
3. Case 2: (a) Assume Q(c) is true.
(b) Prove R(c) is true.
4. By logical equivalence, conclude (P(c) ∨Q(c)) = ⇒R(c).
5. By universal generalization, conclude that ∀x, (P(x) ∨Q(x)) = ⇒R(x).
Remark. P(x) and Q(x) do not have to be distinct/disjoint!
If P(x) is “x is prime” and Q(x) is “x is even”, then there will be overlap when x = 2.
Remark. You may have to come up with P(x) and Q(x) on your own.
Remark. You may have to come up with far more than 2 cases. Famously, Kenneth Appel and Wolfgang Haken proved the Four Color Theorem using more than 1500 cases. (Neil Robertson has since proven that 663 cases is sufficient.) Exercise 4.6.1 Prove the following theorem.
Theorem. For all integers n, n2 −3n is even.
NOTE TO INSTRUCTOR: In future iterations of these notes, turn this into an example and uncomment work.
Exercise 4.6.2 Prove the following claim.
Theorem. For every integer n, either 3|n2 or 3|(n2 −1).
NOTE TO INSTRUCTOR: In future iterations of these notes, turn this into an example and uncomment work.
4.6. PROOF BY CASES 81 Proof by cases and the absolute value function Definition: absolute value The absolute value of a real number x, denoted |x|, is defined as follows: |x| = ( x when x ≥0 x when x < 0 Because the definition has two cases, often times proofs involving absolute values will require two (or more) cases.
Exercise 4.6.3: triangle inequality Prove the following claim.
Theorem. For all real numbers x and y, |x + y| ≤|x| + |y|.
NOTE TO INSTRUCTOR: In future iterations of these notes, turn this into an example and uncomment work.
82 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF 4.7 Indirect Argument: Contradiction and Contraposition 4.7.1 Contradiction Proof By Contradiction 1. Assume that the claim is false.
2. Show that this assumption leads logically to a contradiction.
3. Conclude that the claim must actually be true.
Recall that the negation of a universal conditional is logically equivalent to ¬ (∀x, P(x) = ⇒Q(c)) ≡∃x, P(x) ∧¬Q(x).
Proof By Contradiction for a Universal Statement ∀x, P(x) ⇒Q(x) 1. Assume that there exists x0 in the domain for which P(x0) is true and Q(x0) is false.
2. Show that this assumption leads logically to a contradiction.
3. Conclude that P(x0) = ⇒Q(x0) must be true.
4. Universal Generalization implies that ∀x, P(x) = ⇒Q(x) must be true.
Remark. Althought it isn’t necessary, it’s often polite to clue the reader in on the fact that you’re about to prove this by contradiction.
Example 4.7.1: Infinity Integers Prove the following claim.
Theorem. There is no largest integer.
Proof. Tending towards a contradiction, suppose that there is a largest integer, call it N.
Choose M = N + 1, another integer. Then we have that M > N. But since N was assumed to be the largest, then we also have that N ≥M. Clearly N cannot be the largest and not the largest integer.
N ≥M > N + 1 Ridiculous. Contradiction.
Therefore there is no largest integer.
Example 4.7.2 Prove the following claim.
Theorem. The sum of any nonzero rational number and irrational number is irrational.
Proof. Suppose there is a rational number m, and an irrational number n for which m + n is rational. Since the rationals are closed under addition and multiplication, it follows that 4.7. INDIRECT ARGUMENT: CONTRADICTION AND CONTRAPOSITION 83 (m + n) −m is rational. But (m + n) −m = n, so this contradicts our initial assumption that n was irrational.
Therefore the sum of any rational number and irrational number is irrational.
Example 4.7.3 Prove or disprove the following claim: Claim. 7p + 21q = 1 for some integers p and q.
Disproof. The claim is false - there are no integers p, q for which 7p + 21q = 1. To see this, suppose to the contrary that there are integers p, q for which this is true. Then we have that 1 = 7p + 21q = 7(p + 3q).
As p + 3q is an integer, then 7 must divide 1. But this contradicts the previously-proven fact that all divisors of 1 must be less than (or equal to) 1.
4.7.2 Contraposition Proof By Contrapositive: ∀x, P(x) = ⇒Q(x).
1. Let x0 be a fixed, but arbitrary, element of the domain.
2. Suppose that Q(x0) is false.
3. Show that P(x0) is false by using definitions and previously established rules.
4. By contrapositive, conclude that P(x0) = ⇒Q(x0).
5. By universal generalization, conclude that ∀x ∈D, P(x) = ⇒Q(x).
Remark. Althought it isn’t necessary, it’s polite to clue the reader in on the fact that you’re about to prove this by contraposition. Maybe with a phrase like “We approach by contraposition” or “We establish the contrapositive.” Example 4.7.4 Prove the following claim.
Theorem. For all integers n, if n2 is even, then n is even.
Notice that the contrapositive is: For all integers n, if n is odd, then n2 is odd. That proof seems to write itself.
Proof. We approach via contraposition. Let n be an arbitrary integer and suppose that n is odd. Then there is some integer k for which n = 2k + 1. Then n2 = (2k + 1)2 = 2(2k2 + 2k) + 1 Since 2k2 + 2k is an integer, then n2 is odd.
84 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF This verifies the claim.
Example 4.7.5 Prove the following claim.
Theorem. For all integers n, if 9 ∤n2, then 6 ∤n.
We approach with the contrapositive. Let n be an integer and suppose that 6|n. Then there is an integer k for which n = 6k. Now, n2 = 36k2 = 9(4k2). As the integers are closed under addition, 4k2 is an integer, whence 9|n2.
Example 4.7.6 Prove the following claim.
Theorem. For all integers m, n, if m + n is even, then either m and n are both even, or m and n are both odd.
Proof. We approach by contraposition. Suppose that one of m and n is even, and the other is odd. We’ll prove only the case that m is even and n is odd (the opposite case is identical).
Then there are integers k and ℓfor which m = 2k and n = 2ℓ+ 1.
Then m + n = 2k + 2ℓ+ 1 = 2(k + ℓ) + 1, and since the integers are closed under addition, k + ℓis an integer, whence m + n is odd.
Contrapositive – Multiple Hypotheses Observe that (p ∧q) = ⇒r ≡ ¬r = ⇒¬(p ∧q) (contrapositive) ≡¬¬r ∨¬(p ∧q) (conditional identity) ≡¬¬r ∨(¬p ∨¬q) (DeMorgan’s Law) ≡(¬¬r ∨¬p) ∨¬q (associativity) ≡¬(¬r ∧p) ∨¬q (DeMorgan’s Law) ≡ (¬r ∧p) = ⇒¬q (conditional identity) or equivalently ≡ (¬r ∧q) = ⇒¬p 4.7. INDIRECT ARGUMENT: CONTRADICTION AND CONTRAPOSITION 85 Proof By Contrapositive: ∀x, (P(x) ∧Q(x)) = ⇒R(x).
1. Let x0 be a fixed, but arbitrary, element of the domain.
2. Suppose that R(x0) is false and P(x0) is true.
3. Show that Q(x0) is false by using definitions and previously established rules.
4. By logical equivalence, conclude that (P(x0) ∧Q(x0)) = ⇒R(x0).
5. By universal generalization, conclude that ∀x ∈D, (P(x) ∧Q(x)) = ⇒R(x).
Example 4.7.7 Prove the following claim.
Theorem. For all integers x, y,z, if both x + z and y + z are even, then x + y is even.
Scratch: Let E(a, b) mean “a + b is even”.
We then have the predicates P = E(x, z), Q = E(y, z), R = E(x, y), and thus the form of the claim above is ∀x∀y∀z [(P ∧Q) = ⇒R] .
We’ll apply the contrapositive and prove ∀x∀y∀z [(¬R ∧Q) = ⇒P] .
Proof. Approaching via contraposition, let x, y, z be arbitrary integers and suppose that x + y is odd. We further suppose that y + z is even, and intend to show that x + z is odd. By definition of even and off, there are integers k, ℓso that x + y = 2k + 1 and y + z = 2ℓ.
Now, using our basic algebra, we have that x + z = x −z + 2z = x + y −y −z + 2z = (x + y) −(y + z) + 2z = (2k + 1) −(2ℓ) + 2z = 2(k −ℓ+ z) + 1.
Since the integers are closed under additional mutiplication, k −ℓ+ z is an integer, and therefore x + z is odd, as desired.
The above can also be proven directly, by noticing that x + y = (x + z) −2z + (z + y) and all of these terms are even.
86 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Exercise 4.7.8 Prove the following claim.
Theorem. Let C be a circle with center O and let ℓbe a line tangent to the circle at a point P. Then ℓis perpendicular to the segment OP.
We begin by looking at the picture below, which will serve as the guide for the proof.
C ℓ β γ α a b c O P Q Proof. Let C be a circle with center O and radius c, and let ℓbe a tangent line, intersecting C at point P. Tending toward a contradiction, suppose the angle formed by OP and ℓis not a right angle. The on one side of OP, the angle must be less than π 2, so let α be this angle.
Now, there must be some other point, Q say, along ℓwhere OQ meets ℓat a right angle; let a be the length of the segment OQ. Form the triangle △OPQ and let γ be the angle at Q.
Q must lie on the same side of c as α, since otherwise the angle sum of △OPQ would be at least γ + (π −α) = π 2 + (π −α) > π 2 + π 2 = π.
Since the angle sum of △OPQ must be π, it follows that γ is the largest angle, whence c > a (this can be observed using the Law of Sines, if you wish). In turn, this ensures that Q lies somewhere in the interior of circle C, and thus ℓmust intersect C in a second point, contradicting the fact that it was a tangent line.
Therefore OP and ℓmust be perpendicular.
4.8. INDIRECT ARGUMENT: Two THREE FAMOUS THEOREMS 87 4.8 Indirect Argument: Two Three Famous Theorems 4.8.1 √ 2 is Irrational To prove Theorem 4.8.2, we’ll first have to prove a helper result about rational numbers.
Lemma 4.8.1: Rational Numbers – Simplified Form For every rational number x, there exist integers a, b with b ̸= 0 such that x = a b and the only positive integer dividing both a and b is 1.
Proof. Let x be an arbitrary rational number. We have three cases: x = 0, x > 0, and x < 0. Since 0 = 0 1, the first case is obvious. We now prove only the case that x > 0, noting that the proof of the negative case follows identically by replacing x with −x.
Suppose that x is positive. There are integers k0, ℓ0 with ℓ0 ̸= 0 such that x = k0 ℓ0 We may presume that k0, ℓ0 are both positive. If they were both negative, we could take x = −k0 −ℓ0 and continue with the remainder of the proof.
If there are no positive integers other than 1 dividing both k0, ℓ0, then choose a = k0 and b = ℓ0.
Otherwise, let d1 > 1 be an integer dividing k0, ℓ0. Then (by definition of division) there are integers k1, ℓ1 such that k0 = d1k1 where 1 ≤k1 < k0 ℓ0 = d1ℓ1 where 1 ≤ℓ1 < ℓ0 and thus x = k0 ℓ0 = d1k1 d1ℓ1 = k1 ℓ1 .
If there are no positive integers other than 1 dividing both k1, ℓ1, then choose a = k1 and b = ℓ1.
Otherwise, let d2 > 1 be an integer dividing k1, ℓ1. Then (by definition of division) there are integers k2, ℓ2 such that k1 = d2k2 where 1 ≤k2 < k1 < k0 ℓ1 = d2ℓ2 where 1 ≤ℓ2 < ℓ1 < ℓ0 and thus x = k0 ℓ0 = k1 ℓ1 = d2k2 d2ℓ2 = k2 ℓ2 .
Continue to repeat this procedure. As there are only finitely many integers x, y satisfying 1 ≤x < k0 and 1 ≤y < ℓ0, this procedure must terminate after only finitely-many steps. Suppose the nth step is the terminal step. Then we take a = kn and b = ℓn.
88 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Theorem 4.8.2: Irrationality of √ 2 √ 2 is irrational.
Proof. Seeking a contradiction, suppose that √ 2 is rational. Then, from Lemma ?? there are integers a, b with b ̸= 0 and a, b not both even such that √ 2 = a b Squaring both sides of this equation yields 2 = a2 b2 and rearranging produces b2 = 2a2 (4.4) whence we see that b2 is even. By Example 4.7.4, it follows that b is an even number, and thus Lemma ?? implies that a is odd.
Since b is even, there is some integer k for which b = 2k. Substituting this into Equation 4.4, we get (2k)2 = 2a2 4k2 = 2a2 2k2 = a2 and thus a2 is even. By Example 4.7.4, it follows that a is an even number, which contradicts the fact that a was odd.
Therefore, we conclude that √ 2 is, in fact, irrational.
4.8.2 The Infinitude of Primes Theorem 4.8.3: Infinitude of Prime Numbers There are infinitely-many prime numbers.
Proof. Tending toward a contradiction, suppose that there are only finitely-many primes, with P the largest prime. Let N be the product of all primes, i.e.
N = 2 · 3 · 5 · 11 · · · · · P Since P is the largest prime, then N + 1 must be composite, and in particular, divisible by one of the primes on the list. Let q be such a prime. Since q is a prime, it is one of the factors of N, so q divides N. By definition of division, we must have integers k and ℓfor which N = qk and N + 1 = qℓ.
We then have that 1 = (N + 1) −N = qℓ−qk = q(ℓ−k).
4.8. INDIRECT ARGUMENT: Two THREE FAMOUS THEOREMS 89 Because 1 and q are both positive, ℓ−k is also positive, and the only positive divisor of 1 is 1. It follows that (ℓ−k) = q = 1 which contradicts our assumption that q is prime. We conclude, then, that there cannot be finitely-many primes.
4.8.3 Area of a Circle Archimedes famously proved that the area of a circle was equal to the familiar πr2 using contradiction and proof by cases.
We first need to know what the Greeks knew.
Lemma 4.8.4 The circumference C of a circle of radius r is given by C = 2πr.
Lemma 4.8.5 The area A of a regular polygon with apothem h and perimeter P is given by A = 1 2hP.
h Figure 4.2: A regular polygon (all edge lengths are the same, all internal angles are the same) with apothem, of length h.
Lemma 4.8.6 A circle can be approximated (to any desired accuracy) by both inscribed regular polygons or circumscribed regular polygons.
h h Figure 4.3: To the left, approximating a circle with an inscribed regular polygon. To the right, approximating the circle with a circumscribed regular polygon. Pay special attention to the relationship between the circle radius and the apothem.
90 CHAPTER 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Theorem 4.8.7: A = πr2 The area of a circle with radius r is given by πr2.
Proof. We first note that, by Lemma 4.8.4 above, πr2 = 1 2rC, where C is the circumference of the circle. Tending toward a contradiction, suppose that the area of the circle is not 1 2rC. There are two cases: either Area(circle) > 1 2rC or Area(circle) < 1 2rC.
Case 1. Suppose that Area(circle) > 1 2rC. Applying Lemma 4.8.6, we find an inscribed regular polygon X satisfying Area(circle) > Area(X) > 1 2rC If X has apothem h and perimiter P, it’s clear that h < r and that P < C (see Figure 4.3), hence Lemma 4.8.5 yields Area(X) = 1 2hP < 1 2rC, which contradicts our assumption.
Case 2. Suppose that Area(circle) < 1 2rC. Applying Lemma 4.8.6, we find a circumscribed regular polygon X satisfying Area(circle) < Area(X) < 1 2rC If X has apothem h and perimiter P, we observe that h > r and that P > C (see Figure 4.3), hence Lemma 4.8.5 yields Area(X) = 1 2hP > 1 2rC, which contradicts our assumption.
Therefore, the area of the circle must be precisely 1 2rC = πr2.
Remark. This proof is certainly convincing, but by modern standards the lemmas are a bit too imprecise to allow one to meet the minimum criteria for a proof.
Chapter 5 Sequences, Mathematical Induction, and Recursion 5.1 Sequences One of the most important tasks in mathematics is to recognize and categorize regular patterns.
Definition: sequence, index A sequence is a (possibly-infinite) ordered list of objects (or events): {a1, a2, .., ak} Each individual element ak is called a (kth) term. The k in “ak” is called the index and indicates its position in the sequence. Notationally, we often write {an}k n=1 or (an)k n=1 to denote a sequence.
Remark. Note that a sequence’s index doesn’t have to start at 1 (in computer science, these often start at 0). One can always perform an index shift to rewrite a sequence starting at k = 1, so there’s not loss of generality in using this convention.
Definition: length of sequene The number of ordered elements in a sequence is called its length. An infinite sequence has an initial term, but continues indefinitely having no final term. A finite sequence has a final term am, where m is some natural number.
Definition: explicit formula An explicit or general formula for a sequence {an} is a rule that shows how the value of ak depends on k.
Example 5.1.1 Write the first four terms of the following sequences.
1. The sequence {ak} where ak = k 10 + k for all integers k ≥1.
2. The sequence {bj} where bj = 5 −j 5 + j for all integers j ≥3.
3. The sequence {ci} where ci = (−1)i 3i for all integers i ≥0.
91 92 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION 1.
1 11, 2 12, 3 13, 4 14 2.
2 8, 1 9, 0 10, −1 11 3.
1 1, −1 3 , 1 9, −1 27 Remark. The third sequence above is known as an alternating sequence.
Example 5.1.2 Given the first few terms of the sequence, find an explicit formula for it.
1. The sequence {an}∞ n=0 given by 1, 1 2, 1 4, 1 8, 1 16, 1 32, 1 64, . . .
2. The sequence {bj}∞ n=2 given by 1, 4, 9, 16, 25, 36, 49, 64, . . .
1. We have 1 = 1 20, 1 2 = 1 21, 1 4 = 1 22, 1 8 = 1 23, 1 16 = 1 24, . . .
a0 a1 a2 a3 a4 .
so the explicit formula appears to be an = 1 2n 2. We have 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52, . . .
b2 b3 b4 b5 b6 .
so the explicit formula appears to be bj = (j −1)2 Definition: sum of a sequence If m and n are integers and m ≤n, the symbol n X k=m ak is the sum of all terms from am to an.
That is, n X k=m ak = am + am+1 + ... + an−1 + an.
Remark. It’s important to observe that one can rewrite sums by “splitting off” some number of terms.
n X k=m ak = am + am+1 + · · · + an−1 + an = am + n X k=m+1 ak !
| {z } split off first term = n−1 X k=m ak !
+ an | {z } split off last term .
This strategy may come in handy when we get to induction.
5.1. SEQUENCES 93 Example 5.1.3 Calculate the following: 1.
3 X k=0 1 2k 2.
2 X j=1 (−1)j j + 1 1.
3 X k=0 1 2k = 1 20 + 1 21 + 1 22 + 1 23 == 15 8 .
2.
2 X j=1 (−1)j j + 1 = Definition: product of a sequence If m and n are integers and m ≤n, the symbol n Y k=m ak is the product of all terms from am to an. That is, n Y k=m ak = am · am+1 · · · an−1 · an.
Remark. It’s important to observe that one can rewrite products by “splitting off” some number of terms.
n Y k=m ak = am · am+1 · · · an−1 · an = am · n Y k=m+1 ak !
| {z } split off first term = n−1 Y k=m ak !
· an | {z } split off last term .
This strategy may come in handy when we get to induction.
Example 5.1.4 Calculate the following: 1.
3 Y k=0 1 2k 2.
3 Y j=1 (−1)j j + 1 1.
3 Y k=0 1 2k = 1 20 · 1 21 · 1 22 1 23 = 1 26 = 1 64 94 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION 2.
3 Y j=1 (−1)j j + 1 = −1 2 · 1 3 · −1 4 = 1 24 Definition: factorial For each positive integer n, the quantity n factorial, denoted n!, is defined to be the product of all integers from 1 to n: n! = n · (n −1) · (n −2) · · · 3 · 2 · 1, where we define 0! = 1.
Remark. It’s important to observe that one can rewrite factorials by “splitting off” some number of terms.
n!
= n · (n −1) · (n −2) · · · 3 · 2 · 1 = n · (n −1)!
.
This strategy may come in handy when we get to induction.
Example 5.1.5 Simplify the following expressions.
1.
5!
2!3!
2. (n + 1)!
n!
3.
n!
(n −3)!
4. n + (n −1) 2!
+ (n −2) 3!
+ (n −3) 4!
+ .... + 1 n!
1.
5!
2!3! = 1 · 2 · 2 · 3 · 4 · 5 1 · 2 · 1 · 2 · 3 = 4 · 5 1 · 2 = 10 2. (n + 1)!
n!
= n 3.
n!
(n −3)! = n(n −2)(n −1) 4. n + (n −1) 2!
+ (n −2) 3!
+ (n −3) 4!
+ .... + 1 n! = 5.1. SEQUENCES 95 Writing Sequences and Summations Explicitly Example 5.1.6 Condense the following using notation from above.
1.
1, −1 4, 1 9, −1 16, 1 25, −1 36 2.
1 n 2 n + 1 3 n + 2 · · · n + 1 2n 3. The sum of the first n odd integers 4. The sum of the first n2 −1 factorials (starting with 1!) 1. {an}∞ n=1 where an = (−1)n+1 1 n2 2.
n Y k=0 k + 1 n + k 3.
n−1 X k=0 2k + 1 4.
n2−1 X k=1 k!
96 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION 5.2 Mathematical Induction I: Proving Formulas Definition: Principle of Mathematical Induction Let P(n) be a predicate that is defined for integers n, and let N0 be a fixed integer. Suppose the following two statements are true.
1. P(N0) is true.
2. For all integers k ≥N0, if P(k) is true, then P(k + 1) is true.
Then P(n) is true for all n ≥N0.
Base P(0) · · · P(N0 −1) P(N0) P(N0 + 1) P(N0 + 2) P(N0 + 3) · · · Figure 5.1: A visual interpretation of mathematical induction. Once you show that P(k) ⇒P(k + 1), and you know that P(N0) is true, then it must be that P(N0 + 1) is true. From this it follows that P(N0 + 2) is true. From this it follows that P(N0 + 3) is true. And so on.
Proof By Induction Consider a statement of the form For every integer n ≥N0, the property P(n) is true.
To show this 1. [Base Step] Show that P(N0) is true.
2. [Inductive Step] Show that, for every integer k ≥N0, if P(k) is true, then P(k + 1) is true. Notice that this is a standard universal conditional. So to actually do it: • Let k be an arbitrary integer with k ≥N0.
• Suppose P(k) is true (this is the “induction hypothesis”).
• Deduce that P(k + 1) must be true.
Example 5.2.1: Sum of the first n integers.
Show that, for every n ≥1, n X i=1 i = 1 + 2 + 3 + 4 + · · · + n = n(n + 1) 2 First we identify the predicate P(n) as meaning n X i=1 i = n(n + 1) 2 . The base case will be to verify that P(1) is true. When we get to the induction step, we 5.2. MATHEMATICAL INDUCTION I: PROVING FORMULAS 97 keep in mind that we want to see that P(k + 1) is true: k+1 X i=1 i = (k + 1)(k + 2) 2 .
Proof. Let the P(n) denote the equation n X i=1 i = n(n + 1) 2 .
Base Step. To see that P(1) is true, note that the left-hand side of the equation is 1 and the right-hand side is 1(1 + 1) 2 = 2 2 = 1. Hence P(1) is true.
Inductive Step. Let k be an arbitrary integer with k ≥1. Suppose that P(k) is true, that is k X i=1 i = 1 + 2 + 3 + 4 + · · · + k = k(k + 1) 2 .
Now, we have that k+1 X i=1 = 1 + 2 + 3 + 4 + · · · + k + (k + 1) = k X i=1 i !
+ (k + 1) = k(k + 1) 2 + (k + 1) (apply induction hypothesis) = k(k + 1) 2 + 2(k + 1) 2 = (k + 1)(k + 2) 2 hence P(k + 1) is true.
Therefore P(n) holds for all n ≥1.
Example 5.2.2: Sum of a geometric sequence.
Let a, r be real numbers with r ̸= 1. A geometric sequence is a sequence of the form a, ar, ar2, ar3, ar4, . . . Show that, for every n ≥0, n X i=0 ari = a(1 −rn+1) 1 −r 98 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Scratch Work.
First we identify the predicate P(n) as meaning n X i=0 ari = a + ar + ar2 + ar3 + · · · + arn = a(1 −rn+1) 1 −r The base case will be to verify that P(0) is true. When we get to the induction step, we keep in mind that we want to see that P(k + 1) is true: k+1 X i=0 ari = a(1 + rk+2) (1 −r) .
Proof. Let P(n) denote the equation n X i=0 ari = a + ar + ar2 + ar3 + · · · + arn = a(1 −rn+1) 1 −r Base Step. To see that P(0) is true, note that the left-hand side of the equation is a and the right-hand side is a(1 −r0+1) 1 −r = a.
Inductive Step. Let k be an arbitrary integer with k ≥0. Suppose that P(k) is true, that is k X i=0 ari = a(1 −rk+1) 1 −r .
Now, we have that k+1 X i=0 ari = a + ar + ar2 + ar3 + · · · + ark + ark+1 = k X i=0 ari !
+ ark+1 = a(1 −rk+1) 1 −r + ark+1 (apply induction hypothesis) = a(1 −rk+1) 1 −r + ark+1(1 −r) 1 −r = a(1 −rk+1) 1 −r + a(rk+1 −rk+2) 1 −r = a(1 −rk+1 + rk+1 −rk+2) 1 −r = a(1 −rk+2) 1 −r hence P(k + 1) is true.
Therefore P(n) holds for all n ≥0.
5.2. MATHEMATICAL INDUCTION I: PROVING FORMULAS 99 Example 5.2.3: Sum of the first n odd integers Prove that, for every n ≥1, n X i=1 (2i −1) = 1 + 3 + 5 + · · · + (2n −1) = n2.
Proof. Let P(n) denote the equation n X i=1 (2i −1) = 1 + 3 + 5 + · · · + (2n −1) = n2.
Base Step. To see that P(1) is true, note that the left-hand side of this equation is 2(1) −1 = 1 and the right-hand side of the equation is (1)2 = 1.
Inductive Step. Let k be an arbitrary integer with k ≥1. Suppose that P(k) is true, that is k X i=1 (2i −1) = k2.
Now, we have that k+1 X i=1 (2i −1) = 1 + 3 + 5 + · · · + (2k −1) + (2(k + 1) −1) = k X i=1 (2i −1) !
+ (2(k + 1) −1) = k2 + (2(k + 1) −1) (apply induction hypothesis) = k2 + 2k + 1 = (k + 1)2 hence P(k + 1) is true.
Therefore P(n) holds for all n ≥1.
Exercise 5.2.4: Trees A tree, T, is a nonempty set of vertices, V , together with a set of edges E ⊂V × V with the following properties: 1. For all v ∈V , the edge (v, v) not in E.
2. Between any two distinct vertices u, w ∈V , there is a unique sequence of vertices v0 = u, v1, v2, . . . , vk−1, vk = w so that the edges (u = v0, v1), (v1, v2), . . . , (vk−1, vk = w) are all contained in E. (Call u, v1, . . . , vk, w a path.) Prove the following claim: If T is a tree with |V | = p and |E| = q, then p = q + 1.
100 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Exercise 5.2.5: Binary tree A tree T is a binary tree if there is one vertex v0 which is adjancet to precisely two vertices (that is, there are exactly two vertices v11 and v12 for which (v0, v11) and (v0, v12) are in E), and every other vertex is adjance to at most three vertices. Notably, a binary tree can always be organized into the following shape.
depth 0 depth 1 depth 2 depth 3 The depth, d, of a vertex v is number of edges in the path/sequence from v0 to v. [See figure above] Prove the following claim: If |V | = n, then n ≤2D + 1, where D is the maximum depth of all verices.
5.3. MATHEMATICAL INDUCTION II: APPLICATION 101 5.3 Mathematical Induction II: Application Example 5.3.1: Proving Divisibility Prove that n3 −n is divisible by 3 for all integers n ≥0.
Proof. Let P(n) denote the sentence “n3 −n is divisible by 3.” Base Step. We first show that P(1) is true. Note that (1)3 −1 = 1 −1 = 0 and 0 is divisible by any nonzero number.
Inductive Step. Let k be an arbitrary integer with k ≥1. Suppose that P(k) is true, i.e. suppose that k3 −k is divisible by 3. By definition of divisiblity, this means that there exists an integer ℓsuch that k3 −k = 3ℓ. Now, we have that (k + 1)3 −(k + 1) = (k3 + 3k2 + 3k + 1) −(k + 1) = k3 −k + 3k2 + 3k = (k3 −k) | {z } apply I.H.
+3k2 + 3k = 3ℓ+ 3k2 + 3k and so this quantity is divisible by 3. Therefore P(k + 1) is true.
Therefore P(n) is true for all n ≥1.
Example 5.3.2: Proving Inequalities Prove that n! > 2n for all integers n ≥4.
Proof. Let P(n) denote the inequality n! > 2n.
Base Step. To see that P(4) is true, note that 4! = 24 > 16 = 24.
Inductive Step. Let k be an arbitrary integer with k ≥4. Suppose that P(k) is true, that is k! > 2k.
We’ll also remark that, since k ≥4, then of course k + 1 ≥3 > 2. Now we have that (k + 1)! = (k!)(k + 1) > 2k(k + 1) (apply induction hypothesis) > 2k(2) (by our remark) = 2k+1 hence P(k + 1) is true.
Therefore, P(n) is true for all n ≥4.
102 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Example 5.3.3: Covering a Board with L-Shaped Trominoes An L-shaped tromino is comprised of three squares arranged in an L-shape, like so: Show that, for all integers n ≥1, any 2n × 2n checkerboard with one square removed can be covered by L-shaped trominoes.
Scratch work n = 1 n = 2 Q1 Q2 Q3 Q4 n = 3 Proof. Let P(n) be the property Any 2n × 2n checkerboard with one square removed can be covered by L-shaped trominoes.
Base Step. To see that P(1) is true, Inductive Step. Let k be an arbitrary integer with k ≥1. Suppose that P(k) is true, that is Any 2k × 2k checkerboard with one square removed can be covered by L-shaped trominoes.
Consider a 2k+1 × 2k+1 checkerboard with one square removed. Notice that this checkerboard is comprised of four 2k ×2k checkerboards. Let Q1 the quadrant containing the missing square and label the other quadrants Q2, Q3, Q4. By the induction hypothesis, quadrant Q1 can be covered by L-shaped trominoes. Now place an L-shaped tromino so that one square lies in each of the remaining quadrants Q2, Q3, Q4 (see figure below).
5.3. MATHEMATICAL INDUCTION II: APPLICATION 103 Q1 Q2 Q3 Q4 2k 2k 2k 2k In this way, the uncovered tiles form three more 2k × 2k checkerboards, each with a single square removed. Applying the induction hypothesis to each of these quadrants, we must have that all of Q2, Q3, Q4 can be covered by L-shaped trominoes. That is to say, P(k + 1) is true.
Therefore P(n) is true for every integer n ≥1.
104 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Exercise 5.3.4: T-shaped tetrominoes An T-shaped tetromino is comprised of four squares arranged in an T-shape, like so: Prove that, for all m ≥1 and for all n ≥1, a 4m × 4n checkerboard can be tiled by T-shaped tetrominoes.
5.4. STRONG MATHEMATICAL INDUCTION AND THE WELL-ORDERING PRINCIPLE FOR THE INT 5.4 Strong Mathematical Induction and the Well-Ordering Principle for the Integers Motivation Suppose you were trying to prove a statement P(n) for all n ≥N0, and you figured out that P(k) = ⇒P(k + 3). What base case(s) would you have to prove?
P(0) · · · P(N0 −1) P(N0) · · · P(N1) P(N1 + 1) P(N1 + 2) · · · Base Figure 5.2: A visual interpretation of strong mathematical induction. Once you show that P(k) ⇒P(k + 3), and you know that P(N0) is true, then it must be that P(N0 + 3) is true. From this it follows that P(N0 + 6) is true. From this it follows that P(N0 + 9) is true. And so on. This, however, does not prove the statement for all integers. You need to also show that P(N0 + 1) and P(N0) + 2) are true to get that.
Suppose instead you were trying to prove a statement P(n) for all n ≥N0, and you figured out that P(k) ∧P(k + 1) ∧P(k + 2) = ⇒P(k + 3). What base case(s) would you have to prove?
P(0) · · · P(N0 −1) P(N0) · · · P(N1) P(N1 + 1) P(N1 + 2) · · · Figure 5.3: A visual interpretation of strong mathematical induction. Once you show that P(k) ⇒P(k + 3), and you know that P(N0) is true, then it must be that P(N0 + 3) is true. From this it follows that P(N0 + 6) is true. From this it follows that P(N0 + 9) is true. And so on. This, however, does not prove the statement for all integers. You need to also show that P(N0 + 1) and P(N0) + 2) are true to get that.
Definition: Principle of Strong Mathematical Induction Let P(n) be a predicate that is defined for integers n, and let N0, N1 be fixed integers with N0 ≤N1. Suppose the following two statements are true.
1. P(N0), P(N0 + 1), P(N0 + 2), . . . , P(N1) are true.
2. For all integers k ≥N1, if P(N0), P(N0 +1), . . . , P(N1), . . . , P(k) are true, then P(k +1) is true.
Then P(n) is true for all n ≥N0.
106 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Remark. In words, the “base step” is actually a finite number of cases (not just a singular case), and the “inductive step” requires that all preceding cases (from the basis steps onward) are true. In practice, you may not actually need to use all preceding cases to verify the inductive step – it’s overkill, sure, but you lose nothing by assuming more.
Remark. Regular ol’ induction is sometimes called weak induction. Paradoxically, one is not actually stronger than the other (any statement provable with strong induction can be proven with weak induction).
Example 5.4.1: Recursive Sequence/Explicit Sequence Let {an}∞ n=0 be the recursively-defined sequence a0 = 0, a1 = 4, and an = 6an−1 −5an−2 for all n ≥2.
Show that, for all n ≥0, an = 5n −1.
For scratch, we compute the first few terms of the sequence to confirm: a0 = 0 = 50 −1 a1 = 4 = 51 −1 a2 = 6a1 −5a0 = 6(4) −5(0) = 24 = 52 −1 a3 = 6a2 −5a1 = 6(24) −5(4) = 124 = 53 −1 a4 = 6a3 −5a2 = 6(124) −5(24) = 624 = 54 −1 The claim seems reasonable. Since the recursive sequence begins with two terms, our base case should reflect this similarly: using N0 = 0, N1 = 1.
Proof. Let n be an arbitrary natural number, {an} the recursive sequence defined above, and let P(n) be the predicate “an = 5n −1.” Base steps. Take N0 = 0 and N1 = 1. We see that aN0 = a0 = 0 = 50 −1 andaN1 = 4 = 51 −1 hence P(N0) and P(N1) are true.
Inductive step. Let k ≥1 and suppose that P(0), P(1), . . . , P(k) are all true, that is aj = 5j −1 for j = 0, 1, . . . , k. We then have that ak+1 = 6ak −5ak−1 = 6(5k −1) −5(5k−1 −1) = 6(5k) −6 −5k + 5 = 5(5k) −1 = 5k+1 −1.
Therefore P(k + 1) is true.
5.4. STRONG MATHEMATICAL INDUCTION AND THE WELL-ORDERING PRINCIPLE FOR THE INT Example 5.4.2: Tribonacci Sequence Let {Tn}∞ n=0 be the recursively-defined sequence T0 = 0, T1 = 0, T2 = 1 and Tn = Tn−1 + Tn−2 + Tn−3 for all n ≥3.
Show that, for all n ≥0, Tn < 2n−1.
Proof. Let n be a natural number and let {Tn} be the recursively-defined sequence above.
Base. Take N0 = 0 and N1 = 2. We see that T0 = 0 < 1 2 = 20−1 T1 = 0 < 1 = 21−1 T2 = 1 < 2 = 22−1 Induction. Let k ≥2 be an arbitrary natural number and suppose that, for all m ∈N satisfying 0 ≤m ≤k, Tm < 2m−1. Then Tk+1 = Tk + Tk−1 + Tk−2 < 2k−1 + 2k−2 + 2k−3 = 2k−3(4 + 2 + 1) < 2k−3 · 23 (since 4 + 2 + 1 < 23) = 2k = 2(k+1)−1.
Therefore, for all k ≥0, we have that Tk < 2k−1.
Example 5.4.3: 2-adic representation Show that every positive integer n can be uniquely written as a sum of powers of 2. Explicitly, for every n, there are natural numbers b0, . . . , br with b0 < b1 < · · · < br so that n = 2b0 + 2b1 + · · · + 2br.
Proof.
Base. Note that 1 = 20.
Induction. Let k ≥1 be an arbitrary natural number and suppose that, for all m ∈N satisfying 1 ≤m ≤k, m has a unique representation as a sum of powers of two.
Case 1. Suppose k + 1 is odd. Then k is even, and by our inductive hypothesis, there are unique natural numbers b0, . . . , br with b0 < b1 < . . . < br for which k = 2b0 + · · · + 2br.
As k is even, then 0 < b0, hence k + 1 = 20 + 2b0 + · · · + 2br.
108 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Case 2. Suppose that k+1 is even. Writing m = k+1 2 , we must have that 1 ≤m ≤k.
By our inductive hypothesis, there are then unique natural numbers b0, . . . , br with b0 < b1 < . . . < br for which m = 2b0 + 2b1 + · · · + 2br.
Then k + 1 = 2m = 2 2b0 + 2b1 + · · · + 2br = 2b0+1 + 2b1+1 + · · · + 2br+1.
In either case above, the uniqueness of the exponents is immediate, but one could easily formalize this in Case 1 using ˜ b0 = 0, and ˜ bi = bi−1 for i = 1, . . . , r + 1 or in Case 2 using ˜ bi = bi + 1 for i = 0, . . . , r.
Remark. The above extends naturally to negative integers (just multiply everything by −1), but extending to all rational numbers requires some real thought (search term “2-adic numbers”).
Theorem 5.4.4: Divisibility by Primes Any integer greater than 1 is divisible by a prime.
The strategy is this: if a number is composite, then it is the product of two numbers that are smaller, so as long as one of those numbers is divisible by a prime, then this composite number will be as well.
Proof. Let n be an arbitrary integer and let P(n) be the predicate “n is divisible by a prime.” Base step. Since N0 = N1 = 2 is prime, then N0 is divisible by 2. Hence P(N0) is true.
Inductive step. Let k ≥2 be some integer an suppose that P(2), P(3), . . . , P(k) are all true. We have two cases for k + 1: (k + 1 is prime). If k + 1 is prime, then k + 1 is divisible by itself, hence P(k + 1) is true.
(k + 1 is composite). If k + 1 is composite, then there are two integers, a, b for which k = ab and 1 ≤a ≤k and 1 ≤b ≤k. By the inductive hypothesis, a (and b) is divisible by a prime p.
Since p|a implies that p|(ab), then p|k + 1, hence P(k + 1) is true.
5.4.1 Well-Ordering Principle for the Integers Well-Ordering Principle Let N0 be a fixed integer and let S be a nonempty set of integers, all of which are greater than N0. Then S has a least element.
5.4. STRONG MATHEMATICAL INDUCTION AND THE WELL-ORDERING PRINCIPLE FOR THE INT Remark. The Well-Ordering Principle is actually equivalent to the principle of mathematical induction. Since we’ve assumed Induction already, we could reasonably call the above a theorem and prove it. But your instructor doesn’t want to, so the proof is left as an exercise to the reader.
Theorem 5.4.5: Quotient-Remainder Theorem Let n be an integer and let d be a positive integer (d ≥1). Then there are unique integers q, r with 0 ≤r < d satisfying n = qd + r.
The strategy for proving this is going to be to collect all possible nonnegative remainders r, use the well-ordering principle to see that there is a smallest r, and then verify that 0 ≤r < d. Lastly, we’ll show that q and r are unique.
Proof. Let n be an arbitrary integer and let d be a positive integer.
Existence.
Let X be the set of nonnegative integers of the form n −dq (so n −dq ≥0, or rather n ≥dq), where q is some integer. We first show that X contains at least one integer.
Case 1 (n ≥0). In this case, taking q = 0, then n −dq = n ≥0, so n is contained in X.
Case 2 (n < 0). In this case, take q = n. We then have that n −dq = n −nd = n(1 −d).
Since d > 0, then (1 −d) ≤0, so since both n and (1 −d) are nonpositive, their product is nonnegative, hence X contains n −nd.
By the ??, there must be a smallest natural number r ≥0 satisfying n = dq + r.
Now we show that r < d. To do this, we write n −dq = r. Tending toward a contradiction, suppose that r ≥d. We then have that n −dq ≥d n −dq −d ≥0 n −d(q + 1) ≥0 In other words, we have that n −d(q + 1) is contained in X, and that r = n −dq > n −d(q + 1), which contradicts that r was the least element.
Uniqueness.
At long last, we just need to show that q, r are unique. To do so, suppose that n = dq1 + r1 = dq2 + r2 (5.1) where q1, q2, r1, r2 are integers and 0 ≤r1 < d and 0 ≤r2 < d. Without loss of generality, suppose that r2 ≥r1. Rearranging Equation 5.1, we have (r2 −r1) = d(q1 −q2) So r2 −r1 is a multiple of d. Since r1, r2 were assumed to be less than d, their difference also satisfies 0 ≤r2 −r1 < d.
110 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION The only multiple of d in this range is 0, hence r2 −r1 = 0 = ⇒r1 = r2. It follows then that q1 = n −r1 d = n −r2 d = q2.
Therefore q and r are unique, as desired.
Definition: quotient, remainder The variables q and r in the Quotient-Remainder Theorem are called the quotient and remainder, respectively. One might see these written as n div d = q and n mod d = r.
The proof of the quotient–remainder theorem actually suggests an algorithm to find the quotient and remainder. When n is negative, keep adding multiples of d until your number is greater than 0.
When n is nonnegative, keep subtracting multiples of d until your number less than or equal to d.
Explicitly Algorithm 5.4.6: The Division Algorithm To find the quotient q = n div d and and remainder r = n mod d, break it into cases and employ the following Case 1: n ≥0 Case 2: n < 0 q := 0 q := 0 r := n r := n while r ≥d do while r < 0 do q := (q + 1) q := (q −1) r := (r −d) r := (r + d) end end Example 5.4.7 Use the division algorithm to find the quotient and remainder for...
1. ... n = 7, d = 2 2. ... n −8, d = 3 INCOMPLETE Exercise 5.4.8 Prove that, for all integers x, y, and for any positive integer d, (x mod d) + (y mod d) mod d = (x + y) mod d and (x mod d)(y mod d) mod d = (x + y) mod d.
5.6. DEFINING SEQUENCES RECURSIVELY 111 5.6 Defining Sequences Recursively Definition: recurrence relation, recursive sequence A recurrence relation for a sequence {an} is a formula that relates each term ak to certain predecessorts ak−1, ak−2, etc. A recursive sequence is a sequence {an} where each term satisfies a recurrence relation.
Example 5.6.1: Fibonacci Sequence Let {Fn}∞ n=0 be the infinite recursive sequence defined by F0 = 0, F1 = 1 (the initial conditions) and Fk = Fk−1 + Fk−2 for k ≥2.
(the recursive formula).
Find F2, F3, F4, F5 F2 = F1 + F0 = 1 + 0 = 1 F3 = F2 + F1 = 1 + 1 = 2 F4 = F3 + F2 = 2 + 1 = 3 F5 = F4 + F3 = 3 + 2 = 5 Example 5.6.2 Define a sequence b0, b1, b2, .. recursively as follows: b0 = 2 (the initial condition) and bk = 1 −(bk−1)k for k ≥1 (the recursive formula).
Find b1, b2, b3.
b1 = 1 −(b0)1 = 1 −21 = 1 −2 = −1 b2 = 1 −(b1)2 = 1 −(−1)2 = 1 −1 = 0 b3 = 1 −(b2)3 = 1 −(0)3 = 1 −0 = 1 112 CHAPTER 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Chapter 6 Set Theory 6.1 Set Theory: Definitions and the Element of Proof Definition: set,element A set is an unordered list of items, called elements. If A is a set and a is an element of the set, we write a ∈A. If a is not an element of the set, we write a / ∈A.
Remark. An element of a set cannot appear twice.
We typically write a set using curly braces and listing the elements.
Example 6.1.1 Examples of sets.
1. S = {1, 3, 19} 2. [natural numbers] N = {0, 1, 2, 3, 4, 5, . . .} 3. [empty set] ∅= {} 4. T = {1, 2, {1, 2}} Remark. The last example above appears to violate the remark that an element cannot appear twice, but in fact the elements of T are 1, 2, and {1, 2}. A set can be an element of another set (but we musn’t be na¨ ıve – there famously cannot be a set containing all sets). You can make sense of this by thinking about a situation in which one is carrying a bucket of balls, and within that bucket is a smaller container which also contains a couple of balls. If you reach into the large bucket and grab an object at random, you can grab either a ball or a small container.
Sometimes sets are infinite (and writing it out explicitly is impossible) or sometimes sets are finite but we really want to communicate the pattern of the elements within the set.
113 114 CHAPTER 6. SET THEORY Definition: set-builder notation Let X be some set (called the universe) and let P1(x), P2(x), . . . be some properties (or predicates) with domain X. One can define a set A using set-builder notation A = {x ∈X : P1(x) is true, P2(x) is true . . .} .
This is read “A is the set of all elements in X such that P1(x), P2(x), . . . are all true” or “A is the set of all elements in X with properties P1(x), P2(x), . . .” Remark. If this feels like a truth set, it’s because it is.
Remark. The universe is almost never directly stated, as it is either unimportant for the statement, or it is clear from context.
Example 6.1.2 Define P(x) : x is prime, Q(x) : x < 20, and R(x) : x is even. Determine the elements of the following sets: 1. A = {x ∈Z+ | Q(x)} 2. B = {x ∈Z+ | P(x) ∧Q(x)} 3. C = {x ∈Z+ | P(x) ∧R(x)} 1. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19} 2. {2, 3, 5, 7, 11, 13, 17, 19} 3. {2} Definition: subset Let B be a set (inside of some universe X). We say that A is a subset of B, denoted A ⊆B, if and only if ∀x ∈X, if x ∈A then x ∈B A subset A is called a proper subset of B, denoted A ⊊B, if and only if A ⊆B and ∃x ∈B such that x / ∈A.
In other words, A is a proper subset of B if and only if (1) A is a subset of B and (2) A ̸= B.
In either case, B is sometimes called the superset of A.
Remark. There are some competing notations about subsets. Some authors will prefer to make the subset notation look the same as ≤and < notations, using ⊆for subset and ⊂for a proper subset.
Others will use ⊂for a subset and ⊊for a proper subset. We shy away from using the ⊂symbol at all because is very similar to the letter C when hand-written.
6.1. SET THEORY: DEFINITIONS AND THE ELEMENT OF PROOF 115 Example 6.1.3 Let A = {1, 3, 5, 9}, B = {1, 2, 3, 4, 5, 7, 9}.
1. Is A a subset of B?
2. Is A a proper subset of B?
1. Yes. Every element of A is also contained in B.
{1, 2, 3, 4, 5, 7, 9} 2. Yes. There are elements of B that are not contained in A.
{1, 2, 3, 4, 5, 7, 9} If we treat A and B as regions of the plane, we can think about a Venn diagram which relates these sets.
A B Figure 6.1: B ̸⊆A and A ̸⊆B A B Figure 6.2: B ̸⊆A and A ̸⊆B A B Figure 6.3: B ⊊A and A ̸⊆B A B Figure 6.4: A ⊆B and B ⊆A For small sets, it is easy to just compare elements explicitly and check whether one is a subset of another. But for large sets – especially ones described in terms of set-builder notation – this can become more complicated. Look at the definition of subset more closesly B ⊆A if and only if [∀x ∈A, x ∈B = ⇒x ∈A] This yields the following technique Element Method for Subsets Let A, B be sets. To show that B ⊆A.
1. Suppose x ∈B, where x is arbitrary.
2. Show that x ∈A.
Example 6.1.4 For any integer k, we define the set kZ := {x ∈Z : x is a multiple of k} .
Show that 6Z ⊆2Z.
In order to show that this is true, we need to show that any multiple of 6 is also a multiple of 2.
Proof. Let x ∈Z be arbitrary and suppose that x ∈6Z, i.e., x is a multiple of 6. Then, by 116 CHAPTER 6. SET THEORY definition, there is some integer n for which x = 6n = 2(3n). Since 3n is an integer, we see that such an x is also a multiple of 2, and therefore x ∈2Z.
Figure 6.4 suggests to us the following definition Definition: set equality set-equality Two sets A and B are said to be equal if and only if both are subsets of each other. Symbolically, A = B ⇐ ⇒(A ⊆B) ∧(B ⊆A).
Example 6.1.5 Using the notation from Example 6.1.4, let A be the set A = {x ∈Z : x ∈2Z ∧x ∈3Z} .
Show that A = 6Z.
Showing 6Z ⊆A. Let x be an arbitrary integer and suppose x ∈6Z. Then by definition x = 6k for some integer k. But x = 6k = 2(3k) showing both that x ∈2Z. Moreover, x = 6k = 3(2k) so x ∈3Z. Therefore x ∈A and thus 6Z ⊆A.
[Showing A ⊆6Z] Let x be an arbitrary integer and suppose x ∈A. Then, by definition of A, x ∈2Z and x ∈3Z. Since x ∈3Z, there is some integer k for which x = 3k. Since x ∈2Z, then x is even, and since 3 is not even, it must be that k is even, i.e., that there is an integer ℓfor which x = 3(2ℓ) = 6ℓ. Therefore x ∈6Z and thus A ⊆6Z Since A ⊆6Z and 6Z ⊆A, then A = 6Z.
Exercise 6.1.6: Subset Transitivity Prove the following: For all sets A, B, C, if A ⊆B and B ⊆C, then A ⊆C.
6.1.1 Set Operations Definition Let A and B be subsets of the same universal set X.
1. The intersection of A and B, denoted A ∩B, is the set of all elements common to both A and B.
A ∩B = {x ∈X : x ∈A and x ∈B} 6.1. SET THEORY: DEFINITIONS AND THE ELEMENT OF PROOF 117 X A B 2. The union of A and B, denoted A ∪B, is the set of all elements contained in either A or B.
A ∪B = {x ∈X : x ∈A or x ∈B} X A B 3. The difference, A minus B (sometimes called the relative complement of B in A), denoted A −B, is the set of all elements in A that are not contained in B.
A −B = {x ∈X : x ∈A and x / ∈B} X A B 4. The symmetric difference of A and B, denoted A∆B, is the set of all elements in either A or B, but not in both.
A∆B = {x ∈X : (x ∈A) ⊕(x ∈B)} 118 CHAPTER 6. SET THEORY X A B 5. The complement of A in X, denoted Ac, is the set of all elements not in A.
Ac = {x ∈X : x / ∈A} X A B Example 6.1.7 Let A = {1, 2, 3, 4}, B = {2, 4, 6, 8}, C = {x ∈Z | x is prime}, and X = Z. Find each of the following: 1. A ∩C 2. B −A 3. Ac 4. Ac ∩B 5. (A ∩C) ∪B 1. A ∩C = {2, 3} 2. B −A = {6, 8} 3. Ac = {. . . , −3, −2, −1, 0, 5, 6, 7, 8, . . .} 4. Ac ∩B = {6, 8} 5. (A ∩C) ∪B = {2, 3} ∪B = {2, 3, 4, 6, 8} When taking the union or intersection of a large number of sets A1, A2, . . . , An, it’s common to 6.1. SET THEORY: DEFINITIONS AND THE ELEMENT OF PROOF 119 simplify notation n [ i=1 Ai = A1 ∪A2 ∪· · · ∪An n \ i=1 Ai = A1 ∩A2 ∩· · · ∩An To be completely clear about this notation, if A1, . . . , An are sets in the same universe X, then n [ i=1 Ai = {x ∈X : x ∈A1 ∨x ∈A2 ∨. . . ∨x ∈An} , n \ i=1 Ai = {x ∈X : x ∈A1 ∧x ∈A2 ∧. . . ∧x ∈An} .
One can extend the above to unions and intersections of a (countably) infinite number of sets A1, A2, . . . , An, . . .
[ i Ai or ∞ [ i=1 Ai = A1 ∪A2 ∪· · · ∪An ∪· · · \ i Ai or ∞ \ i=1 Ai = A1 ∩A2 ∩· · · ∩An ∪· · · To be completely clear about this notation, if A1, A2, . . . , An, . . . are sets in the same universe X, then ∞ [ i=1 Ai = {x ∈X : ∃i ∈N such that x ∈Ai} , ∞ \ i=1 Ai = {x ∈X : ∀i ∈N, x ∈Ai} .
Remark.
1. The above really only makes sense if we know that unions and intersections are associative operations. But don’t worry, we’ll see this in Theorem 6.2.1.
2. If you’re really feeling bold, you can even extend the definition of the above to uncountable sets (like R or C).
Definition: disjoint sets Two sets A and B are called disjoint if and only if they have no elements in common, i.e., if and only if A∩B = ∅. A possibly-infinite collection of sets {A1, A2, . . .} is said to be pairwise disjoint if Ai and Aj are disjoint whenever i ̸= j.
Definition: partition Let A be a set and let {B1, B2, . . .} be a possibly-infinite collection of nonempty subsets of A.
This collection is a partition of A if and only if 1. The Bi are pairwise disjoint, and 2. A = [ i Bi.
120 CHAPTER 6. SET THEORY Remark. In light of the following exercise, the second item above can be replaced with 2. A ⊆ [ i Bi which is faster to prove than set equality.
Exercise 6.1.8 If A1, . . . , An, . . . are (possibly an infinite number of) subsets of B, then [ i Ai is also a subset of B.
Example 6.1.9 Let A = {2, 3, 5, 7, 11, 13}, B1 = {2, 3, 5}, B2 = {7, 11}, and B3 = {13}. Is {B1, B2, B3} a partition of A?
Yes. By inspection, we see that B1, B2, B3 are all mutually disjoint. It is equally straightforward to see that A = B1 ∪B2 ∪B3.
Example 6.1.10 Give an example of a partition of Z (using proper subsets). Any partition at all. Have fun with it. You can give multiple examples. You can give infinitely-many examples! The possibilities are (literally) endless!
Consider the possible remainders when dividing an integer by 3. The possible options are o, 1, 2. In other words, every integer has one of the following forms: 3k, 3k + 1, or 3k + 2 Moreover, after dividing by 3, no integer can have two different remainders, so any integer of the form 3k cannot be written as an integer of the form 3k + 1, etc. With this in mind, we defined the following subsets of integers: T0 = {n ∈Z : n = 3k for k ∈Z} T1 = {n ∈Z : n = 3k + 1 for k ∈Z} T2 = {n ∈Z : n = 3k + 2 for k ∈Z} By our discussion above 1. T0 ∩T1 = T0 ∩T2 = T1 ∩T2 = ∅ 2. Z = T0 ∪T1 ∪T2 and therefore {T0, T1, T2} is a partition of Z.
Definition: power set Given a set A, the power set of A, denoted, P(A), is the set of all subsets of A. That is, if X is the universe, P(A) = {sets B in the universe X : B ⊆A} 6.1. SET THEORY: DEFINITIONS AND THE ELEMENT OF PROOF 121 Example 6.1.11 Find P(A) given A = {x, y, z}.
The possible subsets of A are ∅, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}. Thus P(A) = {∅, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, A} .
Example 6.1.12 Let A and B be sets. Prove that, if A ⊆B, then P(A) ⊆P(B).
Our goal is to show that, S ∈P(A) = ⇒S ∈P(B). Looking at the definition of the power set, this means proving that, for every subset S ⊆A, we have that S ⊆B.
We’ll provide two proofs, depending on whether or not an exercise above has been proven in class/on homework.
Proof. Suppose A ⊆B and let S ∈P(A) be arbitrary. By definition of the power set, we must have that S ⊆A.
(Assuming we’ve proven Subset Transitivity...) Since A ⊆B, it follows from Ex-ercise 6.1.6 that S ⊆B, and therefore S ∈P(B).
(Assuming we’ve not proven Subset Transitivity...) Since S ⊆A, then for every x ∈S we must have that x ∈A. But since A ⊆B, then we must further have that x ∈B. Thus S ⊆B, and therefore S ∈P(B).
Therefore, P(A) ⊆P(B).
Exercise 6.1.13 Prove the converse of the previous example. That is, prove that, if P(A) ⊆P(B), then A ⊆B.
6.1.2 Arithmetic Operations and Set Theory - Some Interesting History This set framework ends up being quite a bit more powerful than one might wthink. Not only does it encode logic (see the next section), but some smart folks noticed that we could come up with a correspondence between some natural numbers and sets: 0 ↭∅, 1 ↭{∅} 2 ↭{∅, {∅}} , . . .
Using the union operation and this correspondence, we actually get a notion of addition of natural numbers: “1 + 1” = “1” ∪{“1”} = {∅} ∪{{∅}} = {∅, {∅}} = “2” Russel and Whitehead, in their Principia Mathematica (published in 3 volumes from 1910-1913) famously spend a few hundred pages getting to this point.1 Their work builds the rest of the natural numbers through the use of a successor function, S, as follows: For any natural number n S(n) = n ∪{n} 1Yes, the proof above is the proof that 1 + 1 = 2 that you may have heard about.
122 CHAPTER 6. SET THEORY What this does is extends the correspondence between N and sets 0 ↭∅ 1 ↭S(0) = ∅∪{∅} = {∅} 2 ↭S(1) = {∅} ∪{{∅}} = {∅, {∅}} 3 ↭S(2) = {∅, {∅, {∅}}} .
.
.
Moreover, notice the successor function allows us to define “addition” between two natural numbers in a recursive fashion: m + 0 = m m + S(n) = S(m + n) For example 2 + 2 = 2 + S(1) = S(2 + 1) = S(2 + S(0)) = S(S(2 + 0)) = S(S(2)) = S(3) = 4 So what this shows us is that sets with the union/intersection/complement operations are enough to encode basic logic as well as arithmetic. One of the goals of Russel and Whitehead was to completely formalize set theory and make it the logical basis for all modern mathematics. In a sense, the idea was that every true math statement could be thusly distilled down to set theory wherein it could be proven.
In 1931, work of Godel showed that this logical system actually contains true statements which are unprovable (in fact, any logical system which could encode basic arithmetic suffers from this feature).
So Russel and Whitehead were sort of doomed to failure in creating a system which would be able to prove everything. However, they did lay the groundwork for a specific branch of mathematics -Type Theory (which your instructor knows absolutely nothing about, but it’s definitely a thing).
6.2. PROPERTIES OF SETS 123 6.2 Properties of Sets The definitions of union, intersection, and complements appear to play the roles of ∨, ∧, and ¬. In fact, the connection is so strong that one can essentially restate the table of logical equivalences in terms of these operations. (Note that the empty set corresponds to a contradiction, and a tautology corresponds to a tautology; which makes sense if one thinks about truth sets).
Theorem 6.2.1 Let P, Q, and R be sets in the same universe X. We then have the following table of set equalities: Commutative Laws P ∩Q = Q ∩P P ∪Q = Q ∪P Associative Laws (P ∩Q) ∩R = P ∩(Q ∩R) (P ∪Q) ∪R = P ∪(Q ∪R) Distributive Laws P ∩(Q ∪R) = (P ∩Q) ∪(P ∩R) P ∪(Q ∩R) = (P ∪Q) ∩(P ∪R) Identity Laws P ∩X = P P ∪∅= P Complement Laws P ∪P c = X P ∩P c = ∅ Double Comple-ment Laws (P c)c = P Idempotent Laws P ∩P = P P ∪P = P Universal Bound Laws P ∪X = X P ∩∅= ∅ De Morgan’s Laws (P ∩Q)c = P c ∪Qc (P ∪Q)c = P c ∩Qc Absorption Laws P ∪(P ∩Q) = P P ∩(P ∪Q) = P Complement of X and ∅ Xc = ∅ ∅c = X Now, we can – and probably should – prove all of these using the element method, but coloring pictures to see the intuition is more fun.
Exercise 6.2.2: Proof of Commutative Laws Color the diagram below to show the Commutative Laws in Theorem 6.2.1.
124 CHAPTER 6. SET THEORY P ∩Q X P Q Q ∩P X P Q P ∪Q X P Q Q ∪P X P Q Exercise 6.2.3: Proof of Associative Laws Color the picture below to show the Associative Laws in Theorem 6.2.1.
(P ∩Q) ∩R X P Q R P ∩(Q ∩R) X P Q R (P ∪Q) ∪R X P Q R P ∪(Q ∪R) X P Q R 6.2. PROPERTIES OF SETS 125 Exercise 6.2.4: Proof of Distributive Laws Color the picture below to show the Distributive Laws in Theorem 6.2.1.
P ∩(Q ∪R) X P Q R (P ∩Q) ∩(P ∩R) X P Q R P ∪(Q ∩R) X P Q R (P ∪Q) ∩(P ∪R) X P Q R Exercise 6.2.5: Proof of Identity Laws Color the picture below to show the Identity Laws in Theorem 6.2.1.
P ∩X X P P X P 126 CHAPTER 6. SET THEORY P ∪∅ X P P X P Exercise 6.2.6: Proof of Complement Laws Color the picture below to show the Complement Laws in Theorem 6.2.1.
P ∪P c X P X X P P ∩P c X P ∅ X P Exercise 6.2.7: Proof of Double Complement Law Color the picture below to show the Double Complement Law in Theorem 6.2.1.
6.2. PROPERTIES OF SETS 127 (P c)c X P P X P Exercise 6.2.8: Proof of Idempotent Laws Color the picture below to show the Idempotent Laws in Theorem 6.2.1.
P ∩P X P P X P P ∪P X P P X P Exercise 6.2.9: Proof of Universal Bound Laws Color the picture below to show the Universal Bound Laws in Theorem 6.2.1.
128 CHAPTER 6. SET THEORY P ∪X X P X X P P ∩∅ X P ∅ X P Exercise 6.2.10: Proof of DeMorgan’s Laws Color the diagram below to show DeMorgan’s Laws in Theorem 6.2.1.
(P ∩Q)c X P Q P c ∪Qc X P Q (P ∪Q)c X P Q P c ∩Qc X P Q 6.2. PROPERTIES OF SETS 129 Exercise 6.2.11: Proof of Absorption Laws Color the diagram below to show the Absorption Laws in Theorem 6.2.1.
P ∪(P ∩Q) X P Q P X P Q P ∩(P ∪Q) X P Q P X P Q Exercise 6.2.12: Proof of Complement of X and ∅ Color the picture below to show the Complement of X and ∅Laws in Theorem 6.2.1.
Xc X P ∅ X P 130 CHAPTER 6. SET THEORY ∅c X P X X P 6.2.1 Properties of Subsets Because subsets are defined by an implication “ = ⇒”, there is a natural ordering to reading statements about them so that we may omit extranneous parentheses. “A ∩B ⊆C” means “(A ∩B) ⊆C”.
Proposition 6.2.13: Some Subset Relations Let A, B be sets (in the same universe). Then 1. A ∩B ⊆A and A ∩B ⊆B 2. A ⊆A ∪B and B ⊆A ∪B 3. if A ⊆B and B ⊆C then A ⊆C 4. [Set Difference Law] A −B = A ∩Bc Proof. We use the Element Method of Proof.
1. Suppose x ∈A ∩B. Then, x ∈A and x ∈B. Therefore A ∩B ⊆A and A ∩B ⊆B.
2. Let A, B be sets. If x ∈A, then x ∈A ∨x ∈B, hence x ∈A ∪B. Therefore A ⊆A ∪B. If x ∈B, then x ∈A ∨x ∈B, hence x ∈A ∪B. Therefore B ⊆A ∪B.
3. Let A, B, C be sets wth A ⊆B and B ⊆C. Suppose that x ∈A. Then, since A ⊆B, x ∈B.
Moreover, since B ⊆C, then x ∈C. We have shown that x ∈A = ⇒x ∈C and therefore A ⊆C.
4. Let A, B be sets. We have to show both that A −B ⊆A ∩Bc and that the opposite containment is true.
[⊆] Suppose that x ∈A −B. Then x ∈A and x / ∈B, hence x ∈A and x ∈Bc, and thus x ∈A ∩Bc. Therefore A −B ⊆A ∩Bc.
[⊇] Suppose now that x ∈A ∩Bc. Then x ∈A and x ∈Bc, that is, x ∈A and x / ∈B, and thus x ∈A −B. Therefore A ∩Bc ⊆A −B.
6.3. DISPROOFS AND ALGEBRAIC PROOFS 131 6.3 Disproofs and Algebraic Proofs An “algebraic” proof is the language your author uses to indicate that the proof is handled by a string of equalities coming from Theorem 6.2.1 and Proposition 6.2.13.
Example 6.3.1 Prove the following statement using the element method, and then again with the Table of Set Equalities and Proposition 6.2.13: For all sets A, B, C, (A ∩B) −(A ∩C) = A ∩(B −C).
Element Method. Proof. Let A, B, C be arbitrary sets.
[⊆] Suppose x ∈(A ∩B) −(A ∩C). Then x ∈A ∩B and x / ∈A ∩C. That x ∈A ∩B implies that x ∈A and x ∈B. As well, x / ∈A ∩C is equivalent to the statement ̸ [x ∈A ∩C] ≢ [x ∈A ∧x ∈C], which, by DeMorgan’s Law (for logical statements), implies x / ∈A or x / ∈C.
At this point we have that x ∈A and x ∈B, and that x / ∈A or x / ∈C. We can’t have that x / ∈A, so it can only be that x / ∈C. Thus x ∈B −C, and therefore x ∈A ∩(B −C).
[⊇] Suppose x ∈A ∩(B −C). Then x ∈A and x ∈B −C, hence x ∈B and x / ∈C. Since x ∈A and x ∈B, then x ∈A ∩B. Also, since x / ∈C, then in particular, x / ∈(A ∩C). It follows that x ∈(A ∩B) −(A ∩C) and therefore A ∩(B −C) ⊆(A ∩B) −(A ∩C).
Table of Set Equalities Proof. Let A, B, C be sets. Then (A ∩B) −(A ∩C) = (A ∩B) ∩(A ∩C)c (Set Difference Law) = (A ∩B) ∩(Ac ∪Cc) (DeMorgan’s) = (A ∩B ∩Ac) ∪(A ∩B ∩Cc) (distributive) = (A ∩Ac ∩B) ∪(A ∩B ∩Cc) (commutative) = ∅∪(A ∩B ∩Cc) (Complement law) = A ∩B ∩Cc (identity) = A ∩(B ∩Cc) (associative) = A ∩(B −C) (Set Difference Law) Since A ∩(B −C) = (A ∩B) −(A ∩C), then in particular A ∩(B −C) ⊆(A ∩B) − (A ∩C).
Example 6.3.2 Prove or disprove the following claim.
Claim. For all sets A, B, C, then (A −B) ∪(B −C) = A −C.
132 CHAPTER 6. SET THEORY A B C (A −B) ∪(A −C) A B C A −C This is false. Observe that any example which contains an element of B which is not shared by A or C (or similarly, an element of A ∩C which is not also in B) will be a counter-example to the claim.
Disproof. Consider A = {1, 2, 3, 4}, B = {2, 4, 5, 6}, and C = {3, 4, 6, 7}.
Then A −B = {1, 3}, B −C = {2, 5}, and A −C = {1, 2}, and so (A −B) ∪(B −C) = {1, 2, 3, 5} ̸= {1, 2} = A −C.
Example 6.3.3 Prove or disprove the following claim.
Claim. For all sets A, B, C, then (A ∪B) −B = A −(A ∩B).
A B C (A ∪B) −B A B C A −(A ∩B) Proof. We have that (A ∪B) −B = (A ∪B) ∩Bc (Set Difference Law) = Bc ∩(A ∪B) (Commutative Law) = (Bc ∩A) ∪(Bc ∩B) (Distributive Law) = (Bc ∩A) ∪∅ (Complement Law) 6.3. DISPROOFS AND ALGEBRAIC PROOFS 133 = Bc ∩A (Identity Law) = A ∩Bc (Commutative Law) Also, A −(A ∩B) = A ∩(A ∩B)c (Set Difference Law) = A ∩(Ac ∪Bc) (DeMorgan’s Law) = (A ∩Ac) ∪(A ∩Bc) (Distributive Law) = ∅∪(A ∩Bc) (Complement Law) = A ∩Bc (Identity Law) Therefore (A ∪B) −B = A ∩Bc = A −(A ∩B).
Power Sets Let A = {1, 2, 3} and B = {1, 2}. Looking at the power set of A and the power set of B, we have P(A) = ∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} P(B) = ∅, {1}, {2}, {1, 2} We have that P(B) ⊊P(A). Morover, notice that the remaining sets in P(A) −P(B) can be cleverly written in terms of the subsets of B.
{3} = ∅∪{3} {1, 3} = {1} ∪{3} {2, 3} = {2} ∪{3} {1, 2, 3} = {1, 2} ∪{3} This observation provides us with the inutition for the inductive step in the proof of the following.
Theorem 6.3.4 For every integer n ≥0, if a set A has n elements, then P(A) has 2n elements.
Proof. Let A be a set. We approach by inducting on the number of elements in A.
Base Step. Suppose A has zero elements. Then A = ∅. Moreover, P(A) = P(∅) = {∅}, which is a set with 1 = 20 element (a symbol called “∅”).
Inductive Step. Suppose now A = {a1, a2, . . . , ak} is any set with k elements and that P(A) has 2k elements. Let B = A ∪{ak+1} so that B has k + 1 elements. Notice the following: • P(A) and P(B) −P(A) are disjoint sets (on the left are subsets of A, on the right are subsets of B which are not subsets of A).
• P(B) = P(A) ∪ P(B) −P(A) (this is just the complement law) 134 CHAPTER 6. SET THEORY so P(A) and P(B) −P(A) provide a partition of P(B). In this way, we should be able to count both sets separately and add them up.
By the inductive hypothesis, P(A) has 2k elements.
Observe that ak+1 is the only element of B which is not in A, hence every set in P(B)−P(A) must be a set of the form X ∪{ak+1} where X ∈P(A). Since P(A) has 2k elements, then there must be 2k sets of the form X ∪{ak+1}, whence P(B) −P(A) also has 2k elements.
It follows then that P(B) has 2k |{z} P(A) + 2k |{z} P(B)−P(A) = 2(2k) = 2k+1 elements.
Therefore, for all n ≥0, if A has n elements, then P(A) has 2n elements.
Example 6.3.5 Prove or disprove the following claim.
Claim. For all sets A, B, then P(A) ∩P(B) = P(A ∩B).
Proof. Let X be an arbitrary set. We have the following string of biconditionals X ∈P(A) ∩P(B) ⇐ ⇒X ⊆A and X ⊆B ⇐ ⇒X ⊆A ∩B ⇐ ⇒X ∈P(A ∩B) This shows that P(A) ∩P(B) ⊆P(A ∩B) and P(A ∩B) ⊆P(A) ∩P(B). Therefore, P(A) ∩P(B) = P(A ∩B).
Example 6.3.6 Prove or disprove the following claim.
Claim. For all sets A, B, then P(A) ∪P(B) ⊆P(A ∪B).
Proof. Let X be an arbitrary set. We have the following string of coniditionals X ∈P(A) ∪P(B) ⇐ ⇒X ⊆A or X ⊆B = ⇒X ⊆A ∪B 6.3. DISPROOFS AND ALGEBRAIC PROOFS 135 ⇐ ⇒X ∈P(A ∩B) This shows that P(A) ∩P(B) ⊆P(A ∩B).
136 CHAPTER 6. SET THEORY 6.4 Boolean Algebras, Russell’s Paradox, and the Halting Problem We saw that set theory and logic both seemed to contain some certain structure. So since the same structure appears in multiple separate contexts, that must mean that it’s important. As such, we give a name to this feature, which is named after the 19th century English mathematician George Boole.
Definition: boolean algebra A Boolean algebra is a set B together with two operations, generally denoted ⊕and ⊗, such that for all a, b ∈B, both a ⊕b ∈B, a ⊗b ∈B and the following properties hold: 1. Commutative Laws: For all a, b ∈B, (a) a ⊕b = b ⊕a (b) a ⊗b = b ⊗a 2. Associative Laws: For all a, b, c ∈B, (a) (a ⊕b) ⊕c = a ⊕(b ⊕c) (b) (a ⊗b) ⊗c = a ⊗(b ⊗c) 3. Distributive Laws: For all a, b, c ∈B, (a) a ⊕(b ⊗c) = (a ⊕b) ⊗(a ⊕c) (b) a ⊗(b ⊕c) = (a ⊗b) ⊕(a ⊗c) 4. Identity Laws: There exist distinct elements Id⊕and Id⊗in B such that for all b ∈B, (a) b ⊕Id⊕= b (b) b ⊗Id⊗= b 5. Complement Laws: For each b ∈B, there exists an element in B, denoted b and called the complement or negation of b, such that (a) b ⊕b = Id⊗ (b) b ⊗b = Id⊕ A Boolean algebra is sometimes shortened to the ordered triple (B, ⊕, ⊗).
Remark. Many use the symbols +, ×, 0, 1 instead of ⊕, ⊗, Id⊕, Id⊗, respectively. While this makes a lot of sense, I’m going to continue with my chosen notation so as not to confuse symbols (since it could be that B is a set of numbers in which case distinguishing between “+” in the Boolean algebra operation vs. “+” in the usual set operation.
Example 6.4.1 logical statements , ∨, ∧ is a Boolean algebra where the complement/negation is given by p = ¬p, and the “identity elements” are Id∨= c (contradiction) and Id∧= t (tautology).
6.4. BOOLEAN ALGEBRAS, RUSSELL’S PARADOX, AND THE HALTING PROBLEM 137 It is straightforward to check the Boolean algebra properties.
• Commutative Laws: • Associative Laws: • Distributive Laws: • Identity Laws: • Complement Laws: Example 6.4.2 subsets of a universe X , ∪, ∩ is a Boolean algebra where the complement/negation is given by A = Ac, and the identity elements are Id∪= ∅and Id∩= X.
• Commutative Laws: • Associative Laws: • Distributive Laws: • Identity Laws: • Complement Laws: Example 6.4.3 Let B = {0, 1} and define the following operations on B: x ⊕y = xy + x + y (mod 2) x ⊗y = xy (mod 2) x = x + 1 (mod 2) Id⊕= 0 (mod 2) Id⊗= 1 (mod 2) Is (B, ⊕, ⊗) a Boolean algebra?
It is! We check the properties in the definition of a Boolean algebra. By commutativity and associativity of the real numbers yield commutativity of ⊕and ⊗almost immediately. The others can be checked with a table.
• Distributive Laws: 138 CHAPTER 6. SET THEORY a b c a ⊕(b ⊗c) (a ⊕b) ⊗(a ⊕c) 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 a b c a ⊗(b ⊕c) (a ⊗b) ⊕(a ⊗c) 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 • Identity Laws: b Id⊕ b ⊕Id⊕ 0 0 0 1 0 1 b Id⊗ b ⊗Id⊗ 0 1 0 1 1 1 • Complement Laws: b b b ⊕b Id⊗ 0 1 1 1 1 0 1 1 b b b ⊗b Id⊕ 0 1 0 1 1 0 0 1 Theorem 6.4.4: Properties of a Boolean Algebra Let B be any Boolean algebra.
1. Uniqueness of the Complement Law For all a and x in B, if a ⊕x = Id⊗and a ⊗x = Id⊕, then x = a.
2. Uniqueness of Id⊕and Id⊗ If there exists x ∈B such that a ⊕x = a ∀a ∈B, then x = Id⊕.
If there exists y ∈B such that a ⊗y = a ∀a ∈B, then y = Id⊗.
3. Double Complement Law For all a ∈B, (a) = a.
4. Idempotent Law For all a ∈B, (a) a ⊕a = a (b) a ⊗a = a 5. Universal Bound Law For all a ∈B, (a) a ⊕Id⊗= Id⊗ (b) a ⊗Id⊕= Id⊕ 6. De Morgan’s Laws For all a, b ∈B, (a) a ⊕b = a ⊗b (b) a ⊗b = a ⊕b 6.4. BOOLEAN ALGEBRAS, RUSSELL’S PARADOX, AND THE HALTING PROBLEM 139 7. Absorption Laws For all a, b ∈B, (a) (a ⊕b) ⊗a = a (b) (a ⊗b) ⊕a = a 8. Complements of Id⊕and Id⊗ (a) Id⊕= Id⊗ (b) Id⊗= Id⊕ Proof.
1.
2.
3.
4. Let a ∈B. Then a ⊕a = (a ⊕a) ⊗Id⊗ (Identity Law) = (a ⊕a) ⊗(a ⊕a) (Complement Law) = a ⊕(a ⊗a) (Distributive Law) = a ⊕Id⊕ (Complement Law) = a (Identity Law) (and the proof that a ⊗a = a is nearly identical with symbols swapped).
5.
Example 6.4.5 Let B = {1, 2, 3, 4, 6, 8, 12, 24} and define the following operations on B: x ⊕y = lcm(x, y) x ⊗y = gcd(x, y) x = 24 x Is (B, lcm, gcd) a Boolean algebra?
This is much harder to tell by simply looking at it, and checking with tables – while not unreasonable for a computer, is not practical by hand (each distributive law will use 216 rows, for example).
The first task might be to try to better understand this collection. What are Id⊕and Id⊖?
This should be inferrable from the Complement Law, since a ⊕a = Id⊗. But writing out this relatively short table, we have 140 CHAPTER 6. SET THEORY b b b ⊕b 1 24 24 2 12 12 3 8 24 4 6 12 6 4 12 8 3 24 12 2 12 24 1 24 But the third column is not constant, so there cannot be a unique choice of Id⊗, which violates one of the Boolean Algebra Properties. Therefore (B, lcm, gcd) is not a Boolean algebra.
Chapter 7 Properties of Functions 7.1 Functions Defined on General Sets Definition Let A, B be sets. The Cartesian product of A and B is the set A × B = {(a, b) : a ∈A and b ∈B} Example 7.1.1: Cartesian Plane The Cartesian plane, usually denoted R2 or R × R is the collection of all ordered pairs of real numbers (x, y).
Example 7.1.2 Let A = {1, 2, 3} and B = {2, 3, 4}. What are 1. A × B 2. (A × A) ∩(B × B) 1. INCOMPLETE 2. INCOMPLETE Exercise 7.1.3 Prove the following claim: Claim. If A1 ⊆A2 and B1 ⊆B2, then (A1 × B1) ⊆(A2 × B2).
You’ve probably always thought about a function f via its graph y = f(x). But what is the graph of the function other than ordered pairs of the form (x, f(x)) where the inputs x ∈R and the outputs f(x) ∈R. This is the motivation to keep in mind as we work on introducing the formal definition of a limit.
Definition: function A function or map, f, from a set A (the domain) to a set B (the codomain), denoted f : A →B, is a subset of A × B with the following properties: 1. [uses the whole domain] For every a ∈A, “f(a)” is defined.
∀a ∈A, ∃b ∈B such that (a, b) ∈f 141 142 CHAPTER 7. PROPERTIES OF FUNCTIONS 2. [well-defined] For every a ∈A, the set “{f(a)}” contains exactly one element.
∀a ∈A and ∀b1, b2 ∈B, if (a, b1) ∈f and (a, b2) ∈f, then b1 = b2.
Since the codomain element b is uniquely associated with the domain element a, one usually writes f(a) = b.
We also sometimes say that a is “mapped” to b if f(a) = b.
Remark. You probably know “well-definedness” as “passing the vertical line test”.
Example 7.1.4 Let f ⊆{1, 2, 3, 4} × {5, 6, 7, 8} be given by f = {(1, 5), (2, 7), (3, 7), (4, 6)} Is f a function?
Yes. We see that f(1), f(2), f(3), f(4) are all defined, and moreover, each of 1, 2, 3, 4 are paired with exactly one element of {5, 6, 7, 8}.
Example 7.1.5 Let f ⊆N × R be given by f = (n, y) : n2 −y = 0 Let g ⊆N × R be given by g = (n, y) : n2 + y2 = 10 Are either of f or g functions?
7.1. FUNCTIONS DEFINED ON GENERAL SETS 143 1 2 3 4 N 1 2 3 4 5 6 7 8 9 R (n,f(n)) 1 2 3 4 N −4 −3 −2 −1 1 2 3 4 R (n,g(n)) f is a function. Its defining equation can be rewritten as x = n2, so every unique n ∈N is mapped to only one x ∈R.
g is not a function. Both (5, √ 75) and (5, − √ 75) are elements of g.
Notably the defining equation can be rearranged to x = √ 100 −n2 or x = − √ 100 −n2, so there are two real numbers paired with n when n ̸= 0, 10.
Example 7.1.6 Let f : Q →Z be given by f p q = p.
Is f a function?
No, f is not a function as it fails to be well-defined. Notice that 1 3 = 2 6, but 1 = f 1 3 ̸= f 2 6 = 2.
144 CHAPTER 7. PROPERTIES OF FUNCTIONS 7.1.1 Arrow Diagram Definition: arrow diagram Given a function f : A →B, an arrow diagram is formed by drawing an arrow from a ∈A to b ∈B if and only if f(a) = b.
Example 7.1.7: Arrow Diagrams Let A = {a1, a2, a3, a4} and B = {b1, b2, b3, b4, b5}. Draw the arrow diagram for each of the following sets: 1. R1 = {(a1, b1), (a2, b3), (a4, b5)} Function 2. R2 = {(a1, b2), (a2, b1), (a3, b3), (a3, b4), (a4, b5)} Not a function 3. R3 = {(a1, b2), (a2, b3), (a3, b1), (a4, b3)} Function Which of these are functions?
1. R1 = {(a1, b1), (a2, b3), (a4, b5)} A R1 B a1 a2 a3 a4 b1 b2 b3 b4 b5 2. R2 = {(a1, b2), (a2, b1), (a3, b3), (a3, b4), (a4, b5)} A R2 B a1 a2 a3 a4 b1 b2 b3 b4 b5 3. R3 = {(a1, b2), (a2, b3), (a3, b1), (a4, b3)} A R3 B a1 a2 a3 a4 b1 b2 b3 b4 b5 7.1. FUNCTIONS DEFINED ON GENERAL SETS 145 Exercise 7.1.8: Function or Not?
Determine which of the following are functions by looking at the arrow diagram.
1. f A B 2. g A B 3. h A B 7.1.2 Range, Preimage Definition: range, preimage Let f : A →B be a function. The range of f, sometimes denoted f(A), is the set Range(f) = {b ∈B : b = f(a) for some a ∈A} .
Given a subset S ⊆B, the preimage of S under f is the set Preim(S) = {a ∈A : f(a) = s for some s ∈S} .
Remark. The preimage of S under f is quite commonly written as f −1(S), but so as to avoid confusion, we’ll not utilize such notation here.
146 CHAPTER 7. PROPERTIES OF FUNCTIONS Example 7.1.9 Let A = {a1, a2, a3, a4, a5} and B = {b1, b2, b3, b4} and let f : A →B be the function f = {(a1, b2), (a2, b3), (a3, b2), (a4, b4), (a5, b2)} .
Find each of the following.
1. Range(f) 2. Preim({b1}) 3. Preim({b2}) 4. Preim({b3}) 5. Preim({b3, b4}) A f B a1 a2 a3 a4 a5 b1 b2 b3 b4 1. Range(f) = {b2, b3, b4} 2. Preim({b1}) = ∅ 3. Preim({b2}) = {a1, a3, a5} 4. Preim({b3}) = {a2} 5. Preim({b3, b4}) = {a2, a4} Definition Two functions f : X →Y and g : X →Y are equal if and only if f(x) = g(x) for every x ∈X.
Remark. This is implicit in the definition, but f and g must have the same domain and codomain in order to be equal.
Remark. Note that this is equivalent to saying that the sets {(x, y) ∈X × Y : f(x) = y} and {(x, y) ∈X × Y : g(x) = y} are equal Example 7.1.10 Determine which of the following pairs of functions, f and g, are equal: 1. f : Q →Q given by f(x) = 1 x2 + 1, 7.1. FUNCTIONS DEFINED ON GENERAL SETS 147 g : R →R given by β(x) = 1 x2 + 1.
2. f : Z −{1} →Z, f(x) = x2 −1 x −1 , g : Z →Z given by g(x) = x + 1.
3. f : Z →Z given by f(x) = x3 + x x2 + 1, g : Z →Z given by g(x) = x.
4. f : Z →Z given by f(x) = x2, g : Z →N given by g(x) = x2.
1. f and g are not equal because f(π) is undefined, but g(π) is defined.
2. f and g are not equal because f(1) is undefined, but g(1) is defined.
3. Noting that x3 + x x2 + 1 = x(x2 + 1) x2 + 1 = x, we see that f(x) = g(x) for all x ∈Z. Thus f = g.
4. Even though both sets have the same domains and seem to have the same ranges (because x2 ≥0), the codomains are different so the functions are not equal.
148 CHAPTER 7. PROPERTIES OF FUNCTIONS 7.2 One-to-One, Onto, and Inverse Functions Definition: one-to-one, onto, bijection Let f : A →B be a function. f is said to be one-to-one or injective if and only if for all a1, a2 ∈A, if f(a1) = f(a2) then a1 = a2.
The tagline is that “two distinct inputs cannot have the same output.” f is said to be onto or surjective if and only if for all b ∈B, there is some a ∈A for which f(a) = b.
The tagline is that the range of f is all of B.
f is called bijective or a bijection if and only if it is both one-to-one and onto.
Remark. A bijection is sometimes called a one-to-one correspondence. We will be avoiding this term for obvious reasons.
Remark. One-to-one and onto are universal statements and should be proven according to the usual strategies from 4.1 (note that “onto” is even a statement with nested quantifiers.) Example 7.2.1 Use arrow diagrams to give examples of functions f : A →B that are all combinations of one-to-one (or not) and onto (or not).
A B f Figure 7.1: f is both one-to-one and onto.
A f B Figure 7.2: f is one-to-one, but not onto.
A f B Figure 7.3: f is not one-to-one, but is onto.
A f B Figure 7.4: f is neither one-to-one nor onto.
7.2. ONE-TO-ONE, ONTO, AND INVERSE FUNCTIONS 149 Example 7.2.2 Give examples of functions R →R that are all combinations of one-to-one (or not) and onto (or not).
−2 −1 1 2 R −2 −1 1 2 R f(x)=x3 Figure 7.5: f is both one-to-one and onto.
−2 −1 1 2 R −2 −1 1 2 R f(x)=arctan(x) Figure 7.6: g is one-to-one, but not onto.
−2 −1 1 2 R −2 −1 1 2 R f(x)=(x−1)x(x+1) Figure 7.7: f is not one-to-one, but is onto.
−2 −1 1 2 R −2 −1 1 2 R f(x)=x2 Figure 7.8: f is neither one-to-one nor onto.
Let’s look at Example 7.2.2 and for every function f, draw the reverse arrows (call them g).
A B f g Figure 7.9: f is both one-to-one and onto.
g is both one-to-one and onto.
A B f g Figure 7.10: f is one-to-one, but not onto.
g is not a function because it fails the first condition in the definition.
150 CHAPTER 7. PROPERTIES OF FUNCTIONS A B f g Figure 7.11: f is not one-to-one, but is onto.
g is not a function.
A B f g Figure 7.12: f is neither one-to-one nor onto.
g is not a function.
Na¨ ıvely, we would want to define an inverse function by just reversing the arrows. What we see is that the “inverse” only exists in the case that f is one-to-one and onto. Explicitly, Theorem 7.2.3: Existence of an inverse Let f : A →B be a function and define the following subset of B × A: f −1 = {(b, a) ∈B × A : f(a) = b} f −1 is a function if and only if f is a bijection.
Proof. Let f : A →B be a function.
Uses whole domain. For every b ∈B, there exists a ∈A so that f −1(b) = a if and only if there exists a ∈A so that f(a) = b (i.e., if and only if f is surjective).
Well-defined. For every a1, a2 ∈A and b ∈B, f −1(b) = a1 ∧f −1(b) = a2 = ⇒a1 = a2 is true if and only if f(a1) = b ∧f(a2) = b = ⇒a1 = a2 (i.e., if and only if f is one-to-one).
Definition: inverse Let f : A →B be a function. The function f −1 : B →A defined in Theorem 7.2.3 is called an inverse of f. If f −1 exists, then f is called invertible.
Example 7.2.4 Let f : Z →Z be given by f(x) = 3|x| −5.
1. Prove or disprove: f is one-to-one.
2. Prove or disprove: f is onto.
3. If f is both one-to-one and onto, find an inverse for f.
1. INCOMPLETE 2. INCOMPLETE 7.2. ONE-TO-ONE, ONTO, AND INVERSE FUNCTIONS 151 3. INCOMPLETE 4. INCOMPLETE Example 7.2.5 Let g : R →R be given by g(x) = 3 √ x5 + 1.
1. Prove or disprove: g is one-to-one.
2. Prove or disprove: g is onto.
3. If g is both one-to-one and onto, find an inverse for f.
1. INCOMPLETE 2. INCOMPLETE 3. INCOMPLETE Example 7.2.6 Let h : R × R →R × R be given by h(x, y) = (x + y, x −y).
1. Prove or disprove: h is one-to-one.
2. Prove or disprove: h is onto.
3. If h is both one-to-one and onto, find its inverse.
1. h is one-to-one. To see this, suppose that there are pairs (x1, y1), (x2, y2) ∈R × R for which h(x1, y1) = (x1 + y1, x1 −y1) = (x2 + y2, x2 −y2) = h(x2, y2).
Then x1 + y1 = x2 + y2 rearranges to x1 = x2 −y1 + y2, and thus x1 −y1 = x2 −y2 = ⇒x2 −y1 + y2 −y1 = x2 −y2 = ⇒−2y1 = −2y2 = ⇒y1 = y2 Then y1 = y2 implies that x1 = x2 −y1 + y2 = x2 −0 = x2.
So it follows that (x1, y1) = (x2, y2).
2. h is onto. To see this, let (w, z) ∈R × R be a point in the codomain. We seek a pair (x, y) so that h(x, y) = (w, z). This can be achieved by solving a linear system (with standard techniques): h(x, y) = (x + y, x −y) = (w, z) = ⇒ x + y = w x −y = z = ⇒ x = w+z 2 y = w−z 2 Therefore, for every (w, z), we have that h w+z 2 , w−z 2 = (w, z).
3. We secretly found the inverse in the previous part.
h(x, y) = x + y 2 , x −y 2 .
152 CHAPTER 7. PROPERTIES OF FUNCTIONS Remark. The astute reader will realize that the last example was a linear transformation, given by multiplication by 1 1 1 −1 , and whose inverse transformation is given by multipliation by 1 1 1 −1 −1 = " 1 2 1 2 1 2 −1 2 # Example 7.2.7 Let j : R →R be given by j(x) = ex.
1. Prove or disprove: j is one-to-one.
2. Prove or disprove: j is onto.
3. If j is both one-to-one and onto, find its inverse.
1. INCOMPLETE 2. INCOMPLETE 3. INCOMPLETE 4. INCOMPLETE 7.3. COMPOSITION OF FUNCTIONS 153 7.3 Composition of functions Definition: function composition Let f : A →B and g : B →C be functions. Define a new function g ◦f : A →C as follows: (a, c) ∈g ◦f iff ∃b ∈B s.t. (a, b) ∈f and (b, c) ∈g The function g ◦f is called the composition of f and g and is usually said aloud as “g of f.” In the above definition, the function definition requires that b ∈range(F), so we can freely use familiar notation: (g ◦f)(a) = g(f(a)) for all a ∈A.
Example 7.3.1 Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8, 9}, and C = {10, 11, 12}. Let f and g be functions given by • f = { (1, 6), (2, 7), (3, 9), (4, 6) } • g = { (5, 12), (6, 10), (7, 11), (8, 12), (9, 10) } Determine g ◦f.
g ◦f = { (1, 10), (2, 11), (3, 10), (4, 10) } .
We can also draw the arrow diagram to see this – composition is just following the arrow paths from A to C.
A f B g C 1 2 3 4 5 6 7 8 9 10 11 12 Theorem 7.3.2: Inverses and Composition Let f : A →B and g : B →A be two functions. Then g = f −1 if and only if, for all a ∈A and forall b ∈B, (g ◦f)(a) = a and (f ◦g)(b) = b.
That is, g ◦f and f ◦g are both the identity functions.
Proof. Let f : A →B and g : B →A be functions.
Case (⇒). Suppose that g = f −1 and let (a, b) ∈f be arbitrary. By definition of the inverse (b, a) ∈g, hence (g ◦f)(a) = a and (f ◦g)(b) = b.
154 CHAPTER 7. PROPERTIES OF FUNCTIONS Case (⇐). Suppose now that, for all a ∈A and forall b ∈B, we have (g ◦f)(a) = a and (f ◦g)(b) = b.
Let (a0, b0) ∈f be arbitrary but fixed. (g ◦f)(a0) = a0 implies that there exists some b1 ∈B for which (a0, b1) ∈f and (b1, a0) ∈g. However, since f is well-defined, then b0 = b1, i.e, (b0, a0) ∈g.
Similarly, let (b0, a0) ∈g be arbitrary but fixed. (f ◦g)(b0) = b0 implies that there exists some a1 ∈A for which (a1, b0) ∈f and (b0, a1) ∈g. However, since g is well-defined, then a1 = a0, i.e, (b0, a0) ∈g.
We now that have (a0, b0) ∈f ⇐ ⇒(b0, a0) ∈g, whence g = f −1 as desired.
Example 7.3.3 Let f : A →B and g : B →C be functions.
1. Draw an arrow diagram where both f and g are one-to-one. What do you observe about g ◦f?
2. Draw an arrow diagram where both f and g are onto. What do you observe about g ◦f?
1. Let f and g be the one-to-one functions shown below.
A f B g C a1 a2 a3 a4 b1 b2 b3 b4 b5 c1 c2 c3 c4 c5 We see that g ◦f is also one-to-one.
2. Let f and g be the onto functions shown below.
A f B g C a1 a2 a3 a4 a5 b1 b2 b3 b4 b5 c1 c2 c3 c4 Theorem 7.3.4 Let f : A →B and g : B →C be functions.
1. If f, g are both one-to-one, then so is g ◦f.
7.3. COMPOSITION OF FUNCTIONS 155 2. If f, g are both onto, then so is g ◦f.
Proof. Let f : A →B and g : B →C be functions.
1. Suppose f and g are both one-to-one. Let a1, a2 ∈A and suppose that g ◦f(a1) = g(f(a1)) = g(f(a2)) = g ◦f(a2).
Since g is one-to-one, then it must follow that f(a1) = f(a2), and since f is one-to-one, it follows that a1 = a2. Therefore g ◦f is one-to-one.
2. Suppose f and g are both onto and let c ∈C. Since g is onto, there must be some b ∈B such that g(b) = c. Since f is onto, there must be some x ∈A such that f(x) = b, i.e., that g ◦f(x) = g(f(x)) = g(b) = c. Therefore g ◦f is onto.
Corollary 7.3.5 If f : A →B and g : B →C are bijections, then g ◦f is a bijection as well.
Corollary 7.3.5 is actually very useful in practice. It can be hard to construct a bijection from a set A to a set D. But it may be straightforward to construct a bijection f : A →B, a bijection g : B →C, and a bijection h : C →D. The result then tells us that h ◦g ◦f : A →D is a bijection.
156 CHAPTER 7. PROPERTIES OF FUNCTIONS 7.4 Cardinality Consider the following three situations: A f B Figure 7.13: f is one-to-one, but not onto.
A f B Figure 7.14: f is both one-to-one and onto.
A f B Figure 7.15: f is not one-to-one, but is onto.
When A and B have the same number of elements, f is a bijection, but when the sets have different numbers of elements, then f can only be one-to-one or onto (and not both). If we let |A| and |B| denote the number of elements in each of A and B, the observation can be stated as follows: • |A| ≤|B| if and only if there is a one-to-one function f : A →B.
• |A| ≥|B| if and only if there is an onto function f : A →B.
• |A| = |B| if and only if there is a bijective function f : A →B.
This allows us to reframe and formalize a notion of “size” of sets via functions.
Definition: cardinality Let A, B be sets. A and B are said to have the same cardinality if there is a bijection between A and B. We write “|A|” or “#A” to denote the cardinality of A.
Example 7.4.1: Shift-by-m map Let m and n be positive integers. Show that the sets {1, . . . , n} and {m + 1, . . . , m + n} have the same cardinality by proving that the function f :{1, . . . , n} →{m + 1, . . . , m + n} f(x) = x + m is a bijection.
Lemma 7.4.2 For all finite sets A and B, |A ∪B| ≤|A| + |B|.
The proof of Lemma 7.4.2 will be reserved for an appendix. The proof isn’t hard, it’s just obnoxiously tedious given how obvious the result is.
7.4. CARDINALITY 157 Example 7.4.3 Prove that, for all natural numbers n ≥2, |A1 ∪· · · ∪An| ≤|A1| + · · · + |An|.
We approach via induction, using Lemma 7.4.2 for the base case and, along with associativity of the union operation, using it again for the inductive step. INCOMPLETE Definition: types of cardinality A set A is called...
• finite if it is either empty or is in one-to-one correspondence with the set {0, 1, 2, 3, . . . , n} for some n ∈N.
• infinite if it is not finite.
• countable (or countably-infinite) if it has the same cardinality as N.
• uncountable if it is neither finite nor countable.
Remark. Some authors allow “countable” to include finite sets. In this class we’ll only use the term in the case of infinite sets, so such conventional discrepancies won’t matter.
Example 7.4.4 Find a bijection f : (0, 1) →R.
1 −2 2 4 The function f : (0, 1) →R f(x) = tan πx −π 2 is a bijection with inverse f −1(x) = 1 π arctan(x) + 1 2.
158 CHAPTER 7. PROPERTIES OF FUNCTIONS Example 7.4.5 Find a bijection f : [0, 1] →R.
The motivation for this is to take the tangent function (which can be modified simply to send (0, 1) to R) and to define it in a piecewise way so as to involve the interval endpoints {0, 1}).
1 −1 1 2 3 4 Let g = tan πx −π 2 . Then the bijection f : [0, 1] →R is given by f(x) = 0 when x = 0 1 when x = 1 g(x) when g(x) / ∈N 2 + g(x) otherwise.
Example 7.4.6: Z is countable Find a bijection f : N →Z.
−3 −2 −1 0 1 2 f(0) = 0 f(1) = −1 f(2) = 1 f(3) = −2 f(4) = 2 .
.
.
f(n) = n 2 when n is even, −n + 1 2 when n is odd.
Proof. We now prove that f is bijective.
One-to-one. To see that f is one-to-one, we consider three separate cases. Let x, y ∈N.
7.4. CARDINALITY 159 • x and y are both even. Suppose f(x) = f(y) where x and y are both even.
Then x 2 = y 2 = ⇒ x = y.
• x and y are both odd. Suppose f(x) = f(y) where x and y are both odd. Then −x + 1 2 = −y + 1 2 = ⇒ x = y.
• x is even and y is odd. (We use the contrapositive of the one-to-one definition here.) Suppose That x is even and y is odd (so in particular we are supposing x ̸= y). Then x 2 = −y + 1 2 = ⇒ x = −(y + 1) Since x ≥0 and −(y + 1) < 0, it follows that f(x) ̸= f(y).
Onto. To see that f is onto, we consider two separate cases. Let y ∈Z be arbitrary.
• y < 0. Suppose y < 0. Then we choose x = −2y −1 and see that f(x) = −−2y −1 + 1 2 = y.
• y ≥0. Suppose y ≥0. Then we choose x = 2y and see that f(x) = 2y 2 = y.
Example 7.4.7: N × N is countable.
Find a bijection f : N →N × N.
Visually, we’ll define f to iterate through all pairs of natural numbers as shown below.
0 0 1 1 2 2 3 3 0 0 1 1 2 2 3 3 f(0) f(1) f(2) f(3) f(4) f(5) f(6) f(7) f(8) f(9) We’re going to define the function 160 CHAPTER 7. PROPERTIES OF FUNCTIONS f : N →N × N f(x) = (x −ℓx, ℓx) but we’ll need to explain a bit about this ℓx.
Looking at the x-axis in the figure above, see that every point on this axis is f k(k+1) 2 for some k ∈N. This leads us to the first observation: for every x ∈N, there exists a unique n ∈N such that n(n + 1) 2 ≤x < (n + 1)(n + 2) 2 .
As such, we x we define a number ℓx by ℓx = x −n(n + 1) 2 where n is the particular number in the above inequality.
Notice, in particular, that 0 ≤ℓx ≤n and that ℓx attains every value between 0 and n as x varies.
0 1 2 3 4 5 6 7 8 9 10 x = 0(0+1) 2 1(1+1) 2 2(2+1) 2 3(3+1) 2 4(4+1) 2 ℓ0 = 0 ℓ1 = 0 ℓ2 = 1 ℓ3 = 0 ℓ4 = 1 ℓ5 = 2 ℓ6 = 0 ℓ7 = 1 ℓ8 = 2 ℓ9 = 3 ℓ10 = 0 Now we prove that this function f is a bijection.
Proof. To see that f is injective, suppose that f(x) = f(y) for some x, y ∈N: (x −ℓx, ℓx) = (y −ℓy, ℓy).
Since ℓx = ℓy, then x −ℓx = y −ℓy implies that x = y.
To see that f is surjective, we first make the following observation (x, y) (x + y, 0) 7.4. CARDINALITY 161 The point (n, 0) is obtained via f n(n+1) 2 , so we must have that (x + y, 0) = f (x + y)(x + y + 1) 2 .
whence (x, y) = f (x + y)(x + y + 1) 2 + y .
In fact, f −1(x, y) = (x + y)(x + y + 1) 2 + y.
Example 7.4.8 Show that Q is countable.
Note that every rational number can be written in the form p q with p ∈Z and q ∈Z+. In this way, we draw out a grid for Z × Z+ and label the point (p, q) with the fraction p q. There will be several fractions that are not written in lowest terms, so cross out each of these points. We then start counting outwards from 0 = 0 1 according to the diagram below (which mimics the strategy which shows that N × N is countable).
−3 1 −3 2 −3 3 −3 4 −2 1 −2 2 −2 3 −2 4 −1 1 −1 2 −1 3 −1 4 0 1 0 2 0 3 0 4 1 1 1 2 1 3 1 4 2 1 2 2 2 3 2 4 3 1 3 2 3 3 3 4 7.4.1 Infinity Infinities: “To Infinity and Beyond” Theorem 7.4.9: Cantor, 1891 R is uncountable.
The proof of this theorem uses a technique which is now called a “Diagonal Argument” or “Cantor’s Diagonal Argument.” We note that, from Example 7.4.4, it suffices to show that (0, 1) is 162 CHAPTER 7. PROPERTIES OF FUNCTIONS uncountable.
Proof. We approach by contradiction. Suppose that (0, 1) is countable. Then there is some way to list all real numbers {x0, x1, x2, x3, . . .} in (0, 1). Writing these down in binary, we would have a table that could look something like the one below (which was randomly generated by Mathematica): x1 = 0 .
4 8 2 9 1 3 2 3 . . .
x2 = 0 .
3 2 8 2 0 5 8 9 . . .
x3 = 0 .
4 6 5 1 9 9 3 8 . . .
x4 = 0 .
5 8 1 0 8 1 3 1 . . .
x5 = 0 .
3 5 5 4 7 2 3 3 . . .
x6 = 0 .
5 3 8 2 3 1 2 6 . . .
x7 = 0 .
6 3 1 3 3 6 9 1 . . .
x8 = 0 .
0 1 4 3 3 3 8 5 . . .
.
.
.
...
For each natural number m in the range 0 ≤m ≤9, we define the “swap of m,” denoted ◦ m, as 9 −m. Now we have that 0 ≤ ◦ m≤9 and there are no numbers for which m = ◦ m.
Let dk represent the kth digit in xk, and let x = 0.
◦ d1 ◦ d2 ◦ d3 ◦ d4 . . . be the decimal formed from these digits, after “swapping.” With the example table above, we have that x = 0.
◦ 4 ◦ 2 ◦ 5 ◦ 0 ◦ 7 ◦ 1 ◦ 9 ◦ 5 . . . = 0.58492804 · · · Such an x is specifically designed to disagree with xk at the kth digit, for every natural number k.
We have thus constructed a real number which was not counted. (If it did appear on the table somewhere, we could write x = xn for some n, but then x and xn would agree at their nth digit).
This contradicts the assumption that (0, 1) was countable.
This blew the minds of many mathematicians at the time (and stirred up some controversy), because it implied that there were different sizes of infinity, and the cardinality of R was a “larger infinity” than the cardinality of N.
It also leads to an interesting question: how many different infinities are there? The answer is infinitely-many (actually, uncountably-many!) This discussion is quite heavy, but there is a way to always get a “larger infinity.” Example 7.4.10 Show that there is no bijection between {a, b} and P({a, b}) using only functions (i.e. without appealing to simply counting.) This is pretty obvious from a counting perspective: {a, b} has 2 elements and its power set has 22 = 4 elements. Nevertheless, let’s draw a random function f : {a, b} →P({a, b}) 7.4. CARDINALITY 163 f a b ∅ {a} {b} {a, b} Observe that a ∈f(a) = {a, b} and b / ∈f(b) = {a} Define the set T = x ∈{a, b} : x / ∈f(x) , which in our particular case is T = {b}. By construction, T is a subset of {a, b} and is not in the range of f. You can try playing around with different functions, and every time you’ll find that T is not in the range of f. It’s precisely this way that one proves the above theorem.
Theorem 7.4.11 For every set X, there is no bijection between X and the power set P(X).
Proof of Theorem 7.4.11. Let X be a set and P(X) its power set. Let f : X →P(X) be any function, and define the (possibly empty) set: T = x ∈X : x / ∈f(x) .
Tending toward a contradiction, assume that f is bijective. In particular, f is surjective, so there is some y ∈X for which f(y) = T.
• If y ∈T, then since f(y) = T it must be that y ∈f(y). But by definition of T, y / ∈f(y).
Contradiction.
• If y / ∈T, then since f(y) = T, it must be that y / ∈f(y). But by definition of T, y ∈T.
Contradiction.
Therefore f cannot be a surjective map, hence there cannot be any bijection f : X →P(X).
Since there is an obvious injection from X to P(X) f(x) = {x} this means that the cardinality of P(X) must be strictly larger than the cardinality of X.
By repeatedly applying this result, one can find an infinite sequence of infinite sets with larger and larger cardinalities N, P(N) , P(P(N)) , P(P(P(N))) , . . .
164 CHAPTER 7. PROPERTIES OF FUNCTIONS To simplify symbols, the cardinalities of the above sets are usually defined using the symbols ℵ0, 2ℵ0, 22ℵ0, 222ℵ0 , . . .
which matches how we think about the size of the power set in the finite case.
Corollary 7.4.12 P(N) is uncountable.
Exercise 7.4.13 Find a bijection f : R →P(N).
Exercise 7.4.14 Let X be any set and f, g both one-to-one functions for which N f − − →X g − →P(N).
Show that one of f or g must be an onto function.
Exercise 7.4.14 is really just here as a joke. The relevant Google search phrase is “Continuum Hypothesis”, and the long-and-short of it is that it’s not something you can prove.
7.4.2 New Bijections from Old Recall that Theorem 7.3.4 allows us to compose bijections and retain a bijection - this is extremely convenient because it allows us to come up with more “obvious” bijections and compose them.
Example 7.4.15 Let E = ek : k ∈Z . Find a bijection E →N.
Let f : N →Z be the bijection from Example 7.4.6. We consider a composition of bijections: δ : E ln − − →Z f−1 − − − →N where δ(x) = f −1(ln(x)).
Since f −1 is a bijection, we just need to prove that ln is a bijection.
Proof.
ln is one-to-one. This fact easily follows from calculus - ln(x) is strictly increas-ing on (0, ∞). But to see it with out current methods, let that ek1, ek2 be arbitrary elements of E and suppose that ln(ek1) = ln(ek2). Then k1 = k2, whence ek1 = ek2.
ln is onto. Let k ∈Z be arbitrary. Choosing ek ∈E, one see that ln(ek) = k.
7.4. CARDINALITY 165 Explicitly, the bijection δ is given by δ(x) = f −1(ln(x)) = ( 2 ln(x) when ln(x) ≥0 −2 ln(x) −1 when ln(x) < 0 = ( 2 ln(x) when x ≥1 −2 ln(x) −1 when x < 1 Proposition 7.4.16: Cartesian product of bijections If f1 : A →B and f2 : C →D are bijections, then ˜ f :A × C →B × D ˜ f(x, y) = f1(x), f2(y) Is also a bijection.
Proof. Let A, B, C, D, f1, f2, and ˜ f be as in the proposition.
One-to-one. Let (a1, c1), (a2, c2) ∈A × C and suppose that ˜ f(a1, c1) = ˜ f(a2, c2), i.e., suppose that f1(a1), f2(c1) = f1(a2), f2(c2) which, by definition of the Cartesian product, yields f1(a1) = f1(a2) f2(c1) = f2(c2) and since f1 and f2 are bijections (in particular, are one-to-one), it follows that a1 = a2 and c1 = c2.
Therefore (a1, c1) = (a2, c2).
Onto. Let (b, d) ∈B × D. Since f1 and f2 are bijections (and in particular, are onto), then there must be some a ∈A and c ∈C so that f1(a) = b and f2(c) = d and therefore, there is a pair (a, c) ∈A × C for which ˜ f(a, c) = f1(a), f2(c) = (b, d).
Example 7.4.17 Find a bijection Z × Z →N.
166 CHAPTER 7. PROPERTIES OF FUNCTIONS From previous examples, we know two useful bijections already: α :N × N →N β :Z →N α(x, y) = (x + y)(x + y + 1) 2 + y β(x) = ( 2x when x ≥0, −2x −1 when x < 0 With this in mind, we can make ˜ β as in Proposition 7.4.16, and compose it with α (which will again be a bijection by Corollary 7.3.5.
γ : Z × Z ˜ β − − →N × N α − − →N Explicitly, the bijection γ is given by γ(x) = α(˜ β(x, y)) = α(β(x), β(y)) = (β(x) + β(y))(β(x) + β(y) + 1) 2 + β(y) = (2x + 2y)(2x + 2y + 1) 2 + 2y x ≥0 ∧y ≥0 (2x −2y −1)(2x −2y) 2 −2y −1 x ≥0 ∧y < 0 (−2x −1 + 2y)(−2x + 2y) 2 + 2y x < 0 ∧y ≥0 (−2x −2y −2)(−2x −2y −1) 2 −2y −1 x < 0 ∧y < 0 Example 7.4.18 Use induction to prove that, for every n, the n-fold Cartesian product of N, i.e.
n Y j=1 N = N × N × · · · × N | {z } n times = {(n1, n2, . . . , nk) : each ni ∈N} is countable.
Hint: N × N × · · · × N | {z } k times = N × N × · · · × N | {z } k−1 times INCOMPLETE. Although frankly, I’m not even sure we’ll actually do this in class. Maybe move it to be an “exercise” in a future semester?.
7.4. CARDINALITY 167 Example 7.4.19 Prove that the infinite Cartesian product of N, i.e.
∞ Y j=1 N = N × N × N · · · = {(n1, n2, n3, . . .) : each ni ∈N} is uncountable.
Hint: Consider a diagonal argument, like in the proof of Theorem 7.4.9.
INCOMPLETE.
Although frankly, I’m not even sure we’ll actually do this in class. Maybe move it to be an “exercise” in a future semester?.
Exercise 7.4.20: “weaving” pairs of real numbers Let x, y be real numbers in (0,1). Write out x, y in terms of their decimal digits, which we’ll denote with xi’s and yj’s: x = 0.x1x2x3x4x5 . . .
y = 0.y1y2y3y4y5 . . .
Show that the function w :(0, 1) × (0, 1) →(0, 1) w(x, y) = 0.x1y1x2y2x3y3x4y4x5y5 . . .
is a bijection.
Example 7.4.21 Find a bijection R × R →R.
In Example 7.4.4, we constructed a function f : (0, 1) →R, and its inverse was given by f −1(x) = 1 π arctan(x) + 1 2 Combining this with the “weaving” bijection w from Exercise 7.4.20 and the Cartesian product bijection ˜ f one gets from Proposition 7.4.16, the following composition is a bijection: R × R ˜ f−1 − − − →(0, 1) × (0, 1) w − − →(0, 1) f − − →R Example 7.4.22 Find a bijection Z ∪E →Z where E = ek : k ∈Z as in Example 7.4.15.
We remark that Z = N ∪(−N) (where, for lack of a better notation, −N denotes the set of nonpositive integers {. . . , −2, −1, 0}). We already have bijections f −1 : Z →N (from 168 CHAPTER 7. PROPERTIES OF FUNCTIONS Example 7.4.6) and δ : E →N (from Example 7.4.15), so we naively try β :Z ∪E →N ∪(−N) β(x) = ( f −1(x) when x ∈Z −δ(x) when x ∈E We note that E ∩Z = {1} and N ∩(−N) = {0}, so there should be some concern that this function is not well-defined, or isn’t one-to-one. We observe the following f −1(Z) = {0, 1, 2, . . .} f −1(1) = 2 −δ(E) = {. . . , −2, −1, 0} δ(1) = 0 By simply removing the intersection {1} from both E and Z, we eliminate the overlap at 0, and we can then define β(1) = 2 to plug the hole at 2.
β :Z ∪E →Z β(x) = f −1(x) when x ∈Z −E −δ(x) when x ∈E −Z 2 when x ∈E ∩Z .
Explicitly, our function is doing this: β(Z −E) = {0, 1, 3, 4, 5, . . .} β(E −Z) = {. . . , −3, −2, −1} β(E ∩Z) = {2} Now we prove that β is a bijection.
Proof. Let β, δ, f be as described above.
β is one-to-one. Let x1, x2 ∈Z ∪E. We approach by cases.
• Case 1. [x1, x2 ∈Z −E] Left as an exercise.
• Case 2. [x1, x2 ∈E −Z] Left as an exercise.
• Case 3. [x1, x2 ∈Z ∩E] Left as an exercise.
• Case 4. [x1 ∈Z −E and x2 ∈E −Z] Left as an exercise.
• Case 5. [x1 ∈Z −E and x2 ∈Z ∩E] Left as an exercise.
• Case 6. [x1 ∈E −Z and x2 ∈Z ∩E] Left as an exercise.
β is onto. Let y ∈Z be an integer. We approach by cases.
• Case 1. Suppose that y ∈{0, 1, 3, 4, 5, . . .} – Subcase 1. Suppose that y is even. Left as an exercise.
– Subcase 2. Suppose that y is odd. Left as an exercise.
• Case 2. Suppose that y ∈{. . . , −3, −2, −1}. Left as an exercise.
• Case 3. Suppose that y = 2. Left as an exercise.
Therefore, β is a bijection.
7.4. CARDINALITY 169 Theorem 7.4.23 Let A, B be countable. Then A ∪B is countable.
In order to prove this, we first need the following lemma.
Lemma 7.4.24 If A is countable and B ⊆A, then B is either finite or countable.
Proof. If B is finite, then we’re done. So, suppose B is infinite and let f : A →N be a bijection (in particular an injection). Since B ⊆A, there is the natural inclusion map ι : B →A (where ι(b) = b). This map is clearly injective, so the composition of functions f ◦ι : B →N is also injective (by Theorem 7.3.4). Therefore |B| ≤|N|.
Proof of Theorem 7.4.23. Consider the sets A −B and B, which are disjoint. Since B is countable, there is a bijection β : B →N. Since A is countable and A −B is a subset of A, then A −B is either finite or countable.
Case 1 (A −B is finite). Suppose that A −B is finite. Then there is a bijection α : A −B →{0, . . . , n} for some n ∈N. We thus define γ :A ∪B →Nγ(x) = ( α(x) when x ∈A −B n + 1 + β(x) when x ∈B Left as an exercise - prove that γ is a bijection.
Case 1 (A −B is countable). Suppose that A −B is countable. Then there is a bijection α : A −B →N. So we define γ :A ∪B →Zγ(x) = ( α(x) when x ∈A −B −β(x) −1 when x ∈B Left as an exercise - prove that γ is a bijection.
Since both N and Z are countable, so is A ∪B.
170 CHAPTER 7. PROPERTIES OF FUNCTIONS Chapter 8 Properties of Relations 8.1 Relations on Sets Definition: binary relation A (binary) relation R from a set A to a set B is a subset R ⊆A × B. We write aRb if and only if (a, b) ∈R. If R is a relation from a set A to the same set A, we simply say that R is a deffy on A.
One can think of a relation on a set as a generalization of a function f : A →A, but what we’ll see is that a relation (with certain properties) more appropriately provides a way of comparing elements of a set, and gives rise to generalizations of equality and greater-than/less-than on sets.
Example 8.1.1 Let A = {1, 2, 3, 4} and let R be the relation defined by aRb ⇐ ⇒a −b is even.
Write down R.
R = (1, 1), (1, 3), (2, 2), (2, 4), (3, 1), (3, 3), (4, 2), (4, 4) .
Example 8.1.2 Let R be the relation defined on Z: xRy ⇐ ⇒x −y is even.
Describe R.
Since the difference of two even numbers is again even, and the difference of two odd numbers is even, then R = {(x, y) ∈Z × Z : x and y have the same parity} .
Example 8.1.3 Let A = {1, 2, 3, 4} and let R be the relation defined by aRb ⇐ ⇒a|b.
Write down R.
171 172 CHAPTER 8. PROPERTIES OF RELATIONS R = (1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4) .
Example 8.1.4 Let R be the relation defined on R: xRy ⇐ ⇒x < y Describe R by plotting the pairs (x, y) in the plane.
−3 −2 −1 1 2 3 −3 −2 −1 1 2 3 y = x 8.1.1 Arrow Diagrams/Directed Graphs One can visually represent relations on sets by drawing a point for every element in the set, and an arrow from a to b if and only if aRb. This is a special type of arrow diagram that more commonly is referred to as a directed graph (or digraph for short).
Example 8.1.5 Let A and R be as in Example 8.1.1, that is, A = {1, 2, 3, 4} and aRb ⇐ ⇒a −b is even.
Draw the arrow diagram from this relation.
INCOMPLETE 1 2 3 4 8.1. RELATIONS ON SETS 173 Example 8.1.6 Let R be a relation on the set P({1, 2}) be given by XRY ⇐ ⇒X ⊆Y Draw the directed graph for this relation.
∅ {1} {2} {a, b} Instructor Note: still tidying up below.
Example 8.1.7 Let R be the relation defined on P({a, b}): XRY ⇐ ⇒#X ≥#Y, where #X represents the cardinality of X. Write down all related subsets of {a, b}.
• {a, b}R∅ • {a, b}R{a} • {a, b}R{b} • {a, b}R{a, b} • {a}R∅ • {a}R{a} • {a}R{b} • {b}R∅ • {b}R{a} • {b}R{b} • ∅R∅ 174 CHAPTER 8. PROPERTIES OF RELATIONS 8.1.2 Inverse Relations Definition: inverse relation Given a relation R from A to B, the inverse relation, denoted R−1 is a relation from B to A defined as follows: bR−1a ⇐ ⇒aRb.
Example 8.1.8 Let R be the relation from Example 8.1.4: xRy ⇐ ⇒x < y Find the inverse relation R−1.
R consisted of all ordered pairs (a, b) where a < b. So R−1 consists of all ordered pairs (x, y) with x > y. In other words, −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 y = x Example 8.1.9 Let R be the relation from Example 8.1.2: xRy ⇐ ⇒x −y is even.
Find the inverse relation R−1.
Notice that, for any integers x, y, x −y is even if and only if y −x is even. That means that, whenever xRy, then also yRx, hence xR−1y. It follows that R = R−1.
Example 8.1.10 Let R be the relation from Example 8.1.7: XRY ⇐ ⇒X ⊆Y Find the inverse relation R−1.
8.1. RELATIONS ON SETS 175 • ∅R{a, b} • {a}R{a, b} • {b}R{a, b} • {a, b}R{a, b} • ∅R{a} • {a}R{a} • ∅R{b} • {b}R{b} • ∅R∅ 176 CHAPTER 8. PROPERTIES OF RELATIONS 8.2 Reflexivity, Symmetry, and Transitivity Definition: (anti)reflexive, (anti)symmetric, transitive Let R be a relation on a set A. We say that R is...
• ...reflexive if it has the following property: for all a ∈A, aRa.
• ...anti-reflexive if it has the following property: for all a, b ∈A, if aRb, then a ̸= b.
• ...symmetric if it has the following property: for all a, b ∈A, if aRb then bRa.
• ...anti-symmetric if it has the following property: for all a, b ∈A, if aRb and bRa, then a = b.
• ... transitive if it has the following property: for all a, b, c ∈A, if aRb and bRc, then aRc.
Example 8.2.1 Let R be the relation from Example 8.1.4: For all x, y ∈R, xRy ⇐ ⇒x < y.
Which of the above properties does it have?
Anti-reflexive, Transitive Example 8.2.2 Let R be the relation from Example 8.1.2; For all x, y ∈Z, xRy ⇐ ⇒x −y is even.
Which of the above properties does it have?
Refleexive, Transitive, Symmetric 8.2. REFLEXIVITY, SYMMETRY, AND TRANSITIVITY 177 Example 8.2.3 Let R be the relation from Example 8.1.7: For all X, Y ∈P({1, 2}), XRY ⇐ ⇒X ⊆Y.
Which of the above properties does it have?
Reflexive, antisymmetric, Transitive Example 8.2.4 Let R be a relation on A = {1, 2, 3, 4, 5} be given by the directed graph below.
1 2 3 4 5 Which of the properties does the above relation have?
Symmetric Exercise 8.2.5 Let A = {1, 2, 3, 4, 5}. Draw directed graphs representing relations R on A which have the following properties: 1. R is not reflexive, not neither symmetric, and not transitive.
2. R is reflexive, but neither symmetric nor transitive.
3. R is transitive, but neither reflexive nor symmetric.
4. R is reflexive and symmetric, but not transitive.
5. R is reflexive and transitive, but not symmetric.
6. R is symmetric and transitive, but not reflexive.
7. R is reflexive, symmetric, and transitive.
178 CHAPTER 8. PROPERTIES OF RELATIONS 8.2.1 Proving and disproving properties of binary relations Example 8.2.6 Let n > 0 be an integer. Define a relation R on Z as follows: xRy ⇐ ⇒n|(x −y).
Prove that R is reflexive, symmetric, and transitive.
Proof. Let n > 0 and let x, y, z ∈Z be arbitrary.
Reflexive. Since x −x = 0 and n|0, then xRx for every x ∈Z.
Symmetric. Suppose xRy. Then n|(x −y), that is, there is some k ∈Z for which x −y = nk. It follows that y −x = −(x −y) = −nk = n(−k) and thus n|(y −x) as well. Therefore yRx.
Transitive. Suppose that xRy and that zRy. Then n|(x −y) and n|(y −z), and thus there are integers k, ℓfor which x −y = nk and y −z = nℓ. We then have that x −z = x −y + y −z = nk + nℓ= n(k + ℓ) whence n|(x −z).
Example 8.2.7 Define a relation R on Z × Z as follows: (a1, b1)R(a2, b2) ⇐ ⇒ ( a1 < a2, or a1 = a2 and b1 ≤b2 Prove that R is reflexive, antisymmetric, and transitive.
Proof. Let (a1, b1), (a2, b2), and (a3, b3) be arbitrary ordered pairs in Z × Z.
Reflexive. Observe that a1 = a2 and b1 ≤b1, hence (a1, b1)R(a1, b1).
Symmetric. Suppose (a1, b1)R(a2, b2) and (a2, b2)R(a1, b1).
This implies that both a1 ≤a2 and a2 ≤a2, whence a1 = a2. Since a1 = a2, then this implies both that b1 ≤b2 and b2 ≤b1, and thus b1 = b2. Therefore (a1, b1) = (a2, b2).
Transitive. Suppose (a1, b1)R(a2, b2) and (a2, b2)R(a3, b3). We examine the following cases.
• Case 1 (a1 < a2) – Subcase 1 (a2 < a3) Left as an exercise.
8.2. REFLEXIVITY, SYMMETRY, AND TRANSITIVITY 179 – Subcase 2 (a2 = a3) Left as an exercise.
• Case 2 (a1 = a2) – Subcase 1 (a2 < a3) Left as an exercise.
– Subcase 2 (a2 = a3) Left as an exercise.
180 CHAPTER 8. PROPERTIES OF RELATIONS 8.3 Equivalence Relations Definition: equivalence relation A relation R on a set A is called an equivalence relation if and only if it is • reflexive, • symmetric, and • transitive.
Example 8.3.1: Revisiting Example 8.2.6 The relation R on Z given in Example 8.2.6 is an equivalence relation.
For each fixed positive integer n, xRy ⇐ ⇒n|(x −y).
Example 8.3.2: “Mod 1” (aka, S1) Let R be the following relation on R.
xRy ⇐ ⇒ ∃k ∈Z such that x = y + k.
Show that R is an equivalence relation.
Proof. Let x, y, z ∈R.
Reflexive. Since x = x + 0, then xRx.
Symmetric. Suppose xRy. Then there is an integer k ∈Z so that x = y + k. It follows that y = x + (−k), and since −k ∈Z, then yRx.
Transitive. Suppose xRy and yRz. Then there are integers k, ℓfor which x = y + k and y = z + ℓ. Hence x = y + k = (z + ℓ) + k = z + (ℓ+ k) and since ℓ+ k is an integer, xRz, as desired.
Therefore R is an equivalence relation.
Exercise 8.3.3 Let R be the relation on R × R given by (x1, y1)R(x2, y2) ⇐ ⇒(x1, y1) = (x2 + k, y2 + ℓ) for some k, ℓ∈Z.
Prove that R is an equivalence relation.
8.3. EQUIVALENCE RELATIONS 181 Example 8.3.4 Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and let A = {A1, A2, A3} be the following partition of A: A1 = {1, 3, 5, 7, 9} A2 = {2, 4, 6} A3 = {8, 10} Let R be the relation on A given by xRy ⇐ ⇒∃i such that x, y ∈Ai.
Prove that R is an equivalence relation.
Theorem 8.3.5: Equivalence relation induced by a partition.
Let A be some set and let A = {A1, A2, . . . , An, . . .} be a (possibly-infinite) partition of A.
Define a relation R on A by xRy ⇐ ⇒∃i s.t. x ∈Ai and y ∈Ai.
Then R is an equivalence relation.
Proof. Let x, y, z ∈A be arbitrary and suppose R is the relation described above.
Reflexive. Let Ai ∈A be the set containing x. Then x ∈Ai as well, so xRx.
Symmetric. Suppose xRy. Then there is some set Ai in the partition for which x ∈Ai and y ∈Ai. Since logical conjunction is commutative, this implies y ∈Ai and x ∈Ai. Thus yRx.
Transitivity. Suppose xRy and yRz. Then there is some set Ai in the partition for which x, y ∈Ai, and some set Aj in the partition for which y, z ∈Aj. Since sets in a partition are pairwise disjoint, then y ∈Ai ∩Aj = ⇒Ai = Aj = ⇒i = j. Thus we have that x, z ∈Ai and so xRy.
Therefore R is an equivalence relation.
Definition Given a set A and a partition {A1, . . . , An, . . .}, the relation R on A given by xRy ⇐ ⇒x, y ∈Ai is called the equivalence relation on A induced by a partition.
Example 8.3.6 Let A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and let R be the equivalence relation on A given by xRy ⇐ ⇒4|(x −y) 182 CHAPTER 8. PROPERTIES OF RELATIONS Draw the directed graph representing for this relation.
0 4 8 1 5 9 2 6 3 7 Notice that this graph is naturally clustered into four connected components. Each of these components are related by the fact that they have the same remainder after dividing by 4. This is true more generally with the relation from Example 8.2.6. In general, we can expect n different clusters x = nk and y = nℓ x = nk + 1 and y = nℓ+ 1 x = nk + 2 and y = nℓ+ 2 .
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x = nk + (n −1) and y = nℓ+ (n −1) By collecting integers with the same remainder (after dividing by n), we form a partition of Z.
Letting Aj be the set Aj = {z ∈Z : z = nk + j for some k ∈Z} , we have Z = n−1 [ j=0 Aj.
Clearly these sets must be pairwise disjoint (because numbers cannot have two different remainders), so xRy precisely when x, y are in the same set of this partition.
We know that partitions induce equivalence relations, but it also seems that the converse is true.
On our way to showing this, we introduce the following term: Definition: equivalence class Let A be a set and R an equivalence relation on A. For each a ∈A, the (equivalence) class of a, denoted [a]R (or just [a] when the relation is clear), is the set [a]R = {x ∈A : xRa}.
8.3. EQUIVALENCE RELATIONS 183 The set of all distinct equivalence classes is denoted A/R.
Remark. In the case of integers “mod n,” these may also be called “congruence classes” since two numbers are said to be “congruent modulo n.” Example 8.3.7: Revisiting Example 8.3.6 Let R be the equivalence relation on Z given by xRy ⇐ ⇒4|(x −y) Find all equivalence classes, Z/R.
Looking at the diagram in Example 8.3.6, we see that equivalence classes are determined by the remainder after dividing by 4. So = {x ∈Z : x = 4k + 0 for some k ∈Z} = {. . . , −8, −4, 0, 4, 8, . . .} = {x ∈Z : x = 4k + 1 for some k ∈Z} = {. . . , −7, −3, 1, 5, 9, . . .} = {x ∈Z : x = 4k + 2 for some k ∈Z} = {. . . , −6, −2, 2, 6, 10, . . .} = {x ∈Z : x = 4k + 3 for some k ∈Z} = {. . . , −5, −1, 3, 7, 11, . . .} Example 8.3.8: Revisiting Example 8.3.2 Let R be the equivalence relation on R given by xRy ⇐ ⇒∃k ∈Z such that y = x + k.
Describe all equivalence classes.
INCOMPLETE Exercise 8.3.9 Let R be the equivalence relation on R × R given by (x1, y1)R(x2, y2) ⇐ ⇒(x1, y1) = (x2 + k, y2 + ℓ) for some k, ℓ∈Z.
Describe all equivalence classes.
Lemma 8.3.10 Let R be an equivalence relation on A and let [x], [y] be two equivalence classes. Then either [x] = [y] or [x] ∩[y] = ∅.
Proof. Let [x], [y] be arbitrary equivalence classes. If [x] ∩[y] = ∅, then we’re done, so suppose that z ∈[x] ∩[y]. Then we have that xRz and zRy, and thus, by transitivity, xRy. In turn, this implies that every element of [y] is related to every element of [x], and vice versa. Therefore [x] = [y].
184 CHAPTER 8. PROPERTIES OF RELATIONS Theorem 8.3.11 Let A be some set and let R be an equivalence relation on A. The equivalence classes A/R form a partition of A.
Proof. Lemma 8.3.10 shows that the equivalence classes are disjoint, so all that’s left to prove is that their union is equal to A. Since the equivalence classes are subsets of A, so is their union, so in fact we just need to show that A ⊆S [x]∈A/R[x]. To see this, let a ∈A. By the reflexive property of an equivalence relation, aRa, which means that a ∈[a] ∈A/R.
.1. PROOFS SKIPPED IN CLASS 185 .1 Proofs Skipped In Class Definition: disjoint union Let A1, A2 be sets. The disjoint union of A1 and A2 is the set A1 ⊔A2 := (a, i) ∈ A1 ∪A2) × N : a ∈Ai .
Lemma: Lemma 7.4.2 Prove that, for all finite sets A1 and A2, |A1 ∪A2| ≤|A1| + |A2|.
Proof. The function φ : A1 ⊔A2 →A1 ∪A2 φ(x, i) = x is a surjection, so we get that |A1 ∪A2| ≤|A1 ⊔A2|. As such, we only need to prove that |A1 ⊔A2| = |A1| + |A2|.
By definition of finiteness, there are natural numbers m and n and bijections α, β for which α : A1 →{1, . . . , m} and β : A2 →{1, . . . , n} Let sm be the “shift-by-m” bijection from ??. It follows that γ :A1 ⊔A2 →{1, . . . , m + n} γ(x, i) = ( α(x) if i = 1 sm(β(x)) if i = 2 INCOMPLETE 186 CHAPTER 8. PROPERTIES OF RELATIONS Index absolute value, 81 argument, 21 conclusion, 21 premises, 21 arrow diagram, 144 biconditional, 19 binary relation, 171 boolean algebra, 136 composite number, 59 compound statement, 10 conditional, 16 conclusion, 16 hypothesis, 16 conjunction, 8 constructive proof, 61 Continuum Hypothesis, 164 contradiction, 11 contrapositive, 18, 41 converse, 17, 41 countable set, 157 counterexample, 34, 62 disjunction, 9 divides, 76 divisibility, 76 empty set, 113 English Phrases “but”, 8 “either... or”, 9, 10 “if”, 19 “if... then...”, 16 “implies”, 19 “necessary”, 19 “neither... nor”, 8 “only if”, 19 “sufficient”, 19 equivalence class, 182 equivalence relation, 180 induced by partition, 181 even integer, 58 exclusive or, 9, 10 existential statement, 35 factorial, 94 fallacy, 29 Ambiguous Premises, 29 Circular Reasoning, 29 Converse Error, 30 Inverse Error, 30 Jumping to the Conclusion, 29 finite set, 157 function, 141 bijection, 148 bijective, 148 equality, 146 injective, 148 inverse, 150 one-to-one, 148 onto, 148 preimage, 145 range, 145 surjective, 148 functions composition, 153 implicit quantification, 37 induction, 96 inverse, 17, 41 logical equivalence, 10 Absorpotion Laws, 12 Associative Laws, 12 Commutative Laws, 12 DeMorgan’s Laws, 12 Distributive Laws, 12 Double Negative Laws, 12 Idempotent Laws, 12 Identity Laws, 12 Negation Laws, 12 Negation of Contradiction, 12 Negation of Tautology, 12 Universal Bound laws, 12 logical proof, 26 Method of Exhaustion, 35 method of exhaustion, 63 negation, 9 187 188 INDEX nonconstructive proof, 61 odd integer, 58 power set, 120 predicate, 33 domain, 33 prime number, 59 proper subset, 114 proposition, 7 quantifiers nested, 43 quotient, 110 recurrence relation, 111 recursive sequence, 111 relation, 171 antireflexive, 176 antisymmetric, 176 equivalence relation, 180 equivalence relation induced by partition, 181 inverse, 174 on a set, 171 reflexive, 176 symmetric, 176 transitive, 176 remainder, 110 rule of inference, 23 rules of inference modus ponens, 23 modus tollens, 23 addition, 24 conjunction, 24 contradiction, 23 disjunctive syllogism, 24 division of cases, 23 elimination, 24 existential generalization, 51 existential instantiaion, 51 hypothetical syllogism, 24 resolution, 24 simplification, 24 specialization, 24 transitivity, 24 universal generalization, 51 universal instantiation, 51 sequence, 91 explicit formula, 91 general formula, 91 index of term, 91 length, 91 product of, 93 term, 91 set, 113 complement, 118 difference, 117 disjoint, 119 element of, 113 equality of, 116 intersection, 116 partition, 119 symmetric difference, 117 union, 117 set equality Absorpotion Laws, 123 Associative Laws, 123 Commutative Laws, 123 DeMorgan’s Laws, 123 Distributive Laws, 123 Double Negative Laws, 123 Idempotent Laws, 123 Identity Laws, 123 Negation Laws, 123 Negation of Contradiction, 123 Negation of Tautology, 123 Universal Bound laws, 123 set-builder notation, 114 sets cardinality of, 156 disjoint union, 185 sound argument, 31 statement, 7 strong induction, 105 subset, 114 sum of a sequence, 92 superset, 114 symbol ≡, 10 symbols ∅, 113 ∃, 35 ∀, 34 ∈, 33 INDEX 189 ∧, 7 ∨, 7 N, 34 Q, 34 R, 34 Z, 34 Z+, 34 ¬, 7 Q, 93 ∼, 7 P, 92 Table of Logical Equivalences, 11 Table of Set Equalities, 123 tautology, 11 theorem, 57 corollary, 57 lemma, 57 proposition, 57 truth set, 33 uncountable set, 157 universal conditional, 36 universal statement, 34 vacuously true, 16 valid argument, 21 weak induction, 96 |
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Analysis of a Waveguide Partially Filled with a Dielectric Material
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A rectangular waveguide partially filled with an anisotropic dielectric material has been analysed. The method of analysis has shown to be quite powerful in particular fo...Show More
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A rectangular waveguide partially filled with an anisotropic dielectric material has been analysed. The method of analysis has shown to be quite powerful in particular for high dielectric constant materials. A new configuration with a rectangular waveguide unsymmetrically loaded with a rectangular dielectric insert has been analysed and tested experimentally. As compared e.g. with an image line, the insertion loss of the unsymmetrically loaded waveguide was found to be much lower and the manufacturing problems less troublesome. The unsymmetrically loaded structure has recently been successfully used in a travelling wave maser designed for 30 GHz.
Published in:1980 10th European Microwave Conference
Date of Conference: 08-12 September 1980
Date Added to IEEE Xplore: 19 March 2007
DOI:10.1109/EUMA.1980.332903
Publisher: IEEE
Conference Location: Warszawa, Poland
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Electromagnetic induction
Figure 1. One of Michael Faraday's first devices for demonstrating induction.
Electromagnetic induction is the production of an electromotive force (EMF) being created as a result of relative motion between a magnetic field and a conductor. It was discovered in 1831 by Michael Faraday, and lays the foundation for electrical generation in power plants, electric motors, and AC circuitry which powers the electrical grid, transformers, and many more phenomena.
The equation that mathematically describes electromagnetic induction is Faraday's Law, which states that any change in the magnetic environment of a coiled wire will cause a voltage (EMF) to be induced. Faraday found many ways for this to happen such as changing the magnetic field strength, moving a magnet through a coil of wire, and moving the coil through a magnetic field, just to name a few. The voltage (EMF) generated in a coil of wire can be described by the following equation:
E M F=−N Δ(B A)Δ t
where
N is the number of turns in the wire
Δ(B A) is the change in magnetic flux
Δ t is the change in time
The ways that Faraday found to change this flux as stated above can all be represented in this equation. The reason this equation is negative is because of Lenz's law, which requires any change in magnetic flux to be reproduced in equal strength but opposite direction by the wire.
Faraday's law is important for many electromagnetic applications in the world, including cars. The ignition system in a car's internal combustion engine takes only 12 volts from the battery and ramps that up to 40000 volts! Visit Hyperphysics to learn how.
PhET Simulation of Induction
PhET has graciously allowed us to use their simulations, and the one below demonstrates Faraday's law of electromagnetic induction. The voltage is seen to change as the magnetic flux changes through it.
For Further Reading
For further information please see the related pages below:
Electric current
Electricity
Electromotive force
Battery
Fuel cell
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References
↑Wikimedia Commons [Online], Available:
↑UIUC Physics, A Brief History of The Development of Classical Electrodynamics [Online PDF], Available:
↑ 3.03.1Hyperphysics, Faraday's Law [Online], Available:
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190296 | https://www.northshore.edu/cas/testing/files/fractions-comp-packet.pdf | Fraction Competency Packet Developed by: Nancy Tufo Revised 2004: Sharyn Sweeney Student Support Center North Shore Community College 2 To use this booklet, review the glossary, study the examples, then work through the exercises. The answers are at the end of the booklet. When you find an unfamiliar word, check the glossary for a definition or explanation. Calculators are not allowed when taking the Computerized Placement Test (CPT), nor in Fundamentals of Mathematics, Pre-Algebra, and Elementary Algebra; therefore, do not rely on a calculator when working the problems in this booklet. If you have difficulty understanding any of the concepts, come to one of the Tutoring Centers located on the Lynn, Danvers Main and Danvers Hathorne Campuses. Hours are available at (978) 762-4000 x 5410. Additional Tutoring Center information can be found on the NSCC website at www.northshore.edu/services/tutoring. The Centers are closed when school is not in session, and Summer hours are limited. 3 Table of Contents Glossary 4 General Fraction Information 5 Mixed Numbers 6 Equivalent Fractions with Larger Denominators 7 Equivalent Fractions with Smaller Denominators 8 Improper Fractions 9 Least Common Multiple (Least Common Denominator) 10 Addition and Subtraction of Fractions with Same Denominator 12 Addition and Subtraction of Fractions with Different Denominators 13 Subtraction with Borrowing 14 Multiplication of Fractions 16 Division of Fractions 17 Some Fraction Word Problems 18 Answers to Exercises 20 4 Glossary Boosting: Rewriting a fraction as an equivalent fraction with a higher denominator. Denominator: Bottom number of a fraction indicating how many parts make a whole. Difference: The result when two numbers are subtracted. Divisor: The number after the division sign in a division problem, (i.e. 12÷7); or the bottom number of a fraction, (i.e. 7 12 ); the number "outside" the division house (i.e. 7 12 ). Equivalent Fraction: Fractions that are found by multiplying the numerators and denominators by the same number. Factor: Numbers equal to or less than a given number that divides the number evenly. For example, the factors of 12 are 1, 2, 3, 4, 6, 12. Fraction: Any number written in the form of one whole number over another, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 5 3 , indicating number of parts being considered over the number of parts that make one whole. Fraction Bar: The line separating the numerator and denominator in a fraction, and it indicates division. Greatest Common Factor (GCF): The largest matching factor of two or more given numbers. It is used to reduce fractions. Improper Fraction: Any fraction with the numerator larger than the denominator. Least Common Denominator (LCD): The smallest matching multiple of two or more given numbers. It is used to "boost" fractions. (Also called Least Common Multiple, LCM) Mixed Number: A whole number and a fraction. (It implies addition of wholes and parts; that is, 35 7 is read "three and five sevenths".) Multiple: (Similar to the "times table.") A multiple of a given number is equal to the given number or greater. Multiples are found by multiplying the given number in turn by 1, 2, 3,... For example, multiples of 4 are 4, 8, 12, 16, … Numerator: The top number of a fraction. It indicates how many parts of a certain size are represented. Prime Factor: Factors of a number that are only divisible by 1 and the given number. For example, prime factors of 12 are 1 x 2 x 2 x 3. Some frequently used Prime Numbers are 2, 3, 5, 7, 11, 13. Product: The result when two numbers are multiplied. 5 Proper Fraction: Any fraction when the numerator is less than the denominator. Quotient: The solution to a division problem. Reducing: Dividing the numerator and the denominator by the same number to get an equivalent fraction. Final answers of most fraction problems should be expressed reduced to “simplest terms”; in other words, the numerator and denominator have no more common factors. Remainder: The number left after a whole number division problem is complete. When converting an improper fraction to a mixed number, the remainder is the numerator of the fraction. Sum: the result when two numbers are added. Whole Number: The Numbers system including 0, 1, 2, 3,…. General Fraction Information ¾ The fraction that represents the above picture is 7 5 and is read “five sevenths”. That means that five of the parts are shaded, and it would take seven parts of that size to make a whole. ¾ One whole can be "cut up" into equal size parts; therefore, 1 = 123 123 9 9 13 13 = = , etc. ¾ A whole number can be written as a fraction with a denominator of 1; for example, 2 =2 1 . Zero can be written as a fraction using zero as the numerator and any whole number as the denominator, for example, 0 23 . ¾ Any whole number may be written as a mixed number by using a zero fraction. For example, 42 0 3 3 = . 6 Mixed Numbers To convert a mixed number, 7 2 5 , to an improper fraction, 7 37 : 7 37 7 2 5 = 7 2 5 Work in a clockwise direction, beginning with the denominator, (7). 5 x 7 = 35 Multiply the denominator (7) by the whole number, (5) 35 +2 = 37 Add that product, (35), to the numerator (2) of the fraction. ( ) 7 37 7 2 7 5 = + × The denominator remains the same for the mixed number and the improper fraction. Convert to Improper Fractions: 1) = 5 2 4 6) = 4 3 14 11) 9= Hint: See #10 2) = 8 3 5 7) = 5 3 6 12) = 4 3 7 3) = 9 4 2 8) = 10 1 9 13) = 9 5 12 4) = 7 6 5 9) = 2 1 16 14) = 8 3 10 5) = 8 1 8 10) = 1 0 8 15) = 3 2 28 Denominator Numerator Whole Number 7 Finding Equivalent Fractions with Larger Denominators This process is sometimes called “Boosting” 56 ?
8 5 : = Example 7 8 56 = ÷ Divide the larger denominator by the smaller to find the factor used to multiply the denominator. (Note: The product of the smaller denominator and the factor is the larger denominator) 7 8 7 5 7 7 8 5 × × = × Use this factor to multiply the numerator. 56 35 8 5 = The result is two equivalent fractions. Note: Equal denominators are required for addition and subtraction of fractions. Find the equivalent fractions as indicated: 1) 2 5 = 15 2) 3 8 = 32 3) 4 9 = 54 4) 6 7 = 49 5) 1 8 = 48 6) 3 4 = 44 7) 3 5 = 45 8) 1 10 = 60 9) 1 2 = 28 10) 10 100 = 700 11) 8 9 = 81 12) 3 4 = 68 13) 5 9 = 108 14) 3 8 = 112 15) 2 3 = 462 8 Equivalent Fractions with Smaller Denominators Reducing Fractions Example: Reduce the following fraction to lowest terms 105 90 There are three common methods, DO NOT mix steps of the methods! Method 1: 7 6 15 105 15 90 = ÷ ÷ The Greatest Common Factor for 90 and 105 is 15. Divide the numerator and the denominator by the GCF, 15. Method 2: 21 18 5 105 5 90 = ÷ ÷ 7 6 3 21 3 18 = ÷ ÷ Examine the numerator and denominator for any common factors, divide both numerator and denominator by that common factor. Repeat as needed. ¾ Both 90 and 105 are divisible by 5. ¾ Both 18 and 21 are divisible by 3. Method 3: 5 3 7 5 3 3 2 105 90 × × × × × = Express the numerator and denominator as a product of prime factors. ( ) ( ) 5 3 7 5 3 3 2 105 90 × × × × × = Divide numerator and denominator by common factors, (3x5) 7 6 7 3 2 = × = Multiply remaining factors. Reduce these fractions. 1) = 50 28 2) 8 24 = 3) 30 54 = 4) 18 42 = 5) 32 48 = 6) 36 54 = 7) 14 56 = 8) 18 28 = 9) 36 216 = 10) 35 42 = 11) 12 54 99 = 12) 15 280 320 = 9 Improper Fractions Example: Convert 3 14 to an Improper Fraction 4 3 14 = ÷ Remainder 2 Remember: Dividend ÷Divisor = Quotient Divide the numerator (14) by the denominator (3). 3 2 4 3 14 = Write the mixed number in the form: divisor remainder Quotient Note: Check you answer to see if you can reduce the fraction. Convert these improper fractions to mixed numbers. Be sure to reduce when it’s possible. 1) 8 5 = 2) 18 7 = 3) 37 9 = 4) 127 5 = 5) 32 9 = 6) 114 5 = 7) 128 3 = 8) 401 3 = 9) 36 6 = 10) 235 2 = 11) 15 280 6 = 12) 8 315 3 = 13) 54 8 = 14) 26 8 = 15) 258 9 = #11, 12 Hint: how many wholes will there be? 10 Least Common Multiple (LCM) Used to find the Least Common Denominator (LCD) Example: Find the LCM of 30 and 45 Note: There are four common methods; DO NOT mix the steps of the methods! Method 1 30, 60, 90, 120, … 45, 90, 135, … Remember that multiples are equal to or larger than the given number. List the multiples of each of the given numbers, in ascending order. LCM = 90 The LCM is the first multiple common to both lists. Method 2 45, 90, 135, … 30 45 ÷ remainder 30 90 ÷ no remainder LCM = 90 List the multiples of the larger number. Divide each in turn by the smaller. The LCM is the multiple that the smaller number divides without leaving a remainder. Method 3 6 5 30 = ÷ ; 9 5 45 = ÷ 2 3 6 = ÷ ; 3 3 9 = ÷ Divide both numbers by any common factor, (5 then 3). Continue until there are no more common factors. Note: 2 and 3, the results of the last division have no common factors. LCM = 3 2 3 5 × × × = 90 The LCM equals the product of the factors, (5 and 3) and the remaining quotients, (2 and 3). Method 4 30 45 5 x 6 5 x 9 5 x 2 x 3 5 x 3 x 3 3 2 5 30 × × = 3 3 5 45 × × = Or 2 3 5 45 × = LCM = 5 3 2 2 × × = 90 Find the prime factors of each the given numbers. Write each number as a product of primes using exponents, if required. LCM equals the product of all the factors to the highest power. 11 In each exercise, find the LCM of the given numbers. 1) 4 and 18 2) 16 and 40 3) 20 and 28 4) 5 and 8 5) 12 and 18 6) 12 and 16 7) 50 and 75 8) 24 and 30 9) 36 and 45 10) 8 and 20 11) 16 and 20 12) 28, 35, and 21 12 Addition and Subtraction of Fractions with the Same Denominator To add or subtract fractions, the denominators MUST be the same. Example 1: ?
5 1 5 3 = − 5 1 3 5 1 5 3 − = − 5 2 = Because both fractions have the same denominator, you may subtract the numerators and keep the denominator. Example 2: ?
9 7 9 5 = + 9 7 5 9 7 9 5 + = + 9 12 = 9 3 1 = 3 1 1 = Because both fractions have the same denominator, you may add the numerators and keep the denominator. Always change improper fractions to a mixed number. Reduce, when possible. Add or Subtract as indicated. 1. 8 3 8 4 + 2. 10 1 10 7 − 3. 48 4 48 9 48 7 + + 4. 37 3 37 40 − 5. 13 4 13 10 + 6. 17 17 17 11 17 9 + + 7. 3 6 3 4 3 2 − + 8. 6 1 6 5 6 7 + − 9. 13 9 13 7 + 13 Addition and Subtraction of Fractions with Different Denominators Remember: In order to add or subtract fractions, the denominators MUST be the same. Example: ?
8 3 3 2 = + LCM = 24 Find the LCM 24 16 8 8 3 2 = × + 24 9 3 3 8 3 = × 24 25 Write the problem vertically. Find the equivalent fractions with the LCM as a denominator. Add the fractions with the same denominator. 24 1 1 24 25 = Remember to write as a mixed number and reduce when possible! Add or Subtract: 1) 7 8 + 3 4 2) 7 8 - 3 4 3) 11 12 + 17 18 4) 3 7 + 2 5 5) 15 24 - 10 27 6) 7 12 + 5 16 7) 16 27 - 5 24 8) 11 4 + 3 8 9) 11 4 + 23 18 10) 29 8 + 9 7 11) 213 35 - 1 5 14 12) 2 3 + 1 21 - 2 7 14 Subtraction of Fractions with Borrowing Example 1: Example 2: ?
3 1 1 7 = − ?
6 5 2 3 1 5 = − Note: There are two common methods; DO NOT mix the steps of the methods! Method 1 Example 1 7 = 3 3 6 - 3 1 1 = 3 1 1 3 2 5 Subtraction with Borrowing Write problem vertically Cannot subtract fraction from whole without finding common denominator. Borrow one whole from 7 and express as .
LCD LCD ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= 3 3 1 Subtract numerators and whole numbers. Example 2 6 2 5 3 1 5 = 6 8 4 = - 6 5 2 6 5 2 = = 6 5 2 2 1 2 6 3 2 = Write problem vertically and find LCD Cannot subtract 5 from 2. Borrow one whole from 5, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 6 6 4 and add ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = 6 2 6 4 6 2 5 . Subtract numerators and whole numbers; reduce as needed. Method 2 Example 1: 7 = 3 21 - 3 1 1 = 3 4 3 2 5 3 17 = Subtraction Using Improper Fractions Write the problem vertically. Convert the whole numbers and mixed numbers to improper fractions using the LCD. Subtract ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 3 4 21 and convert improper fraction to mixed number. Example 2: 6 2 5 3 1 5 = 6 32 = - 6 5 2 6 5 2 = = 6 17 2 3 2 6 15 = 2 1 2 2 3 2 = Write problem vertically and find the LCD. Change the mixed numbers to improper fractions. Subtract the numerators. Convert to a mixed number. Reduce. 15 Subtract: 1) 5 - 21 3 2) 7 - 1 1 6 3) 10 - 4 5 6 4) 3 5 8 - 2 7 8 5) 1 1 8 - 3 4 6) 3 5 12 - 115 16 7) 8 - 6 4 5 8) 4 3 8 - 3 5 6 9) 17 - 4 5 9 10) 5 5 18 - 1 3 4 11) 5 2 7 - 3 3 8 12) 18 - 1 7 16 - 7 12 16 Multiplication of Fractions Example: 6 5 3 10 3 × Note: LCD is not needed to multiply fractions. 6 5 ) 3 6 ( 6 5 3 + × = Change mixed numbers to improper fractions 2 10 23 1 6 23 10 3 × × = × Before multiplying, reduce by dividing any numerator with any denominator with a common factor. (3 and 6 have a common factor of 3) 20 23 2 10 23 1 = × × Multiply numerators and denominators 20 3 1 20 23 = Convert improper fractions to mixed numbers. Multiply: 1) 3 2 2 1 4 × 2) 4 1 1 5 1 3 × 3) 9 1 1 6 × 4) 2 1 1 6 1 2 × 5) 15 7 1 11 10 × 6) 15 5 3 4 × 7) 9 2 2 8 3 3 × 8) 17 3 2 34 × 9) 5 4 8 7 9 × 10) 4 1 1 10 9 7 × 11) 15 4 7 3 1 18 × × 12) 8 3 6 5 1 5 1 3 × × 17 Division of Fractions Example: 8 3 2 4 3 2 ÷ OR 8 3 2 4 3 2 Note: One fraction divided by another may be expressed in either way shown above. Also, LCD is not needed to divide fractions. 4 11 4 3 2 = and 8 19 8 3 2 = Convert mixed numbers to improper fractions 19 8 4 11 8 19 4 11 × = ÷ Invert the divisor ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 8 19 . (Turn the fraction after the division sign upside down) 19 1 2 11 19 4 8 11 × × = × × Reduce if possible. (4 and 8 have a common factor) 19 22 19 1 2 11 = × × Multiply numerators and denominators 19 3 1 19 22 = Convert to a mixed number and reduce if needed. Divide these fractions. Reduce to lowest terms! 1) 5 6 ÷ 1 2 2) = ÷ 7 3 4 3 3) 3 ÷ 1 2 5 = 4) 3 1 2 1 = 5) 1 2 ÷ 6 = 6) 2 1 4 ÷ 3 = 7) 3 1 7 ÷ 2 5 14 = 8) 2 5 8 1 7 8 9) 4 1 2 ÷ 1 3 4 = 18 Some Fraction Word Problems Example 1: One day Ashley biked 4 3 of a mile before lunch and 8 7 of a mile after lunch. How far did she cycle that day? Note: this problem is asking you to add the distances traveled. 8 7 4 3 + 8 7 8 6 + 8 5 1 8 13 = To add fractions, find a LCD (8). Add the numerators; keep the denominators. Convert improper fraction to a mixed number; reduce if needed. Ashley cycled 8 5 1 miles that day. Example 2: A tailor needs 4 1 3 yards of fabric to make a jacket. How many jackets can he make with 2 1 19 yards of fabric? Note: this problem is asking you to divide. 4 1 3 2 1 19 ÷ 4 13 2 39 ÷ 1 1 2 3 13 4 2 39 × × = × 3 1 3 = To divide fractions, convert mixed numbers to improper fractions. Invert the divisor and reduce if possible, (39 and 13 have a common factor, as do 2 and 4). Multiply numerators and denominators. The tailor can make 3 jackets from 2 1 19 yards of fabric. 19 Solve the following problems. 1. An empty box weighs 4 1 2 pounds. It is then filled with 3 2 16 pounds of fruit. What is the weight of the box when it is full? 2. Yanni is making formula for the baby. Each bottle contains 5 2 6 scoops of formula. The formula container holds 320 scoops of formula. How many bottles of formula can Yanni make? 3. Miguel bought 4 1 2 pounds of hamburger, 5 1 1 pounds of sliced turkey, and 2 pounds of cheese. What was the total weight of all of his purchases? 4. Sheila had 8 yards of fabric. She used 4 1 2 yards to make a dress. How much fabric does she have left? 5. A father leaves his money to his four children. The first received 3 1 , the second received 6 1 , and the third received 5 2 . How much did the remaining child receive? (Hint: You can think of father’s money as one whole.) 6. Find the total perimeter (sum of the sides) of an equilateral triangle, (triangle with equal sides), if each side measures 4 1 2 inches. 20 Answers to Fractions Competency Packet p. 6 p. 7 p. 8 p. 9 p. 11 1) 22 5 1) 6 1) 14 25 1) 1 3 5 1) 36 2) 43 8 2) 12 2) 3 1 2) 2 4 7 2) 80 3) 9 22 3) 24 3) 5 9 3) 4 1 9 3) 140 4) 41 7 4) 42 4) 3 7 4) 25 2 5 4) 40 5) 65 8 5) 6 5) 2 3 5) 3 5 9 5) 36 6) 4 59 6) 33 6) 2 3 6) 22 4 5 6) 48 7) 33 5 7) 27 7) 1 4 7) 42 2 3 7) 150 8) 91 10 8) 6 8) 14 9 8) 133 2 3 8) 120 9) 33 2 9) 14 9) 1 6 9) 6 9) 180 10) 1 8 10) 70 10) 5 6 10) 117 1 2 10) 40 11) 1 9 11) 72 11) 12 6 11 11) 61 2 3 11) 80 12) 4 31 12) 51 12) 15 7 8 12) 113 12) 420 13) 113 9 13) 60 13) 6 3 4 14) 83 8 14) 42 14) 3 1 4 15) 86 3 15) 308 15) 28 2 3 21 p. 12 p. 13 p. 15 p. 16 p. 17 1) 8 7 1) 1 5 8 1) 2 2 3 1) 3 1) 1 2 3 2) 5 3 2) 1 8 2) 5 5 6 2) 4 2) 4 3 1 3) 12 5 3) 1 31 36 3) 5 1 6 3) 6 2 3 3) 7 1 2 4) 1 4) 29 35 4) 3 4 4) 3 1 4 4) 2 1 1 5) 13 1 1 5) 55 216 5) 3 8 5) 1 1 3 5) 12 1 6) 17 3 2 6) 43 48 6) 1 23 48 6) 69 6) 3 4 7) 3 0 7) 83 216 7) 1 1 5 7) 2 1 7 7) 3 1 1 8) 2 1 8) 1 5 8 8) 13 24 8) 74 8) 5 2 1 9) 13 3 1 9) 36 1 4 9) 12 4 9 9) 10 9 7 9) 7 4 2 10) 4 51 56 10) 3 19 36 10) 8 7 9 11) 1 1 70 11) 1 51 56 11) 7 6 6 12) 3 7 12) 15 48 47 12) 5 1 2 P. 19 1) 12 11 18 pounds 3) 20 9 5 pounds 5) 10 1 of the money 2) 50 bottles 4) 4 3 5 yards 6) 4 3 6 inches |
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Eczema herpeticum
HIV associated eczema herpeticum
Keywords:African ethnicity, HIV infection, Africa, Blister exudate, Eczema herpeticum, Face, Human immunodeficiency virus, Kaposi varicelliform eruption, Lips, Male, Pustule, Skin of colour, Umbilicated papule, Vesicle, Viral infection
What is eczema herpeticum?
Eczema herpeticum is a disseminated viral infection characterised by fever and clusters of itchy blisters or punched-out erosions. It is most often seen as a complication of atopic dermatitis/eczema.
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190298 | https://byjus.com/jee/standing-wave-on-a-string/ | The standing waves are formed by the superposition of two harmonic waves of equal amplitude and frequency travelling through the medium in the opposite direction. The standing waves are also known as stationary waves. These waves are localised and not progressive, hence the name stationary waves.
Standing Waves
Standing waves are characterised by nodes and antinodes. The amplitude of vibration of the particle is maximum at the antinodes and minimum at the nodes. The vibration within a string can produce a variety of patterns. Each pattern corresponds to the vibration that takes place at a particular frequency, and it is called harmonic. The lowest frequency at which the string vibrates to form a standing wave is called the fundamental frequency or the first harmonic. The wavelength and the speed of the wave determine the frequency of each harmonic.
Standing Waves on an Infinite Length String
Let us consider two harmonic waves of the same amplitude and period T and wavelength λ travelling at the same speed in the opposite direction.
y1 = A sin (kx – ωt)
y2 = A sin (kx + wt)
Considering the principle of superposition, the resultant can be calculated as
Resultant y = y1 + y2
= A sin (kx – ωt) + A sin (kx + wt)
(\begin{array}{l}y = A.2sin\left [ \frac{(kx-\omega t)+(kx+\omega t)}{(2)} \right ]cos\left [ \frac{(kx-\omega t)-(kx+\omega t)}{(2)} \right ]\end{array} )
y = 2Asin (kx) cos (ωt)
The equation represents the SHM of the collection of particles. Here, the term 2ASin (kx) is the amplitude of the resultant wave. From the expression of amplitude, it can be concluded that the amplitude of the particles executing SHM depends upon the location of the particles.
Nodes
The amplitude of the wave is zero for all the values of kx that give sin kx = 0. It means for kx = 0, π, 2π …..nπ, the amplitude of vibration of the particles will be zero. Here, n is an integer.
By substituting k = 2π/λ, we get
(2π/λ)x = nπ
⇒ x = nλ /2
Therefore, x = 0, λ /2, λ , 3λ /2……..
These points of zero displacements of the particles are called the nodes.
Antinodes
The amplitude will have a maximum value of 2A for all values of kx that give sin kx = ± 1. This means, for kx = π/2, 3π/2 ——— (n + ½)π, the amplitude of vibration of the particles will be maximum.
By substituting k = 2n+1π/λ, we get
(2π/λ)x = (2n+1)π /2
⇒ x = (2n+1)λ /4
⇒ x = λ /4, λ , 3λ /4,5λ /4……..
These points at which the displacement of the particles is maximum are called antinodes. The nodes and the antinodes in a standing wave are equally spaced, the distance is equal to λ/2, where λ is the wavelength of the wave.
Standing Waves on a String with Two Fixed Ends
Consider a uniform string of length L which is stretched between two fixed ends. A wave that travels in one direction along the string reflects at the end and returns inverted because of the fixed ends. These two identical waves, travelling in the opposite direction, form the standing wave on the string. The length of the string is given as L, so the wavelength of the wave is restricted by the boundary condition.
λ = 2L/n , here n = 1,2,3…..
The standing waves are formed in the string only if the wavelength satisfies the relationship with L. If v is the speed with which the waves travel along the string, then the frequency of the standing wave is
f = v/λ = nv/2L , n = 1,2,3…..
Nodes are formed at the fixed ends. In addition to the nodes, if an antinode exists at the centre of the string, the stretched string is said to vibrate in the fundamental frequency. The lowest resonance frequency, corresponding to n = 1, is the fundamental frequency. The higher frequencies are called the harmonics. Harmonics are the integer multiples of the fundamental frequency.
Standing Wave on a String with One End Fixed
When the string has one of its ends fixed and the other end is free, the node will be formed at the fixed end, and the antinode will be formed at the free end. The simplest standing wave formed in this condition is one-quarter a wavelength long. The next possible standing wave will be formed by adding both node and antinode. The length of this wave will be three-quarters of the wavelength. In general, the wavelength of the wave can be written as
λ = 4L/n , here n = 1,2,3…..
Then, the frequency of the standing wave is restricted to
f = nv/4L
Characteristics of Standing Waves
The standing waves are stationary. The disturbance does not travel in any direction.
Standing waves have points of zero amplitude called nodes and points of maximum amplitude called antinodes.
There will not be any flow of energy across any section of the medium.
The distance between the two consecutive nodes or the antinodes is equal to λ/2, whereas the distance between an antinode and its adjacent node is λ/4.
The pressure change is maximum at the node and minimum at the antinodes.
The medium is divided into a number of segments. All the particles in a particular segment will vibrate in phases. However, the amplitude increases from zero to maximum between the node and the antinode. Particles in neighbouring segments vibrate in opposite phases.
The period of oscillation of any particle in the medium is the same as that of the component waves.
Solved Questions
1. The distance between the nearest node and antinode in a stationary wave is
a. λ/2
b. λ/4
c. 1
d. 0
Answer: b. λ/4
2. Which of the following properties is different in a stationary wave and a progressive wave?
a. Frequency
b. Amplitude
c. Phase of the wave
d. Propagation of energy
Answer: d. Propagation of energy
Solution:
The progressive wave propagates energy, while the stationary wave will not propagate energy.
3. For the given stationary wave,
(\begin{array}{l}y = 4sin(\frac{\pi x}{15})cos(96\pi t)\end{array} )
, the distance between the nearest antinode and node is
a. 15
b. 7.5
c. 25
d. 8.5
Answer: b. 7.5
Solution:
Comparing the equation
(\begin{array}{l}y = 4sin(\frac{\pi x}{15})cos(96\pi t)\end{array} )
with the standard equation y = 2ASin (kx) Cos (ωt)
k = π/15
⇒ 2π/λ = π/15 (Since k = 2π/λ)
⇒ λ = 30
The distance between the nearest antinode and node is λ/4 = 30/4 = 7.5
Wave on String Rapid Revision
Standing Wave on String
Frequently Asked Questions on Standing Wave on a String
Q1
What are standing waves?
A wave in which the crests and troughs or compressions and rarefactions do not change their locations in space is called a stationary wave.
Q2
What are nodes?
The nodes are points on a standing wave at which the amplitude of vibration of the particle is zero.
Q3
What are antinodes?
Antinodes are points on the stationary waves at which the amplitude of vibration of the particle is maximum.
Q4
What is called the first harmonic?
The lowest frequency at which the string vibrates to form a standing wave is called the fundamental frequency or the first harmonic.
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190299 | https://www.brown.edu/Departments/Engineering/Courses/En1750/Notes/Elasticity/Elasticity.htm | EN175: Advanced Mechanics of Solids - Linear Elastic Stress-Strain Relations
Chapter 7
Stress-Strain relations for linear elastic materials
You are probably familiar with the behavior of a linear elastic material from introductory materials courses.
7.1 Isotropic, linear elastic material behavior
If you conduct a uniaxial tensile test on almost any material, and keep the stress levels sufficiently low, you will observe the following behavior:
The specimen deforms reversibly: If you remove the loads, the solid returns to its original shape.
The strain in the specimen depends only on the stress applied to it –– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@it doesn’t depend on the rate of loading, or the history of loading.
For most materials, the stress is a linear function of strain, as shown in the picture above. Because the strains are small, this is true whatever stress measure is adopted (Cauchy stress or nominal stress), and is true whatever strain measure is adopted (Lagrange strain or infinitesimal strain).
For most, but not all, materials, the material has no characteristic orientation. Thus, if you cut a tensile specimen out of a block of material, as shown in the figure, the the stress—— MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa8hfGaaa@3723@strain curve will be independent of the orientation of the specimen relative to the block of material. Such materials are said to be isotropic.
If you heat a specimen of the material, increasing its temperature uniformly, it will generally change its shape slightly. If the material is isotropic (no preferred material orientation) and homogeneous, then the specimen will simply increase in size, without shape change.
7.2 Stress—— MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieqajugybabaaaaaaaaapeGaa8hfGaaa@3724@strain relations for isotropic, linear elastic materials. Young’s Modulus, Poissons ratio and the Thermal Expansion Coefficient.
Before writing down stress—— MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa8hfGaaa@3723@strain relations, we need to decide what strain and stress measures we want to use. Because the model only works for small shape changes
Deformation is characterized using the infinitesimal strain tensor ε i j=(∂u i/∂x j+∂u j/∂x i)/2 ε i j=(∂u i/∂x j+∂u j/∂x i)/2 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabew7aLnaaBaaaleaacaWGPbGaamOAaaqabaGccqGH9aqpdaqadaqaaiabgkGi2kaadwhadaWgaaWcbaGaamyAaaqabaGccaGGVaGaeyOaIyRaamiEamaaBaaaleaacaWGQbaabeaakiabgUcaRiabgkGi2kaadwhadaWgaaWcbaGaamOAaaqabaGccaGGVaGaeyOaIyRaamiEamaaBaaaleaacaWGPbaabeaaaOGaayjkaiaawMcaaiaac+cacaaIYaaaaa@4880@defined in Section 4.6. This is convenient for calculations, but has the disadvantage that linear elastic constitutive equations can only be used if the solid experiences small rotations, as well as small shape changes.
All stress measures are taken to be equal. We can use the Cauchy stress σ i j σ i j MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabeo8aZnaaBaaaleaacaWGPbGaamOAaaqabaaaaa@3434@as the stress measure.
You probably already know the stress—— MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa8hfGaaa@3723@strain relations for an isotropic, linear elastic solid. They are repeated below for convenience.
⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢ε 11 ε 22 ε 33 2 ε 23 2 ε 13 2 ε 12⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥=1 E⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢1−ν−ν 0 0 0−ν 1−ν 0 0 0−ν−ν 1 0 0 0 0 0 0 2(1+ν)0 0 0 0 0 0 2(1+ν)0 0 0 0 0 0 2(1+ν)⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢σ 11 σ 22 σ 33 σ 23 σ 13 σ 12⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥+α Δ T⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢1 1 1 0 0 0⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥[ε 11 ε 22 ε 33 2 ε 23 2 ε 13 2 ε 12]=1 E[1−ν−ν 0 0 0−ν 1−ν 0 0 0−ν−ν 1 0 0 0 0 0 0 2(1+ν)0 0 0 0 0 0 2(1+ν)0 0 0 0 0 0 2(1+ν)][σ 11 σ 22 σ 33 σ 23 σ 13 σ 12]+α Δ T[1 1 1 0 0 0] MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@A385@
Here, E and ν ν MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8skY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabe27aUbaa@322F@are Young’s modulus and Poisson’s ratio, α α MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8skY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabeg7aHbaa@3216@is the coefficient of thermal expansion, and Δ T Δ T MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabfs5aejaadsfaaaa@32A6@is the increase in temperature of the solid. The remaining relations can be deduced from the fact that both σ i j σ i j MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabeo8aZnaaBaaaleaacaWGPbGaamOAaaqabaaaaa@3433@and ε i j ε i j MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabew7aLnaaBaaaleaacaWGPbGaamOAaaqabaaaaa@3417@are symmetric.
The inverse relationship can be expressed as
⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢σ 11 σ 22 σ 33 σ 23 σ 13 σ 12⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥=E(1+ν)(1−2 ν)⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢1−ν ν ν 0 0 0 ν 1−ν ν 0 0 0 ν ν 1−ν 0 0 0 0 0 0(1−2 ν)2 0 0 0 0 0 0(1−2 ν)2 0 0 0 0 0 0(1−2 ν)2⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢ε 11 ε 22 ε 33 2 ε 23 2 ε 13 2 ε 12⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥−E α Δ T 1−2 ν⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢1 1 1 0 0 0⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥[σ 11 σ 22 σ 33 σ 23 σ 13 σ 12]=E(1+ν)(1−2 ν)[1−ν ν ν 0 0 0 ν 1−ν ν 0 0 0 ν ν 1−ν 0 0 0 0 0 0(1−2 ν)2 0 0 0 0 0 0(1−2 ν)2 0 0 0 0 0 0(1−2 ν)2][ε 11 ε 22 ε 33 2 ε 23 2 ε 13 2 ε 12]−E α Δ T 1−2 ν[1 1 1 0 0 0] MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@B6D4@
HEALTH WARNING:Note the factor of 2 in the strain vector. Most texts, and most FEM codes use this factor of two, but not all. In addition, shear strains and stresses are often listed in a different order in the strain and stress vectors. For isotropic materials this makes no difference, but you need to be careful when listing material constants for anisotropic materials (see below).
We can write this expression in a much more convenient form using index notation. Verify for yourself that the matrix expression above is equivalent to
ε i j=1+ν E σ i j−ν E σ k k δ i j+α Δ T δ i j ε i j=1+ν E σ i j−ν E σ k k δ i j+α Δ T δ i j MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabew7aLnaaBaaaleaacaWGPbGaamOAaaqabaGccqGH9aqpdaWcaaqaaiaaigdacqGHRaWkcqaH9oGBaeaacaWGfbaaaiabeo8aZnaaBaaaleaacaWGPbGaamOAaaqabaGccqGHsisldaWcaaqaaiabe27aUbqaaiaadweaaaGaeq4Wdm3aaSbaaSqaaiaadUgacaWGRbaabeaakiabes7aKnaaBaaaleaacaWGPbGaamOAaaqabaGccqGHRaWkcqaHXoqycqqHuoarcaWGubGaeqiTdq2aaSbaaSqaaiaadMgacaWGQbaabeaaaaa@50AA@
The inverse relation is
σ i j=E 1+ν{ε i j+ν 1−2 ν ε k k δ i j}−E α Δ T 1−2 ν δ i j σ i j=E 1+ν{ε i j+ν 1−2 ν ε k k δ i j}−E α Δ T 1−2 ν δ i j MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabeo8aZnaaBaaaleaacaWGPbGaamOAaaqabaGccqGH9aqpdaWcaaqaaiaadweaaeaacaaIXaGaey4kaSIaeqyVd4gaamaacmaabaGaeqyTdu2aaSbaaSqaaiaadMgacaWGQbaabeaakiabgUcaRmaalaaabaGaeqyVd4gabaGaaGymaiabgkHiTiaaikdacqaH9oGBaaGaeqyTdu2aaSbaaSqaaiaadUgacaWGRbaabeaakiabes7aKnaaBaaaleaacaWGPbGaamOAaaqabaaakiaawUhacaGL9baacqGHsisldaWcaaqaaiaadweacqaHXoqycqqHuoarcaWGubaabaGaaGymaiabgkHiTiaaikdacqaH9oGBaaGaeqiTdq2aaSbaaSqaaiaadMgacaWGQbaabeaaaaa@5B07@
The stress-strain relations are often expressed using the elastic modulus tensor C i j k l C i j k l MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaGadeaadaaakeaacaWGdbWaaSbaaSqaaiaadMgacaWGQbGaam4AaiaadYgaaeqaaaaa@3778@or the elastic compliance tensor S i j k l S i j k l MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaGadeaadaaakeaacaWGtbWaaSbaaSqaaiaadMgacaWGQbGaam4AaiaadYgaaeqaaaaa@3788@as
σ i j=C i j k l(ε k l−α Δ T δ k l)ε i j=S i j k l σ k l+α Δ T δ i j σ i j=C i j k l(ε k l−α Δ T δ k l)ε i j=S i j k l σ k l+α Δ T δ i j MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7284@
In terms of elastic constants, C i j k l C i j k l MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaGadeaadaaakeaacaWGdbWaaSbaaSqaaiaadMgacaWGQbGaam4AaiaadYgaaeqaaaaa@3778@and S i j k l S i j k l MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaGadeaadaaakeaacaWGtbWaaSbaaSqaaiaadMgacaWGQbGaam4AaiaadYgaaeqaaaaa@3788@are
C i j k l=E 2(1+ν)(δ i l δ j k+δ i k δ j l)+E ν(1+ν)(1−2 ν)δ i j δ k l S i j k l=1+ν 2 E(δ i l δ j k+δ i k δ j l)−ν E δ i j δ k l C i j k l=E 2(1+ν)(δ i l δ j k+δ i k δ j l)+E ν(1+ν)(1−2 ν)δ i j δ k l S i j k l=1+ν 2 E(δ i l δ j k+δ i k δ j l)−ν E δ i j δ k l MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8CBA@
7.3 Reduced stress-strain equations for plane deformation of isotropic solids
For plane strain or plane stress deformations, some strain or stress components are always zero (by definition) so the stress-strain laws can be simplified.
For a plane strain deformationε 33=ε 23=ε 13=0 ε 33=ε 23=ε 13=0 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabew7aLnaaBaaaleaacaaIZaGaaG4maaqabaGccqGH9aqpcqaH1oqzdaWgaaWcbaGaaGOmaiaaiodaaeqaaOGaeyypa0JaeqyTdu2aaSbaaSqaaiaaigdacaaIZaaabeaakiabg2da9iaaicdaaaa@3E36@. The stress strain laws are therefore
⎡⎣⎢ε 11 ε 22 2 ε 12⎤⎦⎥=(1+ν)E⎡⎣⎢1−ν−ν 0−ν 1−ν 0 0 0 2⎤⎦⎥⎡⎣⎢σ 11 σ 22 σ 12⎤⎦⎥+(1+ν)α Δ T⎡⎣⎢1 1 0⎤⎦⎥[ε 11 ε 22 2 ε 12]=(1+ν)E[1−ν−ν 0−ν 1−ν 0 0 0 2][σ 11 σ 22 σ 12]+(1+ν)α Δ T[1 1 0] MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@6F86@
⎡⎣⎢σ 11 σ 22 σ 12⎤⎦⎥=E(1+ν)(1−2 ν)⎡⎣⎢⎢1−ν ν 0 ν 1−ν 0 0 0 1−2 ν 2⎤⎦⎥⎥⎡⎣⎢ε 11 ε 22 2 ε 12⎤⎦⎥−E α Δ T 1−2 ν⎡⎣⎢1 1 0⎤⎦⎥[σ 11 σ 22 σ 12]=E(1+ν)(1−2 ν)[1−ν ν 0 ν 1−ν 0 0 0 1−2 ν 2][ε 11 ε 22 2 ε 12]−E α Δ T 1−2 ν[1 1 0] MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@7770@
σ 33=E ν(ε 11+ε 22)(1−2 ν)(1+ν)+E α Δ T 1−2 ν,σ 13=σ 23=0 σ 33=E ν(ε 11+ε 22)(1−2 ν)(1+ν)+E α Δ T 1−2 ν,σ 13=σ 23=0 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@622E@
In index notation
ε α β=1+ν E{σ α β−ν σ γ γ δ α β}+(1+ν)α Δ T δ α β σ α β=E 1+ν{ε α β+ν 1−2 ν ε γ γ δ α β}−E α Δ T 1−2 ν δ α β ε α β=1+ν E{σ α β−ν σ γ γ δ α β}+(1+ν)α Δ T δ α β σ α β=E 1+ν{ε α β+ν 1−2 ν ε γ γ δ α β}−E α Δ T 1−2 ν δ α β MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@A201@
where Greek subscripts α,β α,β MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabeg7aHjaacYcacqaHYoGyaaa@3458@can have values 1 or 2.
For a plane stress deformationσ 33=σ 23=σ 13=0 σ 33=σ 23=σ 13=0 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabeo8aZnaaBaaaleaacaaIZaGaaG4maaqabaGccqGH9aqpcqaHdpWCdaWgaaWcbaGaaGOmaiaaiodaaeqaaOGaeyypa0Jaeq4Wdm3aaSbaaSqaaiaaigdacaaIZaaabeaakiabg2da9iaaicdaaaa@3E8A@
⎡⎣⎢ε 11 ε 22 2 ε 12⎤⎦⎥=1 E⎡⎣⎢1−ν 0−ν 1 0 0 0 2(1+ν)⎤⎦⎥⎡⎣⎢σ 11 σ 22 σ 12⎤⎦⎥+α Δ T⎡⎣⎢1 1 0⎤⎦⎥[ε 11 ε 22 2 ε 12]=1 E[1−ν 0−ν 1 0 0 0 2(1+ν)][σ 11 σ 22 σ 12]+α Δ T[1 1 0] MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@6619@
⎡⎣⎢σ 11 σ 22 σ 12⎤⎦⎥=E(1−ν 2)⎡⎣⎢1 ν 0 ν 1 0 0 0(1−ν)/2⎤⎦⎥⎡⎣⎢ε 11 ε 22 2 ε 12⎤⎦⎥−E α Δ T(1−ν)⎡⎣⎢1 1 0⎤⎦⎥[σ 11 σ 22 σ 12]=E(1−ν 2)[1 ν 0 ν 1 0 0 0(1−ν)/2][ε 11 ε 22 2 ε 12]−E α Δ T(1−ν)[1 1 0] MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@6FBC@
ε 33=−ν E(σ 11+σ 22)+α Δ T ε 33=−ν E(σ 11+σ 22)+α Δ T MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaGadeaadaaakeaacqaH1oqzdaWgaaWcbaGaaG4maiaaiodaaeqaaOGaeyypa0JaeyOeI0YaaSaaaeaacqaH9oGBaeaacaWGfbaaamaabmaabaGaeq4Wdm3aaSbaaSqaaiaaigdacaaIXaaabeaakiabgUcaRiabeo8aZnaaBaaaleaacaaIYaGaaGOmaaqabaaakiaawIcacaGLPaaacqGHRaWkcqaHXoqycqqHuoarcaWGubaaaa@48AD@
ε α β=1+ν E(σ α β−ν 1+ν σ γ γ δ α β)+α Δ T δ α β σ α β=E 1+ν{ε α β+ν 1−ν ε γ γ δ α β}−E α Δ T 1−ν δ α β ε α β=1+ν E(σ α β−ν 1+ν σ γ γ δ α β)+α Δ T δ α β σ α β=E 1+ν{ε α β+ν 1−ν ε γ γ δ α β}−E α Δ T 1−ν δ α β MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@9FF3@
7.4 Representative values for density, and elastic constants of isotropic solids
Most of the data in the table below were taken from the excellent introductory text `Engineering Materials,’ by M.F. Ashby and D.R.H. Jones, Pergamon Press. The remainder are from random web pages…
Note the units–– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@values of E are given in G N/m 2 G N/m 2 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4raiaad6eacaGGVaGaamyBamaaCaaaleqabaGaaGOmaaaaaaa@3A24@; the G stands for Giga, and is short for 10 9 10 9 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaicdadaahaaWcbeqaaiaaiMdaaaaaaa@385C@. The units for density are in M g m−3 M g m−3 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiaadEgacaWGTbWaaWbaaSqabeaacqGHsislcaaIZaaaaaaa@3A7E@- that’s Mega grams. One mega gram is 1000 kg.
Material Mass density
ρ/M g m−3 ρ/M g m−3 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdiNaaGPaVlaaykW7caGGVaGaaGPaVlaaykW7caaMc8UaamytaiaadEgacaWGTbWaaWbaaSqabeaacqGHsislcaaIZaaaaaaa@44A8@Youngs Modulus
E/G N m−2 E/G N m−2 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyraiaaykW7caaMc8UaaGPaVlaac+cacaaMc8UaaGPaVlaaykW7caWGhbGaamOtaiaad2gadaahaaWcbeqaaiabgkHiTiaaikdaaaaaaa@451D@Poisson Ratio
ν ν MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyVd4gaaa@37AF@Expansion coeft K−1 K−1 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaadUeadaahaaWcbeqaaiabgkHiTiaaigdaaaaaaa@330D@
Tungsten Carbide 14 –– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@17 450–– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@650 0.22 5×10−6 5×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaiwdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOnaaaaaaa@368D@
Silicon Carbide 2.5 –– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@3.2 450 0.22 4×10−6 4×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaisdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOnaaaaaaa@368C@
Tungsten 13.4 410 0.30 4×10−6 4×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaisdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOnaaaaaaa@368C@
Alumina 3.9 390 0.25 7×10−6 7×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaiEdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOnaaaaaaa@368F@
Titanium Carbide 4.9 380 0.19 13×10−6 13×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaigdacaaIZaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3746@
Silicon Nitride 3.2 320 - 270 0.22 3×10−6 3×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaiodacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOnaaaaaaa@368B@
Nickel 8.9 215 0.31 14×10−6 14×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaigdacaaI0aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3747@
CFRP 1.5-1.6 70 –– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@200 0.20 2×10−6 2×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaikdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOnaaaaaaa@368A@
Iron 7.9 196 0.30 13×10−6 13×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaigdacaaIZaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3746@
Low alloy steels 7.8 200 - 210 0.30 15×10−6 15×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaigdacaaI1aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3748@
Stainless steel 7.5-7.7 190 - 200 0.30 11×10−6 11×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaigdacaaIXaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3744@
Mild steel 7.8 196 0.30 15×10−6 15×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaigdacaaI1aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3748@
Copper 8.9 124 0.34 16×10−6 16×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaigdacaaI2aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3749@
Titanium 4.5 116 0.30 9×10−6 9×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaiMdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOnaaaaaaa@3691@
Silicon 2.5-3.2 107 0.22 5×10−6 5×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaiwdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOnaaaaaaa@368D@
Silica glass 2.6 94 0.16 0.5×10−6 0.5×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaicdacaGGUaGaaGynaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI2aaaaaaa@37F9@
Aluminum & alloys 2.6-2.9 69-79 0.35 22×10−6 22×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaikdacaaIYaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3746@
Concrete 2.4-2.5 45-50 0.3 10×10−6 10×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaigdacaaIWaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3743@
GFRP 1.4-2.2 7-45 10×10−6 10×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaigdacaaIWaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3743@
Wood, parallel grain 0.4-0.8 9-16 0.2 40×10−6 40×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaisdacaaIWaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3746@
Polyimides 1.4 3-5 0.1-0.45 40×10−6 40×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaisdacaaIWaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3746@
Nylon 1.1 –– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@1.2 2 –– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@4 0.25 81×10−6 81×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaiIdacaaIXaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@374B@
PMMA 1.2 3.4 0.35-0.4 50×10−6 50×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaiwdacaaIWaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3747@
Polycarbonate 1.2 –– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@1.3 2.6 0.36 65×10−6 65×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaiAdacaaI1aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@374D@
Natural Rubbers 0.83-0.91 0.01-0.1 0.49 200×10−6 200×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaikdacaaIWaGaaGimaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI2aaaaaaa@37FE@
PVC 1.3-1.6 0.003-0.01 0.41 70×10−6 70×10−6 MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaiEdacaaIWaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@3749@
7.5 Other Elastic Constants –– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieqajugybabaaaaaaaaapeGaa83eGaaa@3723@bulk, shear and Lame modulus.
Young’s modulus and Poisson’s ratio are the most common properties used to characterize elastic solids, but other measures are also used. For example, we define the shear modulus,bulk modulusand Lame modulusof an elastic solid as follows:
Bulk Modulus K=E 3(1−2 ν)Shear Modulus μ=E 2(1+ν)Lame Modulus λ=ν E(1+ν)(1−2 ν)Bulk Modulus K=E 3(1−2 ν)Shear Modulus μ=E 2(1+ν)Lame Modulus λ=ν E(1+ν)(1−2 ν) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@752D@
A nice table relating all the possible combinations of moduli to all other possible combinations is given below. Enjoy!
Lame
Modulus
λ λ MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBaaa@3729@Shear
Modulus
μ μ MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH8oqBaaa@372B@Young’s
Modulus
E E MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGfbaaaa@363F@Poisson’s
Ratio
ν ν MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH9oGBaaa@372D@Bulk
Modulus
K K MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGlbaaaa@3645@
λ,μ λ,μ MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBcaGGSaGaeqiVd0gaaa@398F@μ(3 λ+2 μ)λ+μ μ(3 λ+2 μ)λ+μ MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaeqiVd0MaaiikaiaaiodacqaH7oaBcqGHRaWkcaaIYaGaeqiVd0MaaiykaaqaaiabeU7aSjabgUcaRiabeY7aTbaaaaa@3D97@λ 2(λ+μ)λ 2(λ+μ) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaeq4UdWgabaGaaGOmaiaacIcacqaH7oaBcqGHRaWkcqaH8oqBcaGGPaaaaaaa@388C@3 λ+2 μ 3 3 λ+2 μ 3 MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaG4maiabeU7aSjabgUcaRiaaikdacqaH8oqBaeaacaaIZaaaaaaa@36F9@
λ,E λ,E MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBcaGGSaGaamyraaaa@38A3@Irrational Irrational Irrational
λ,ν λ,ν MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBcaGGSaGaeqyVd4gaaa@3991@λ(1−2 ν)2 ν λ(1−2 ν)2 ν MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaeq4UdWMaaiikaiaaigdacqGHsislcaaIYaGaeqyVd4MaaiykaaqaaiaaikdacqaH9oGBaaaaaa@3A14@λ(1+ν)(1−2 ν)ν λ(1+ν)(1−2 ν)ν MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaeq4UdWMaaiikaiaaigdacqGHRaWkcqaH9oGBcaGGPaGaaiikaiaaigdacqGHsislcaaIYaGaeqyVd4Maaiykaaqaaiabe27aUbaaaaa@3E06@λ(1+ν)3 ν λ(1+ν)3 ν MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaeq4UdWMaaiikaiaaigdacqGHRaWkcqaH9oGBcaGGPaaabaGaaG4maiabe27aUbaaaaa@394E@
λ,K λ,K MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBcaGGSaGaaGPaVlaadUeaaaa@3A34@3(K−λ)2 3(K−λ)2 MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaG4maiaacIcacaWGlbGaeyOeI0Iaeq4UdWMaaiykaaqaaiaaikdaaaaaaa@36BA@9 K(K−λ)3 K−λ 9 K(K−λ)3 K−λ MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaGyoaiaadUeacaGGOaGaam4saiabgkHiTiabeU7aSjaacMcaaeaacaaIZaGaam4saiabgkHiTiabeU7aSbaaaaa@3B02@λ 3 K−λ λ 3 K−λ MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaeq4UdWgabaGaaG4maiaadUeacqGHsislcqaH7oaBaaaaaa@3659@
μ,E μ,E MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH8oqBcaGGSaGaaGPaVlaaykW7caWGfbaaaa@3BBB@μ(2 μ−E)E−3 μ μ(2 μ−E)E−3 μ MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaeqiVd0MaaiikaiaaikdacqaH8oqBcqGHsislcaWGfbGaaiykaaqaaiaadweacqGHsislcaaIZaGaeqiVd0gaaaaa@3BD9@E−2 μ 2 μ E−2 μ 2 μ MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaamyraiabgkHiTiaaikdacqaH8oqBaeaacaaIYaGaeqiVd0gaaaaa@3712@μ E 3(3 μ−E)μ E 3(3 μ−E) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaeqiVd0MaamyraaqaaiaaiodacaGGOaGaaG4maiabeY7aTjabgkHiTiaadweacaGGPaaaaaaa@3937@
μ,ν μ,ν MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH8oqBcaGGSaGaaGPaVlaaykW7cqaH9oGBaaa@3CA9@2 μ ν 1−2 ν 2 μ ν 1−2 ν MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaGOmaiabeY7aTjabe27aUbqaaiaaigdacqGHsislcaaIYaGaeqyVd4gaaaaa@38BD@2 μ(1+ν)2 μ(1+ν) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaikdacqaH8oqBcaGGOaGaaGymaiabgUcaRiabe27aUjaacMcaaaa@3787@2 μ(1+ν)3(1−2 ν)2 μ(1+ν)3(1−2 ν) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaGOmaiabeY7aTjaacIcacaaIXaGaey4kaSIaeqyVd4MaaiykaaqaaiaaiodacaGGOaGaaGymaiabgkHiTiaaikdacqaH9oGBcaGGPaaaaaaa@3DC9@
μ,K μ,K MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH8oqBcaGGSaGaaGPaVlaadUeaaaa@3A36@3 K−2 μ 3 3 K−2 μ 3 MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaG4maiaadUeacqGHsislcaaIYaGaeqiVd0gabaGaaG4maaaaaaa@3620@9 K μ 3 K+μ 9 K μ 3 K+μ MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaGyoaiaadUeacqaH8oqBaeaacaaIZaGaam4saiabgUcaRiabeY7aTbaaaaa@37E5@3 K−2 μ 2(3 K+μ)3 K−2 μ 2(3 K+μ) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaG4maiaadUeacqGHsislcaaIYaGaeqiVd0gabaGaaGOmaiaacIcacaaIZaGaam4saiabgUcaRiabeY7aTjaacMcaaaaaaa@3B9D@
E,ν E,ν MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGfbGaaiilaiaaykW7caaMc8UaeqyVd4gaaa@3BBD@ν E(1+ν)(1−2 ν)ν E(1+ν)(1−2 ν) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaeqyVd4MaaGPaVlaadweaaeaacaGGOaGaaGymaiabgUcaRiabe27aUjaacMcacaGGOaGaaGymaiabgkHiTiaaikdacqaH9oGBcaGGPaaaaaaa@3EA7@E 2(1+ν)E 2(1+ν) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaamyraaqaaiaaikdacaGGOaGaaGymaiabgUcaRiabe27aUjaacMcaaaaaaa@36AB@E 3(1−2 ν)E 3(1−2 ν) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaamyraaqaaiaaiodacaGGOaGaaGymaiabgkHiTiaaikdacqaH9oGBcaGGPaaaaaaa@3773@
E,K E,K MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGfbGaaiilaiaaykW7caWGlbaaaa@394A@3 K(3 K−E)9 K−E 3 K(3 K−E)9 K−E MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaG4maiaadUeacaGGOaGaaG4maiaadUeacqGHsislcaWGfbGaaiykaaqaaiaaiMdacaWGlbGaeyOeI0Iaamyraaaaaaa@39EB@3 E K 9 K−E 3 E K 9 K−E MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaG4maiaadweacaWGlbaabaGaaGyoaiaadUeacqGHsislcaWGfbaaaaaa@3618@3 K−E 6 K 3 K−E 6 K MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaG4maiaadUeacqGHsislcaWGfbaabaGaaGOnaiaadUeaaaaaaa@354B@
ν,K ν,K MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH9oGBcaGGSaGaaGPaVlaaykW7caWGlbaaaa@3BC3@3 K ν(1+ν)3 K ν(1+ν) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaG4maiaadUeacqaH9oGBaeaacaGGOaGaaGymaiabgUcaRiabe27aUjaacMcaaaaaaa@386A@3 K(1−2 ν)2(1+ν)3 K(1−2 ν)2(1+ν) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaamaalaaabaGaaG4maiaadUeacaGGOaGaaGymaiabgkHiTiaaikdacqaH9oGBcaGGPaaabaGaaGOmaiaacIcacaaIXaGaey4kaSIaeqyVd4Maaiykaaaaaaa@3CE3@3 K(1−2 ν)3 K(1−2 ν) MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaaiodacaWGlbGaaiikaiaaigdacqGHsislcaaIYaGaeqyVd4Maaiykaaaa@3769@
7.6 Physical Interpretation of elastic constants for isotropic solids
It is important to have a feel for the physical significance of the two elastic constants E and ν ν MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8skY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabe27aUbaa@322F@.
Young’s modulusE is the slope of the stress—— MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa8hfGaaa@3723@strain curve in uniaxial tension. It has dimensions of stress ( N/m 2 N/m 2 MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaad6eacaGGVaGaamyBamaaCaaaleqabaGaaGOmaaaaaaa@33C8@) and is usually large –– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@for steel, E=210×10 9 N/m 2 E=210×10 9 N/m 2 MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaadweacqGH9aqpcaaIYaGaaGymaiaaicdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaaGyoaaaakiaaykW7caqGobGaae4laiaab2gadaahaaWcbeqaaiaabkdaaaaaaa@3DCE@. You can think of E as a measure of the stiffness of the solid. The larger the value of E, the stiffer the solid. For a stable material, E>0.
Poisson’s ratiois the ratio of lateral to longitudinal strain in uniaxial tensile stress. It is dimensionless and typically ranges from 0.2—— MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa8hfGaaa@3723@0.49, and is around 0.3 for most metals. For a stable material, −1<ν<0.5−1<ν<0.5 MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabgkHiTiaaigdacqGH8aapcqaH9oGBcqGH8aapcaaIWaGaaiOlaiaaiwdaaaa@37FA@. It is a measure of the compressibility of the solid. If ν=0.5 ν=0.5 MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabe27aUjabg2da9iaaicdacaGGUaGaaGynaaaa@3550@, the solid is incompressible –– MathType@MTEF@5@5@+=feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaaaaaaaapeGaa83eGaaa@3722@its volume remains constant, no matter how it is deformed. If ν=0 ν=0 MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8skY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabe27aUjabg2da9iaaicdaaaa@33EF@, then stretching a specimen causes no lateral contraction. Some bizarre materials have ν<0 ν<0 MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8skY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabe27aUjabgYda8iaaicdaaaa@33ED@-- if you stretch a round bar of such a material, the bar increases in diameter!!
Thermal expansion coefficient quantifies the change in volume of a material if it is heated in the absence of stress. It has dimensions of (degrees Kelvin)-1 and is usually very small. For steel,α≈6−10×10−6 K-1 α≈6−10×10−6 K-1 MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabeg7aHjabgIKi7kaaiAdacqGHsislcaaIXaGaaGimaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI2aaaaOGaaGPaVlaabUeadaahaaWcbeqaaiaab2cacaqGXaaaaaaa@4033@
The bulk modulus quantifies the resistance of the solid to volume changes. It has a large value (usually bigger than E).
The shear modulus quantifies its resistance to volume preserving shear deformations. Its value is usually somewhat smaller than E.
7.7 Strain Energy Density for Isotropic Solids
Note the following observations
If you deform a block of material, you do work on it (or, in some cases, it may do work on you…)
In an elastic material, the work done during loading is stored as recoverable strain energy in the solid. If you unload the material, the specimen does work on you, and when it reaches its initial configuration you come out even.
The work done to deform a specimen depends only on the state of strain at the end of the test. It is independent of the history of loading.
Based on these observations, we define the strain energy density of a solid as the work done per unit volume to deform a material from a stress free reference state to a loaded state.
To write down an expression for the strain energy density, it is convenient to separate the strain into two parts
ε i j=ε e i j+ε T i j ε i j=ε i j e+ε i j T MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabew7aLnaaBaaaleaacaWGPbGaamOAaaqabaGccqGH9aqpcqaH1oqzdaqhaaWcbaGaamyAaiaadQgaaeaacaWGLbaaaOGaey4kaSIaeqyTdu2aa0baaSqaaiaadMgacaWGQbaabaGaamivaaaaaaa@3F38@
where, for an isotropic solid,
ε T i j=α Δ T δ i j ε i j T=α Δ T δ i j MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabew7aLnaaDaaaleaacaWGPbGaamOAaaqaaiaadsfaaaGccqGH9aqpcqaHXoqycqqHuoarcaWGubGaeqiTdq2aaSbaaSqaaiaadMgacaWGQbaabeaaaaa@3D8D@
represents the strain due to thermal expansion (known as thermal strain), and
ε e i j=1+ν E σ i j−ν E σ k k δ i j ε i j e=1+ν E σ i j−ν E σ k k δ i j MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabew7aLnaaDaaaleaacaWGPbGaamOAaaqaaiaadwgaaaGccqGH9aqpdaWcaaqaaiaaigdacqGHRaWkcqaH9oGBaeaacaWGfbaaaiabeo8aZnaaBaaaleaacaWGPbGaamOAaaqabaGccqGHsisldaWcaaqaaiabe27aUbqaaiaadweaaaGaeq4Wdm3aaSbaaSqaaiaadUgacaWGRbaabeaakiabes7aKnaaBaaaleaacaWGPbGaamOAaaqabaaaaa@491D@
is the strain due to mechanical loading (known as elastic strain).
Work is done on the specimen only during mechanical loading. It is straightforward to show that the strain energy density is
U=1 2 σ i j ε e i j U=1 2 σ i j ε i j e MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiaadwfacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiabeo8aZnaaBaaaleaacaWGPbGaamOAaaqabaGccqaH1oqzdaqhaaWcbaGaamyAaiaadQgaaeaacaWGLbaaaaaa@3C3F@
You can also re-write this as
U=1+ν 2 E σ i j σ i j−ν 2 E σ k k σ j j U=E 2(1+ν)ε e i j ε e i j+E ν 2(1+ν)(1−2 ν)ε e j j ε e k k U=1+ν 2 E σ i j σ i j−ν 2 E σ k k σ j j U=E 2(1+ν)ε i j e ε i j e+E ν 2(1+ν)(1−2 ν)ε j j e ε k k e MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@7465@
Observe that
ε e i j=∂U∂σ i j σ i j=∂U∂ε e i j ε i j e=∂U∂σ i j σ i j=∂U∂ε i j e MathType@MTEF@5@5@+=feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vipgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaacmqaamaaaOqaaiabew7aLnaaDaaaleaacaWGPbGaamOAaaqaaiaadwgaaaGccqGH9aqpdaWcaaqaaiabgkGi2kaadwfaaeaacqGHciITcqaHdpWCdaWgaaWcbaGaamyAaiaadQgaaeqaaaaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeq4Wdm3aaSbaaSqaaiaadMgacaWGQbaabeaakiabg2da9maalaaabaGaeyOaIyRaamyvaaqaaiabgkGi2kabew7aLnaaDaaaleaacaWGPbGaamOAaaqaaiaadwgaaaaaaaaa@6506@ |
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