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Synchrotron radiation (also known as magnetobremsstrahlung radiation) is the electromagnetic radiation emitted when charged particles are accelerated radially, i.e., when they are subject to an acceleration perpendicular to their velocity (av). It is produced, for example, in synchrotrons using bending magnets, undulators and/or wigglers. If the particle is non-relativistic, then the emission is called cyclotron emission. If, on the other hand, the particles are relativistic, sometimes referred to as ultrarelativistic, the emission is called synchrotron emission. [1] Synchrotron radiation may be achieved artificially in synchrotrons or storage rings, or naturally by fast electrons moving through magnetic fields. The radiation produced in this way has a characteristic polarization and the frequencies generated can range over the entire electromagnetic spectrum which is also called continuum radiation.

Bremsstrahlung, from bremsen "to brake" and Strahlung "radiation"; i.e., "braking radiation" or "deceleration radiation", is electromagnetic radiation produced by the deceleration of a charged particle when deflected by another charged particle, typically an electron by an atomic nucleus. The moving particle loses kinetic energy, which is converted into radiation, thus satisfying the law of conservation of energy. The term is also used to refer to the process of producing the radiation. Bremsstrahlung has a continuous spectrum, which becomes more intense and whose peak intensity shifts toward higher frequencies as the change of the energy of the decelerated particles increases.

In physics, electromagnetic radiation refers to the waves of the electromagnetic field, propagating (radiating) through space, carrying electromagnetic radiant energy. It includes radio waves, microwaves, infrared, (visible) light, ultraviolet, X-rays, and gamma rays.

A synchrotron is a particular type of cyclic particle accelerator, descended from the cyclotron, in which the accelerating particle beam travels around a fixed closed-loop path. The magnetic field which bends the particle beam into its closed path increases with time during the accelerating process, being synchronized to the increasing kinetic energy of the particles. The synchrotron is one of the first accelerator concepts to enable the construction of large-scale facilities, since bending, beam focusing and acceleration can be separated into different components. The most powerful modern particle accelerators use versions of the synchrotron design. The largest synchrotron-type accelerator, also the largest particle accelerator in the world, is the 27-kilometre-circumference (17 mi) Large Hadron Collider (LHC) near Geneva, Switzerland, built in 2008 by the European Organization for Nuclear Research (CERN). It can accelerate beams of protons to an energy of 6.5 teraelectronvolts (TeV).

## History

Syncradiation was named after its discovery in Schenectady, New York from a General Electric synchrotron accelerator built in 1946 and announced in May 1947 by Frank Elder, Anatole Gurewitsch, Robert Langmuir and Herb Pollock in a letter entitled "Radiation from Electrons in a Synchrotron". [2] Pollock recounts:

General Electric Company (GE) is an American multinational conglomerate incorporated in New York and headquartered in Boston. As of 2018, the company operates through the following segments: aviation, healthcare, power, renewable energy, digital industry, additive manufacturing, venture capital and finance, lighting, transportation, and oil and gas.

On April 24, Langmuir and I were running the machine and as usual were trying to push the electron gun and its associated pulse transformer to the limit. Some intermittent sparking had occurred and we asked the technician to observe with a mirror around the protective concrete wall. He immediately signaled to turn off the synchrotron as "he saw an arc in the tube." The vacuum was still excellent, so Langmuir and I came to the end of the wall and observed. At first we thought it might be due to Cherenkov radiation, but it soon became clearer that we were seeing Ivanenko and Pomeranchuk radiation. [3]

Cherenkov radiation is an electromagnetic radiation emitted when a charged particle passes through a dielectric medium at a speed greater than the phase velocity of light in that medium. The characteristic blue glow of an underwater nuclear reactor is due to Cherenkov radiation.

Dmitri Dmitrievich Ivanenko was a Soviet-Ukrainian theoretical physicist who made great contributions to the physical science of the twentieth century, especially to nuclear physics, field theory, and gravitation theory. He worked in the Poltava Gravimetric Observatory of the Institute of Geophysics of NAS of Ukraine, was the head of the Theoretical Department Ukrainian Physico-Technical Institute in Kharkiv, Head of the Department of Theoretical Physics of the Kharkiv Institute of Mechanical Engineering. Professor of University of Kharkiv, Professor of Moscow State University.

Isaak Yakovlevich Pomeranchuk was a Soviet theoretical physicist working in particle physics, quantum field theory, electromagnetic and synchrotron radiation, condensed matter physics and the physics of liquid helium. The Pomeranchuk instability, the pomeron, and a few others phenomena in particle and condensed matter physics are named after him.

1. Broad Spectrum (which covers from microwaves to hard X-rays): the users can select the wavelength required for their experiment;
2. High Flux: high intensity photon beam allows rapid experiments or use of weakly scattering crystals;
3. High Brilliance: highly collimated photon beam generated by a small divergence and small size source (spatial coherence);
4. High Stability: submicron source stability;
5. Polarization: both linear and circular;
6. Pulsed Time Structure: pulsed length down to tens of picoseconds allows the resolution of process on the same time scale.

## Emission mechanism

When high-energy particles are in acceleration, including electrons forced to travel in a curved path by a magnetic field, synchrotron radiation is produced. This is similar to a radio antenna, but with the difference that, in theory, the relativistic speed will change the observed frequency due to the Doppler effect by the Lorentz factor, γ. Relativistic length contraction then bumps the frequency observed by another factor of γ, thus multiplying the GHz frequency of the resonant cavity that accelerates the electrons into the X-ray range. The radiated power is given by the relativistic Larmor formula while the force on the emitting electron is given by the Abraham–Lorentz–Dirac force.

A magnetic field is a vector field that describes the magnetic influence of electrical currents and magnetized materials. In everyday life, the effects of magnetic fields are often seen in permanent magnets, which pull on magnetic materials and attract or repel other magnets. Magnetic fields surround and are created by magnetized material and by moving electric charges such as those used in electromagnets. Magnetic fields exert forces on nearby moving electrical charges and torques on nearby magnets. In addition, a magnetic field that varies with location exerts a force on magnetic materials. Both the strength and direction of a magnetic field varies with location. As such, it is an example of a vector field.

Relativistic speed refers to speed at which relativistic effects become significant to the desired accuracy of measurement of the phenomenon being observed. Relativistic effects are those discrepancies between values calculated by models considering and not considering relativity. Related words are velocity, rapidity, and celerity which is proper velocity. Speed is a scalar, being the magnitude of the velocity vector which in relativity is the four-velocity and in three-dimension Euclidean space a three-velocity. Speed is empirically measured as average speed, although current devices in common use can estimate speed over very small intervals and closely approximate instantaneous speed. Non-relativistic discrepancies include cosine error which occurs in speed detection devices when only one scalar component of the three-velocity is measured and the Doppler effect which may affect observations of wavelength and frequency.

The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. It is named after the Austrian physicist Christian Doppler, who described the phenomenon in 1842.

The radiation pattern can be distorted from an isotropic dipole pattern into an extremely forward-pointing cone of radiation. Synchrotron radiation is the brightest artificial source of X-rays.

The planar acceleration geometry appears to make the radiation linearly polarized when observed in the orbital plane, and circularly polarized when observed at a small angle to that plane. Amplitude and frequency are however focused to the polar ecliptic.

Synchrotron radiation may occur in accelerators either as a nuisance, causing undesired energy loss in particle physics contexts, or as a deliberately produced radiation source for numerous laboratory applications. Electrons are accelerated to high speeds in several stages to achieve a final energy that is typically in the GeV range. In the LHC proton bunches also produce the radiation at increasing amplitude and frequency as they accelerate with respect to the vacuum field, propagating photoelectrons, which in turn propagate secondary electrons from the pipe walls with increasing frequency and density up to 7×1010. Each proton may lose 6.7 keV per turn due to this phenomenon. [4]

Particle physics is a branch of physics that studies the nature of the particles that constitute matter and radiation. Although the word particle can refer to various types of very small objects, particle physics usually investigates the irreducibly smallest detectable particles and the fundamental interactions necessary to explain their behaviour. By our current understanding, these elementary particles are excitations of the quantum fields that also govern their interactions. The currently dominant theory explaining these fundamental particles and fields, along with their dynamics, is called the Standard Model. Thus, modern particle physics generally investigates the Standard Model and its various possible extensions, e.g. to the newest "known" particle, the Higgs boson, or even to the oldest known force field, gravity.

A synchrotron light source is a source of electromagnetic radiation (EM) usually produced by a storage ring, for scientific and technical purposes. First observed in synchrotrons, synchrotron light is now produced by storage rings and other specialized particle accelerators, typically accelerating electrons. Once the high-energy electron beam has been generated, it is directed into auxiliary components such as bending magnets and insertion devices in storage rings and free electron lasers. These supply the strong magnetic fields perpendicular to the beam which are needed to convert high energy electrons into photons.

The Large Hadron Collider (LHC) is the world's largest and most powerful particle collider and the largest machine in the world. It was built by the European Organization for Nuclear Research (CERN) between 1998 and 2008 in collaboration with over 10,000 scientists and hundreds of universities and laboratories, as well as more than 100 countries. It lies in a tunnel 27 kilometres (17 mi) in circumference and as deep as 175 metres (574 ft) beneath the France–Switzerland border near Geneva. Its first data-taking period lasted from March 2010 to early 2013 at an energy of 3.5 to 4 teraelectronvolts (TeV) per beam, about four times the previous world record for a collider and accelerator. Afterwards, the accelerator was taken offline and upgraded over the course of two years. It was restarted in early 2015 for its second research run, reaching 6.5 TeV per beam. At the end of 2018, it entered a second two-year shutdown period.

Synchrotron radiation is also generated by astronomical objects, typically where relativistic electrons spiral (and hence change velocity) through magnetic fields. Two of its characteristics include non-thermal power-law spectra, and polarization. [5]

### History of detection

It was first detected in a jet emitted by Messier 87 in 1956 by Geoffrey R. Burbidge, [6] who saw it as confirmation of a prediction by Iosif S. Shklovsky in 1953, but it had been predicted earlier by Hannes Alfvén and Nicolai Herlofson [7] in 1950. Solar flares accelerate particles that emit in this way, as suggested by R. Giovanelli in 1948 and described critically by J.H. Piddington in 1952. [8]

T. K. Breus noted that questions of priority on the history of astrophysical synchrotron radiation are complicated, writing:

In particular, the Russian physicist V.L. Ginzburg broke his relationships with I.S. Shklovsky and did not speak with him for 18 years. In the West, Thomas Gold and Sir Fred Hoyle were in dispute with H. Alfven and N. Herlofson, while K.O. Kiepenheuer and G. Hutchinson were ignored by them. [9]

Supermassive black holes have been suggested for producing synchrotron radiation, by ejection of jets produced by gravitationally accelerating ions through the super contorted 'tubular' polar areas of magnetic fields. Such jets, the nearest being in Messier 87, have been confirmed by the Hubble telescope as apparently superluminal, travelling at 6 × c (six times the speed of light) from our planetary frame. This phenomenon is caused because the jets are travelling very near the speed of light and at a very small angle towards the observer. Because at every point of their path the high-velocity jets are emitting light, the light they emit does not approach the observer much more quickly than the jet itself. Light emitted over hundreds of years of travel thus arrives at the observer over a much smaller time period (ten or twenty years) giving the illusion of faster than light travel. There is no violation of special relativity. [10]

### Pulsar wind nebulae

A class of astronomical sources where synchrotron emission is important is the pulsar wind nebulae, a.k.a. plerions, of which the Crab nebula and its associated pulsar are archetypal. Pulsed emission gamma-ray radiation from the Crab has recently been observed up to ≥25 GeV, [11] probably due to synchrotron emission by electrons trapped in the strong magnetic field around the pulsar. Polarization in the Crab [12] at energies from 0.1 to 1.0 MeV illustrates a typical synchrotron radiation.

## Formulation

### Liénard–Wiechert Field

{\displaystyle {\begin{aligned}\mathbf {B} (\mathbf {r} ,t)&=-{\frac {\mu _{0}q}{4\pi }}\left[{\frac {c\,{\hat {\mathbf {n} }}\times {\vec {\beta }}}{\gamma ^{2}R^{2}\,(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}+{\frac {{\hat {\mathbf {n} }}\times [\,{\dot {\vec {\beta }}}+{\hat {\mathbf {n} }}\times ({\vec {\beta }}\times {\dot {\vec {\beta }}})]}{R\,(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}\right]_{\mathrm {retarded} },\qquad &(1)\\\mathbf {E} (\mathbf {r} ,t)&={\frac {q}{4\pi \varepsilon _{0}}}\left[{\frac {{\hat {\mathbf {n} }}-{\vec {\beta }}}{\gamma ^{2}R^{2}\,(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}+{\frac {{\hat {\mathbf {n} }}\times [({\hat {\mathbf {n} }}-{\vec {\beta }})\times {\dot {\vec {\beta }}}\,]}{c\,R\,(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}\right]_{\mathrm {retarded} },\qquad &(2)\end{aligned}}}

where R(t) = rr0(t), R(t) = |R(t)|, and n(t) = R(t)/R(t), which is the unit vector between the observation point and the position of the charge at the retarded time, and t is the retarded time.

In equation (1), and (2), the first terms for B and E resulting from the particle fall off as the inverse square of the distance from the particle, and this first term is called the generalized Coulomb field or velocity field. These terms represents the particle static field effect, which is a function of the component of its motion that has zero or constant velocity, as seen by a distant observer at r. By contrast, the second terms fall off as the inverse first power of the distance from the source, and these second terms are called the acceleration field or radiation field because they represent components of field due to the charge's acceleration (changing velocity), and they represent E and B which are emitted as electromagnetic radiation from the particle to an observer at r.

If we ignore the velocity field in order to find the power of emitted EM radiation only, the radial component of Poynting's vector resulting from the Liénard–Wiechert fields can be calculated to be

${\displaystyle [\mathbf {S\cdot } {\hat {\mathbf {n} }}]={\frac {q^{2}}{16\pi ^{2}\varepsilon _{0}c}}\left\{{\frac {1}{R^{2}}}\left|{\frac {{\hat {\mathbf {n} }}\times [({\hat {\mathbf {n} }}-{\vec {\beta }})\times {\dot {\vec {\beta }}}]}{(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}\right|^{2}\right\}_{\text{retarded}}.\qquad \qquad (3)}$

Note that

• The spatial relationship between β and determines the detailed angular power distribution.
• The relativistic effect of transforming from the rest frame of the particle to the observer's frame manifests itself by the presence of the factors (1 βn̂) in the denominator of Eq. (3).
• For ultrarelativistic particles the latter effect dominates the whole angular distribution.

The energy radiated into per solid angle during a finite period of acceleration from t = T1 to t = T2 is

{\displaystyle {\begin{aligned}{\frac {\mathrm {d} P}{\mathrm {d} {\mathit {\Omega }}}}&=R(t')^{2}\,[\mathbf {S} (t')\mathbf {\cdot } {\hat {\mathbf {n} }}(t')]\,{\frac {\mathrm {d} t}{\mathrm {d} t'}}=R(t')^{2}\,\mathbf {S} (t')\mathbf {\cdot } {\hat {\mathbf {n} }}(t')\,[1-{\vec {\beta }}(t')\mathbf {\cdot } {\hat {\mathbf {n} }}(t')]\\&={\frac {q^{2}}{16\pi ^{2}\varepsilon _{0}c}}\,{\frac {|{\hat {\mathbf {n} }}(t')\times \{[{\hat {\mathbf {n} }}(t')-{\vec {\beta }}(t')]\times {\dot {\vec {\beta }}}(t')\}|^{2}}{[1-{\vec {\beta }}(t')\mathbf {\cdot } {\vec {\mathbf {n} }}(t')]^{5}}}.\qquad \qquad (4)\end{aligned}}}

Integrating Eq. (4) over the all solid angles, we get the relativistic generalization of Larmor's formula

${\displaystyle P={\frac {e^{2}}{6\pi \varepsilon _{0}c}}\gamma ^{6}\left[\left|{\dot {\vec {\beta }}}\right|^{2}-\left|{\vec {\beta }}\times {\dot {\vec {\beta }}}\right|^{2}\right].\qquad (5)}$

However, this also can be derived by relativistic transformation of the 4-acceleration in Larmor's formula.

### Velocity perpendicular to acceleration (v ⟂ a): synchrotron radiation

When the charge is in instantaneous circular motion, its acceleration is perpendicular to its velocity β. Choosing a coordinate system such that instantaneously β is in the z direction and is in the x direction, with the polar and azimuth angles θ and φ defining the direction of observation, the general formula Eq. (4) reduces to

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} {\mathit {\Omega }}}}={\frac {q^{2}}{16\pi ^{2}\epsilon _{0}c}}{\frac {|{\dot {\vec {\beta }}}|^{2}}{(1-\beta \cos \theta )^{3}}}\left[1-{\frac {\sin ^{2}\theta \cos ^{2}\phi }{\gamma ^{2}(1-\beta \cos \theta )^{2}}}\right].\qquad (6)}$

In the relativistic limit ${\displaystyle (\gamma \gg 1)}$, the angular distribution can be written approximately as

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} {\mathit {\Omega }}}}\simeq {\frac {2}{\pi }}{\frac {e^{2}}{c^{3}}}\gamma ^{6}{\frac {|{\dot {\mathbf {v} }}|^{2}}{(1+\gamma ^{2}\theta ^{2})^{3}}}\left[1-{\frac {4\gamma ^{2}\theta ^{2}\cos ^{2}\phi }{(1+\gamma ^{2}\theta ^{2})^{2}}}\right].\qquad \qquad (7)}$

The factors (1 βcosθ) in the denominators tip the angular distribution forward into a narrow cone like the beam of a headlight pointing ahead of the particle. A plot of the angular distribution (dP/dΩ vs. γθ) shows a sharp peak around θ = 0.

Integration over the whole solid angle yields the total power radiated by one electron

${\displaystyle P={\frac {e^{2}}{6\pi \epsilon _{0}c}}\left|{\dot {\vec {\beta }}}\right|^{2}\gamma ^{4}={\frac {e^{2}c}{6\pi \epsilon _{0}}}{\frac {\beta ^{4}\gamma ^{4}}{\rho ^{2}}}={\frac {e^{4}}{6\pi \epsilon _{0}m^{4}c^{5}}}B^{2}E^{2}\beta ^{2}={\frac {e^{4}}{6\pi \epsilon _{0}m^{4}c^{5}}}B^{2}(E^{2}-m^{2}c^{4}),\qquad (8)}$

where E is the electron energy, B is the magnetic field, and ρ is the radius of curvature of the track in the field. Note that the radiated power is proportional to 1/m4, 1/ρ2, and B2. In some cases the surfaces of vacuum chambers hit by synchrotron radiation have to be cooled because of the high power of the radiation.

Using

${\displaystyle B={\frac {E\beta }{e\,r\sin(\alpha )}},}$

where α is the angle between the velocity and the magnetic field and r is the radius of the circular acceleration, the power emitted is:

${\displaystyle P={\frac {e^{2}}{6\pi \epsilon _{0}m^{4}c^{5}r^{2}\sin ^{2}(\alpha )}}E^{4}\beta ^{4}={\frac {e^{2}}{6\pi \epsilon _{0}m^{4}c^{5}r^{2}\sin ^{2}(\alpha )}}(E^{2}-m^{2}c^{4})^{2}.}$

Thus the power emitted scales as energy to the fourth, and decreases with the square of the radius and the fourth power of particle mass. This radiation is what limits the energy of an electron-positron circular collider. Generally, proton-proton colliders are instead limited by the maximum magnetic field; this is why, for example, the LHC has a center-of-mass energy 70 times higher than the LEP even though the proton mass is 2000 times the electron mass.

The energy received by an observer (per unit solid angle at the source) is

${\displaystyle {\frac {d^{2}W}{d\Omega }}=\int _{-\infty }^{\infty }{\frac {d^{2}P}{d\Omega }}dt=c\varepsilon _{0}\int _{-\infty }^{\infty }\left|R{\vec {E}}(t)\right|^{2}dt}$

Using the Fourier transformation we move to the frequency space

${\displaystyle {\frac {d^{2}W}{d\Omega }}=2c\varepsilon _{0}\int _{0}^{\infty }\left|R{\vec {E}}(\omega )\right|^{2}d\omega }$

Angular and frequency distribution of the energy received by an observer (consider only the radiation field)

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}=2c\varepsilon _{0}R^{2}\left|{\vec {E}}(\omega )\right|^{2}={\frac {e^{2}}{4\pi \varepsilon _{0}4\pi ^{2}c}}\left|\int _{-\infty }^{\infty }{\frac {{\hat {n}}\times \left[\left({\hat {n}}-{\vec {\beta }}\right)\times {\dot {\vec {\beta }}}\right]}{\left(1-{\hat {n}}\cdot {\vec {\beta }}\right)^{2}}}e^{i\omega (t-{\hat {n}}\cdot {\vec {r}}(t)/c)}dt\right|^{2}.\qquad (9)}$

Therefore, if we know the particle's motion, cross products term, and phase factor, we could calculate the radiation integral. However, calculations are generally quite lengthy (even for simple cases as for the radiation emitted by an electron in a bending magnet, they require Airy function or the modified Bessel functions).

### Example 1: bending magnet

#### Integrating

Trajectory of the arc of circumference is

${\displaystyle {\vec {r}}(t)=\left(\rho \sin {\frac {\beta c}{\rho }}t,\rho \left(1-\cos {\frac {\beta c}{\rho }}t\right),0\right).}$

In the limit of small angles we compute

${\displaystyle {\hat {n}}\times \left({\hat {n}}\times {\vec {\beta }}\right)=\beta \left[-{\vec {\varepsilon }}_{\parallel }\sin \left({\frac {\beta ct}{\rho }}\right)+{\vec {\varepsilon }}_{\perp }\cos \left({\frac {\beta ct}{\rho }}\right)\sin \theta \right]}$
${\displaystyle \omega \left(t-{\frac {{\hat {n}}\cdot {\vec {r}}(t)}{c}}\right)=\omega \left[t-{\frac {\rho }{c}}\sin \left({\frac {\beta ct}{\rho }}\right)\cos \theta \right]}$

Substituting into the radiation integral and introducing

${\displaystyle \xi ={\frac {\rho \omega }{3c\gamma ^{3}}}\left(1+\gamma ^{2}\theta ^{2}\right)^{3/2}}$
${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}={\frac {3e^{2}}{16\pi ^{3}\varepsilon _{0}c}}\left({\frac {2\omega \rho }{3c\gamma ^{2}}}\right)^{2}\left(1+\gamma ^{2}\theta ^{2}\right)^{2}\left[K_{2/3}^{2}(\xi )+{\frac {\gamma ^{2}\theta ^{2}}{1+\gamma ^{2}\theta ^{2}}}K_{1/3}^{2}(\xi )\right]\qquad (10),}$

where the function K is a modified Bessel function of the second kind.

#### Frequency distribution of radiated energy

From Eq.(10), we observe that the radiation intensity is negligible for ${\displaystyle \xi \gg 1}$. Critical frequency is defined as the frequency when ξ = 1/2 and θ = 0. So,

${\displaystyle \omega _{\text{c}}={\frac {3}{2}}{\frac {c}{\rho }}\gamma ^{3},}$

and critical angle is defined as the angle for which ${\displaystyle \xi (\theta _{\text{c}})\simeq \xi (0)+1}$ and is approximately

${\displaystyle \theta _{\text{c}}\simeq {\frac {1}{\gamma }}\left({\frac {2\omega _{\text{c}}}{\omega }}\right)^{1/3}}$ [13]

For frequencies much larger than the critical frequency and angles much larger than the critical angle, the synchrotron radiation emission is negligible.

Integrating on all angles, we get the frequency distribution of the energy radiated.

${\displaystyle {\frac {dW}{d\omega }}=\oint {\frac {d^{3}W}{d\omega d\Omega }}d\Omega ={\frac {{\sqrt {3}}e^{2}}{4\pi \varepsilon _{0}c}}\gamma {\frac {\omega }{\omega _{\text{c}}}}\int _{\omega /\omega _{\text{c}}}^{\infty }K_{5/3}(x)dx}$

If we define

${\displaystyle S(y)\equiv {\frac {9{\sqrt {3}}}{8\pi }}y\int _{y}^{\infty }K_{5/3}(x)dx}$
${\displaystyle \int _{0}^{\infty }S(y)dy=1,}$

where y = ω/ωc. Then

${\displaystyle {\frac {dW}{d\omega }}={\frac {2e^{2}\gamma }{9\varepsilon _{0}c}}S(y).\qquad (11)}$

Note that ${\displaystyle {\frac {dW}{d\omega }}\sim {\frac {e^{2}}{4\pi \varepsilon _{0}c}}\left({\frac {\omega \rho }{c}}\right)^{1/3}}$, if ${\displaystyle \omega \ll \omega _{\text{c}}}$, and ${\displaystyle {\frac {dW}{d\omega }}\approx {\sqrt {\frac {3\pi }{2}}}{\frac {e^{2}}{4\pi \varepsilon _{0}c}}\gamma \left({\frac {\omega }{\omega _{\text{c}}}}\right)^{0.5}e^{-\omega /\omega _{\text{c}}}}$, if ${\displaystyle \omega \gg \omega _{\text{c}}}$

The formula for spectral distribution of synchrotron radiation, given above, can be expressed in terms of a rapidly converging integral with no special functions involved [14] (see also modified Bessel functions ) by means of the relation:

${\displaystyle \int _{\xi }^{\infty }K_{5/3}(x)dx={\frac {1}{\sqrt {3}}}\,\int _{0}^{\infty }\,{\frac {9+36x^{2}+16x^{4}}{(3+4x^{2}){\sqrt {1+x^{2}/3}}}}\exp \left[-\xi \left(1+{\frac {4x^{2}}{3}}\right){\sqrt {1+{\frac {x^{2}}{3}}}}\right]\ dx}$

#### Synchrotron radiation emission as a function of the beam energy

First, define the critical photon energy as

${\displaystyle \varepsilon _{c}=\hbar \omega _{\text{c}}={\frac {3}{2}}{\frac {\hbar c}{\rho }}\gamma ^{3}.}$

Then, the relationship between radiated power and photon energy is shown in the graph on the right side. The higher the critical energy, the more photons with high energies are generated. Note that, there is no dependence on the energy at longer wavelength.

In Eq.(10), the first term ${\displaystyle K_{2/3}^{2}(\xi )}$ is the radiation power with polarization in the orbit plane, and the second term ${\displaystyle {\frac {\gamma ^{2}\theta ^{2}}{1+\gamma ^{2}\theta ^{2}}}K_{1/3}^{2}(\xi )}$ is the polarization orthogonal to the orbit plane.

In the orbit plane ${\displaystyle \theta =0}$, the polarization is purely horizontal. Integrating on all frequencies, we get the angular distribution of the energy radiated

${\displaystyle {\frac {d^{2}W}{d\Omega }}=\int _{0}^{\infty }{\frac {d^{3}W}{d\omega d\Omega }}d\omega ={\frac {7e^{2}\gamma ^{5}}{64\pi \varepsilon _{0}\rho }}{\frac {1}{(1+\gamma ^{2}\theta ^{2})^{5/2}}}\left[1+{\frac {5}{7}}{\frac {\gamma ^{2}\theta ^{2}}{1+\gamma ^{2}\theta ^{2}}}\right].\qquad (12)}$

Integrating on all the angles, we find that seven times as much energy is radiated with parallel polarization as with perpendicular polarization. The radiation from a relativistically moving charge is very strongly, but not completely, polarized in the plane of motion.

### Example 2: undulator

#### Solution of equation of motion and undulator equation

An undulator consists of a periodic array of magnets, so that they provide a sinusoidal magnetic field.

${\displaystyle {\vec {B}}=\left(0,B_{0}\sin(k_{\text{u}}z),0\right)}$

Solution of equation of motion is

${\displaystyle {\vec {r}}(t)={\frac {\lambda _{\text{u}}K}{2\pi \gamma }}\sin \omega _{\text{u}}t\cdot {\hat {x}}+\left({\bar {\beta _{z}}}ct+{\frac {\lambda _{\text{u}}K^{2}}{16\pi \gamma ^{2}}}\cos(2\omega _{\text{u}}t)\right)\cdot {\hat {z}}}$

where

${\displaystyle K={\frac {eB_{0}\lambda _{\text{u}}}{2\pi mc}},}$

and

${\displaystyle {\bar {\beta _{z}}}=1-{\frac {1}{2\gamma ^{2}}}\left(1+{\frac {K^{2}}{2}}\right),}$

and the parameter ${\displaystyle K}$ is called the undulator parameter.

Condition for the constructive interference of radiation emitted at different poles is

${\displaystyle d={\frac {\lambda _{\text{u}}}{\bar {\beta }}}-\lambda _{\text{u}}\cos \theta =n\lambda }$

Expanding ${\displaystyle \cos \theta \approx 1-{\frac {\theta ^{2}}{2}}}$ and neglecting the terms ${\displaystyle O(\theta ^{2})}$ in the resulting equation, one obtains

${\displaystyle \lambda _{n}={\frac {\lambda _{\text{u}}}{2\gamma ^{2}n}}\left({\frac {1+{\frac {K^{2}}{2}}+\gamma ^{2}\theta ^{2}}{\bar {\beta }}}\right)}$

For ${\displaystyle {\bar {\beta }}\rightarrow 1}$, one finally gets

${\displaystyle \lambda _{n}={\frac {\lambda _{\text{u}}}{2\gamma ^{2}n}}\left(1+{\frac {K^{2}}{2}}+\gamma ^{2}\theta ^{2}\right)\qquad (13).}$

This equation is called the undulator equation.

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}={\frac {e^{2}}{4\pi \varepsilon _{0}4\pi ^{2}c}}\left|\int _{-\infty }^{\infty }{\frac {{\hat {n}}\times \left[\left({\hat {n}}-{\vec {\beta }}\right)\times {\dot {\vec {\beta }}}\right]}{\left(1-{\hat {n}}\cdot {\vec {\beta }}\right)^{2}}}e^{i\omega (t-{\hat {n}}\cdot {\vec {r}}(t)/c)}dt\right|^{2}.}$

Using the periodicity of the trajectory, we can split the radiation integral into a sum over ${\displaystyle N}$ terms, where ${\displaystyle 2N}$ is the total number of bending magnets of the undulator.

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}={\frac {e^{2}\omega ^{2}}{4\pi \varepsilon _{0}4\pi ^{2}c}}\left|\int _{-\lambda _{u}/2{\bar {\beta }}c}^{\lambda _{u}/2{\bar {\beta }}c}{\hat {n}}\times \left({\hat {n}}\times {\vec {\beta }}\right)e^{i\omega (t-{\hat {n}}\cdot {\vec {r}}(t)/c)}dt\right|^{2}\left|1+e^{i\delta }+e^{2i\delta }+\cdots +e^{i(N_{u}-1)\delta }\right|^{2},\qquad (14)}$

where 　${\displaystyle {\bar {\beta }}=\beta \left(1-{\frac {K^{2}}{4\gamma ^{2}}}\right)}$

, and ${\displaystyle \delta ={\frac {2\pi \omega }{\omega _{\text{res}}(\theta )}}}$,　${\displaystyle \omega _{\text{res}}(\theta )={\frac {2\pi c}{\lambda _{\text{res}}(\theta )}}}$,　and　${\displaystyle \lambda _{\text{res}}(\theta )={\frac {\lambda _{u}}{2\gamma ^{2}}}\left(1+{\frac {K^{2}}{2}}+\gamma ^{2}\theta ^{2}\right).}$

The radiation integral in an undulator can be written as

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}={\frac {e^{2}\gamma ^{2}N^{2}}{4\pi \varepsilon _{0}c}}L\left(N{\frac {\Delta \omega _{n}}{\omega _{\text{res}}(\theta )}}\right)F_{n}(K,\theta ,\phi )\qquad (15).}$

where ${\displaystyle \Delta \omega _{n}=\omega -n\omega _{\text{res}}}$ is the frequency difference to the n-th harmonic. The sum of δ generates a series of sharp peaks in the frequency spectrum harmonics of fundamental wavelength

${\displaystyle L\left(N{\frac {\Delta \omega _{n}}{\omega _{\text{res}}(\theta )}}\right)={\frac {\sin ^{2}\left(N\pi \Delta \omega _{n}/\omega _{\text{res}}(\theta )\right)}{N^{2}\left(\pi \Delta \omega _{n}/\omega _{\text{res}}(\theta )\right)^{2}}},}$

and Fn depends on the angles of observations and K

${\displaystyle F_{n}(K,\theta ,\phi )\propto \left|\int _{-\lambda _{u}/2{\bar {\beta }}c}^{\lambda _{u}/2{\bar {\beta }}c}{\hat {n}}\times \left({\hat {n}}\times {\vec {\beta }}\right)e^{i\omega (t-{\hat {n}}\cdot {\vec {r}}(t)/c)}dt\right|^{2}.}$

On the axis (θ = 0, φ = 0), the radiation integral becomes

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}={\frac {e^{2}\gamma ^{2}N^{2}}{4\pi \varepsilon _{0}c}}L\left(N{\frac {\Delta \omega _{n}}{\omega _{\text{res}}(0)}}\right)F_{n}(K,0,0)}$

and

${\displaystyle F_{n}(K,0,0)={\frac {n^{2}K^{2}}{1+K^{2}/2}}\left[J_{\frac {n+1}{2}}(Z)-J_{\frac {n-1}{2}}(Z)\right]^{2},}$

where ${\displaystyle Z={\frac {nK^{2}}{4(1+K^{2}/2)}}.}$

Note that only odd harmonics are radiated on-axis, and as K increases higher harmonic becomes stronger.

## Notes

1. Yale Astronomy http://www.astro.yale.edu/vdbosch/astro320_summary27.pdf
2. Elder, F. R.; Gurewitsch, A. M.; Langmuir, R. V.; Pollock, H. C., "Radiation from Electrons in a Synchrotron" (1947) Physical Review, vol. 71, Issue 11, pp. 829-830
3. Iwanenko D., Pomeranchuk I., "On the maximal energy attainable in betatron", (1944) Physical Review vol.65, p. 343
4. Vladimir A. Bordovitsyn, "Synchrotron Radiation in Astrophysics" (1999) Synchrotron Radiation Theory and Its Development , ISBN   981-02-3156-3
5. Burbidge, G. R. "On Synchrotron Radiation from Messier 87. Astrophysical Journal, vol. 124, p. 416"
6. Alfvén, H.; Herlofson, N. "Cosmic Radiation and Radio Stars" Physical Review (1950), vol. 78, Issue 5, pp. 616–616
7. Paddington, J.H., "Thermal Theories of the High-Intensity Component of Solar Radio-Frequency Radiation. (1952) Proceedings of the Physical Society. Section B, vol. 66, Number 2
8. Breus, T. K., "Istoriya prioritetov sinkhrotronnoj kontseptsii v astronomii %t (Historical problems of the priority questions of the synchrotron concept in astrophysics)" (2001) in Istoriko-Astronomicheskie Issledovaniya, Vyp. 26, p. 88 – 97, 262 (2001)
9. Chase, Scott I. "Apparent Superluminal Velocity of Galaxies" . Retrieved 22 August 2012.
10. "Observation of Pulsed {gamma}-Rays Above 25 GeV from the Crab Pulsar with MAGIC", Science 21 November 2008: Vol. 322. no. 5905, pp. 1221–1224"
11. Dean et al.,"Polarized Gamma-Ray Emission from the Crab", Science 29 August 2008: Vol. 321. no. 5893, pp. 1183–1185
12. Jackson, John David (1999). Classical Electrodynamics (3rd ed.). Chichester: Wiley. p. 680. ISBN   0-471-30932-X.
13. M.Kh.Khokonov. Cascade Processes of Energy Loss by Emission of Hard Photons // JETP, V.99, No.4, pp. 690-707 \ (2004).

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[Dev-luatex] Benchmark.

David Kastrup dak at gnu.org
Tue Apr 3 16:00:14 CEST 2007

Taco Hoekwater <taco at elvenkind.com> writes:

> David Kastrup wrote:
>>
>> I am afraid that I am too tired to dig into the allocation routines of
>> TeX right now.
>
> That's what I thought as well. I did some imperical tests to reach
> my conclusion and they could be skewed. But even if this is not the
> cause of all the slowness, it still needs to be resolved. Leaking
> memory is never a good thing.

I had one idea, namely dumping the format file after compiling the
file, and see whether I find large repetitive patterns (in which case
I'd look up their node type).

Unfortunately, I got the following:

luatex -ini -fmt ztex plain "\let\end\dump \input tex"
This is luaTeX, Version 3.141592-snapshot-2007040210 (Web2C 7.5.6) (INITEX)
(/usr/local/texlive/2007/texmf-dist/tex/plain/base/plain.tex
macros, math definitions, output routines, hyphenation
(/usr/local/texlive/2007/texmf/tex/generic/hyphen/hyphen.tex)) (tex.tex
(/usr/local/texlive/2007/texmf-dist/tex/plain/base/webmac.tex) *1 [3] [4]
[5] [6] [7] [8] [9] *17 [10] [11]Segmentation fault (core dumped)

Now the LuaTeX documentation explicitly says that hyphenation in
iniTeX may crash the system.  Any idea how hard it would be to fix?

In the mean time, I'll probably switch off hyphenation and hope that
the situation turns out similar enough for the bug to show.

--
David Kastrup


text:
# Need a hint on this limit

1. Jun 15, 2005

### jdstokes

$\lim_{x\rightarrow 0^+}(1+3x)^{1/x}$

Thanks.

2. Jun 15, 2005

### steven187

hello there

firstly i would make a substitution such as
$x=\frac{1}{u}$
$\lim_{x\rightarrow 0^+}(1+3x)^{1/x}=\lim_{u\rightarrow\infty}(1+\frac{3}{u})^{u}$ you do know what this limit is equal to right?

steven

3. Jun 15, 2005

### HallsofIvy

Staff Emeritus
It might be better to make the substitution $$x= \frac{1}{3u}$$ so that $$1+ 3x= 1+ \frac{1}{u}$$ and $$\frac{1}{x}= 3u$$. Then the limit becomes
$$\lim_{x\rightarrow 0^+}(1+3x)^{1/x}=\lim_{u\rightarrow\infty}(1+\frac{1}{u})^{3u}= \{\lim_{u\rightarrow\infty}(1+ \frac{1}{u})^u\}^3$$.

Last edited: Jun 15, 2005
4. Jun 15, 2005

### jdstokes

I still don't know where to go with this. I seem to keep getting a $(1+ 0)^\infty$ situation.

5. Jun 15, 2005

### dextercioby

Do you know the definition of "e" (Euler's number)...?

Daniel.

6. Jun 15, 2005


text:
# Question on arguments of the Gamma Function

I came accross with this site about Gamma function. I just want to verify, clarify, whatever you may want to call it. It says you can compute for the gamma value for a negative argument using $$\Gamma(-z)=\frac{-\pi}{z\Gamma(z)\sin(\pi z)}$$

Is this true for all? I thought you cant have the negative numbers as argument for the gamma function.

-
Indeed $\Gamma(-n)$ is not defined, for $n \in \Bbb N_{>0}$ (as is apparent from the $\sin(\pi z)$ factor in the denominator). Luckily, there are many more negative numbers than just the negative integers. –  Lord_Farin Oct 23 '12 at 6:26
Hi. thank you for that. Does it follow that $$\Gamma(z)=\frac{\pi}{z\Gamma(z)\sin(\pi z)},z>0,z\not\in\mathbb{Z}$$? Sorry for asking too many questions. Im just really having a tough time on my paper. Thanks again. –  Mynameis Tiara Oct 23 '12 at 6:43
Take a look at Euler's reflection formula. And while you're at it, there is much more information on the gamma function at the link. –  Harald Hanche-Olsen Oct 23 '12 at 7:33


text:
# How to find 1 honest man out of 99 with $100 dollars? 1. Jul 11, 2011 ### hahaputao While hiking through the mountains, you get lost and end up in a village, population 99. You only have$200 left. Everyone is greedy, so they won't do anything for free. The majority of the townsfolk are also honest, but some are troublemakers. You need to hire one of the honest citizens as your guide home, which will cost $100. But first you need to nd one that is honest. For$1, you can ask any villager if another villager is honest. If you ask an honest villager, they tell the truth. If you ask a troublemakers, they might say anything. (The villagers don't understand other kinds of questions.)
Come up with a way to nd an honest villager for sure, and with enough money left to get home.

important:
1. the villagers do NOT understand any other questions
2. the troublemakers may lie OR tell the truth

2. Jul 23, 2011

### 256bits

1. the villagers do NOT understand any other questions
Then how would an honest villager know what you mean when you ask for a guide if they do not understand any other question except "Is he honest?"


text:
# Art Songs

$\begin{array}{l l} \text{Parlo d'amor vegliando,} & \text{Whatsoe'er I am doing,} \\ \text{Parlo d'amor sognando,} & \text{Whatsoe'er I'm pursuing,} \\ \text{All' acqua, all' ombra, ai monti,} & \text{In sunshine or in showers,} \\ \text{Ai fiori, all' erbe, ai fonti, ...} & \text{At home or midst the flowers, ...} \\ \text{All' ecco---all' aria---a venti,} & \text{I sigh---I pant---I languish,} \\ \text{Che il suon de vani accenti, ...} & \text{In bliss that throbs like anguish, ...} \\ \end{array}$

Sung by the pubescent, flirtatious Cherubino (he’s played by a soprano), these words are notable for their arresting meter and rhyme, the heavenly beauty of their melody, and, well, their humor. Mozart’s The Marriage of Figaro—recently performed fantastically by the Peabody Opera Theater—presents such an explosive combination of theatricality, musicality, and hilarity as to make Mozart come across as a supernatural genius.

Only later did I learn that Mozart did not work alone [1]. The Felix Krull-esque French man-of-the-world Pierre Beaumarchais originally wrote the French play Le Mariage de Figaro in 1784; only then did Lorenzo Da Ponte, an equally fascinating Italian librettist, translate the play into Italian, excise a tirade against inherited nobility (thus making the play acceptable to the censors), and set certain of its passages to meter and rhyme. These developments, finally, prepared the way for Mozart to set music to the entire work, which premiered in 1786. The libretto’s rhymed passages became the opera’s arias.

This realization, in fact, placed Mozart into a long tradition within classical music.

Nowhere is this tradition better exemplified than in the world of Lieder, in which historic German poems are set anew to music featuring just a single voice and a single piano. Though Lieder has a long history, it thrived in the Romantic era of the late 18th and early 19th centuries [2]. Robert Schumann, in particular, emerged as a prolific composer of Lieder, and, though many poets appeared in his works, chief among them was the legendary Jewish German Romantic poet Heinrich Heine.

I’ve become especially interested in the Romanzen und Balladen of Schumann’s Liederkreis, Op. 24., a collection of Lieder works composed using poems drawn from Heine’s Buch der Lieder, itself written in 1827. These include Die beiden Grenadiere, in which Schumann concludes Heine’s fiery tribute to French patriotism with a tongue-in-cheek musical quotation of La Marseillaise, as well as Belsatzar, in which Heine’s wrenching rendition of the Old Testament story from the Book of Daniel is made all the more haunting by Schumann’s sensitive musical touch. Also noteworthy are Abends am Strand, where fisherman gaze out to the sea and tell stories, and Die fiendlichen Bruder, a poignant tale of two feuding brothers.

*                *                *

“This kinda sucks,” Josh once told me, after I showed him a Lied set to orchestra. “It’s just not that pleasant to listen to.” While Lob des Hohen Verstandes may be, grantedly, more entertaining than pleasant, Josh has responded with similar encouragement to most of the Lieder I’ve shown him. I often find myself turning to the same response: You have to listen to it with the lyrics.

This exchange, in my view, exhibits something important about Lieder. I haven’t tried listening to Mozart’s Figaro separated from its visual staging, from the lilt of its rhymes, and the from meanings of its words—but I’m not particularly inclined to try. The opera, despite its separate parts and their disparate origins, is fundamentally one work of art. To deprive it of some or all of its parts would be a disservice.

In Lieder, likewise, multiple parts (typically, again, from different sources) come together to yield the final work of art. I can name a few of these:

1. The story. Though sometimes invented by the poet, the stories used in Lieder poems are often based on old folk stories, if not biblical ones.
2. The poetry. The poet—for example, Heine—adds to these stories immaculate and pleasing meter and rhyme.
3. The music. Finally, the composer, in our case Schumann, superimposes music, which itself comprises melody as well as subtler features like dynamics and tempo.

It is the simultaneous combination of these many features, I contend, that makes Lieder exquisitely pleasing. Every word of a Lied’s poem is sung using subtle and shifting dynamics, wrapping together its story into a dramatic and consuming—and entirely musical—experience. These three aspects speak together in a very strong language.

The word Lieder is typically translated into English as Art Songs. No reduction to constituent parts can account for Lieder’s brilliance, however, and in this piece I have used only the original word.

References

1. The Marriage of Figaro, Encyclopedia Brittanica
2. Lied, Encyclopedia Brittanica
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text:
# Order Isomorphism iff Strictly Increasing Surjection

## Theorem

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be totally ordered sets.

A mapping $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ is an order isomorphism if and only if:

$(1): \quad \phi$ is a surjection
$(2): \quad \forall x, y \in S: x \mathop {\prec_1} y \implies \map \phi x \mathop {\prec_2} \map \phi y$

## Proof

### Necessary Condition

Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be an order isomorphism.

Then by definition $\phi$ is a bijection and so a surjection.

Suppose $x \mathop {\prec_1} y$.

That is:

$x \mathop {\preceq_1} y$
$x \ne y$

Then:

$x \mathop {\prec_1} y \implies \map \phi x \mathop {\preceq_2} \map \phi y$

as $\phi$ is an order isomorphism.

But as $\phi$ is a bijection it is also an injection.

Thus:

$\map \phi x = \map \phi y \implies x = y$

and so it follows that:

$x \mathop {\prec_1} y \implies \map \phi x \mathop {\prec_2} \map \phi y$

$\Box$

### Sufficient Condition

Suppose $\phi$ is a mapping which satisfies the conditions:

$(1): \quad \phi$ is a surjection
$(2): \quad \forall x, y \in S: x \mathop {\prec_1} y \implies \map \phi x \mathop {\prec_2} \map \phi y$

From $(2)$ and Strictly Increasing Mapping is Increasing we have:

$x \mathop {\preceq_1} y \implies \map \phi x \mathop {\preceq_2} \map \phi y$

Now suppose $\map \phi x \mathop {\preceq_2} \map \phi y$.

Suppose $y \mathop {\prec_1} x$.

Then from $(2)$ it would follow that $\map \phi y \mathop {\prec_2} \map \phi x$.

So it is not the case that $y \mathop {\prec_1} x$.

So from the Trichotomy Law:

$x \mathop {\preceq_1} y$

Thus it follows that:

$x \mathop {\preceq_1} y \iff \map \phi x \mathop {\preceq_2} \map \phi y$

It follows from Order Isomorphism is Surjective Order Embedding that $\phi$ is an order isomorphism.

$\blacksquare$


text:
Determine PH of this solution

1. Jun 8, 2006

petuniac

Question is:

An unknown solution is 40 times more alkaline than neutral water which has a PH of 7. Determine the PH of the unknown solution.

Here is what I have:

40 = (log base 10 x)/(log base 10 7)
40 = 10^(x-7)
10^1.4 = 10^(x-7)
1.4 = x - 7
x = 8.4

The answer is supposed to be 8.6??? Not sure where my mistake is.

2. Jun 8, 2006

TD

Is the first line correct? Assuming the second is, it's wrong to go from 40 to 10^1.4, recheck that.

3. Jun 8, 2006

dav2008

Like td said, your exponent of 1.4 in that line is wrong.

Think of the following equation:
10y=40

Now you can see how to write y in terms of logarithms. Then when you find what y is you will have 10y=10(x-7).

Note that this is kind of the work-around method. It's easier to look at the line 40=10(x-7) and take the log base 10 of both sides.

Last edited: Jun 8, 2006
4. Jun 8, 2006

petuniac

Thanks! Not sure what I was thinking.


text:
fuzzylite  6.0 A Fuzzy Logic Control Library in C++
fl::Aggregated Class Reference

The Aggregated class is a special Term that stores a fuzzy set with the Activated terms from the Antecedents of a Rule, thereby serving mainly as the fuzzy output value of the OutputVariables. More...

#include <Aggregated.h>

Inheritance diagram for fl::Aggregated:
Collaboration diagram for fl::Aggregated:

## Public Member Functions

Aggregated (const std::string &name="", scalar minimum=fl::nan, scalar maximum=fl::nan, SNorm *aggregation=fl::null)

Aggregated (const Aggregated &other)

Aggregatedoperator= (const Aggregated &other)

virtual ~Aggregated () FL_IOVERRIDE

virtual std::string className () const FL_IOVERRIDE
Returns the name of the class of the term. More...

virtual std::string parameters () const FL_IOVERRIDE
Returns the parameters of the term. More...

virtual void configure (const std::string &parameters) FL_IOVERRIDE
Does nothing. More...

virtual Aggregatedclone () const FL_IOVERRIDE
Creates a clone of the term. More...

virtual Complexity complexity () const FL_IOVERRIDE
Computes the estimated complexity of evaluating the membership function. More...

virtual Complexity complexityOfMembership () const

virtual Complexity complexityOfActivationDegree () const

virtual scalar membership (scalar x) const FL_IOVERRIDE
Aggregates the membership function values of $$x$$ utilizing the aggregation operator. More...

virtual scalar activationDegree (const Term *forTerm) const
Computes the aggregated activation degree for the given term. More...

virtual const ActivatedhighestActivatedTerm () const
Iterates over the Activated terms to find the term with the maximum activation degree. More...

virtual std::string toString () const FL_IOVERRIDE
Returns the representation of the term in the FuzzyLite Language. More...

virtual void setMinimum (scalar minimum)
Sets the minimum of the range of the fuzzy set. More...

virtual scalar getMinimum () const
Gets the minimum of the range of the fuzzy set. More...

virtual void setMaximum (scalar maximum)
Sets the maximum of the range of the fuzzy set. More...

virtual scalar getMaximum () const
Gets the maximum of the range of the fuzzy set. More...

virtual void setRange (scalar minimum, scalar maximum)
Sets the range of the fuzzy set to [minimum, maximum] More...

virtual scalar range () const
Returns the magnitude of the range of the fuzzy set,. More...

virtual void setAggregation (SNorm *aggregation)
Sets the aggregation operator. More...

virtual SNormgetAggregation () const
Gets the aggregation operator. More...

virtual void addTerm (const Term *term, scalar degree, const TNorm *implication)
Adds a new Activated term (from the parameters) to the fuzzy set. More...

virtual void addTerm (const Activated &term)
Adds the activated term to the fuzzy set. More...

virtual const ActivatedgetTerm (std::size_t index) const
Gets the term at the given index. More...

virtual const ActivatedremoveTerm (std::size_t index)
Removes the term at the given index without deleting the term. More...

virtual std::size_t numberOfTerms () const
Returns the number of activated terms. More...

virtual void setTerms (const std::vector< Activated > &terms)
Sets the activated terms. More...

virtual const std::vector< Activated > & terms () const
Returns an immutable vector of activated terms. More...

virtual std::vector< Activated > & terms ()
Returns a mutable vector of activated terms. More...

virtual bool isEmpty () const
Indicates whether the vector of activated terms is empty. More...

virtual void clear ()
Clears and deletes the activated terms. More...

Public Member Functions inherited from fl::Term
Term (const std::string &name="", scalar height=1.0)

virtual ~Term ()

virtual void setName (const std::string &name)
Sets the name of the term. More...

virtual std::string getName () const
Gets the name of the term. More...

virtual void setHeight (scalar height)
Sets the height of the term. More...

virtual scalar getHeight () const
Gets the height of the term. More...

virtual void updateReference (const Engine *engine)
Updates the references (if any) to point to the current engine (useful when cloning engines or creating terms within Importer objects. More...

Protected Attributes inherited from fl::Term
scalar _height

## Detailed Description

The Aggregated class is a special Term that stores a fuzzy set with the Activated terms from the Antecedents of a Rule, thereby serving mainly as the fuzzy output value of the OutputVariables.

The ownership of the activated terms will be transfered to objects of this class, and therefore their destructors will be called upon destruction of this term (or calling Aggregated::clear()).

Antecedent
Rule
OutputVariable
Activated
Term
Since
6.0

Definition at line 47 of file Aggregated.h.

## ◆ Aggregated() [1/2]

 fl::Aggregated::Aggregated ( const std::string & name = "", scalar minimum = fl::nan, scalar maximum = fl::nan, SNorm * aggregation = fl::null )
explicit

## ◆ Aggregated() [2/2]

 fl::Aggregated::Aggregated ( const Aggregated & other )

## ◆ ~Aggregated()

 virtual fl::Aggregated::~Aggregated ( )
virtual

## ◆ activationDegree()

 virtual scalar fl::Aggregated::activationDegree ( const Term * forTerm ) const
virtual

Computes the aggregated activation degree for the given term.

If the same term is present multiple times, the aggregation operator is utilized to sum the activation degrees of the term. If the aggregation operator is fl::null, a regular sum is performed.

Parameters
 forTerm is the term for which to compute the aggregated activation degree
Returns
the aggregated activation degree for the given term

 virtual void fl::Aggregated::addTerm ( const Term * term, scalar degree, const TNorm * implication )
virtual

Adds a new Activated term (from the parameters) to the fuzzy set.

Parameters
 term is the activated term degree is the activation degree implication is the implication operator

 virtual void fl::Aggregated::addTerm ( const Activated & term )
virtual

Adds the activated term to the fuzzy set.

The activated term will be deleted when Aggregated::clear()

Parameters
 term is the activated term

## ◆ className()

 virtual std::string fl::Aggregated::className ( ) const
virtual

Returns the name of the class of the term.

Returns
the name of the class of the term

Implements fl::Term.

## ◆ clear()

 virtual void fl::Aggregated::clear ( )
virtual

Clears and deletes the activated terms.

## ◆ clone()

 virtual Aggregated* fl::Aggregated::clone ( ) const
virtual

Creates a clone of the term.

Returns
a clone of the term

Implements fl::Term.

## ◆ complexity()

 virtual Complexity fl::Aggregated::complexity ( ) const
virtual

Computes the estimated complexity of evaluating the membership function.

Returns
the estimated complexity of evaluating the membership function

Implements fl::Term.

## ◆ complexityOfActivationDegree()

 virtual Complexity fl::Aggregated::complexityOfActivationDegree ( ) const
virtual

## ◆ complexityOfMembership()

 virtual Complexity fl::Aggregated::complexityOfMembership ( ) const
virtual

## ◆ configure()

 virtual void fl::Aggregated::configure ( const std::string & parameters )
virtual

Does nothing.

Parameters
 parameters are irrelevant

Implements fl::Term.

## ◆ getAggregation()

 virtual SNorm* fl::Aggregated::getAggregation ( ) const
virtual

Gets the aggregation operator.

Returns
the aggregation operator

## ◆ getMaximum()

 virtual scalar fl::Aggregated::getMaximum ( ) const
virtual

Gets the maximum of the range of the fuzzy set.

Returns
the maximum of the range of the fuzzy set

## ◆ getMinimum()

 virtual scalar fl::Aggregated::getMinimum ( ) const
virtual

Gets the minimum of the range of the fuzzy set.

Returns
the minimum of the range of the fuzzy set

## ◆ getTerm()

 virtual const Activated& fl::Aggregated::getTerm ( std::size_t index ) const
virtual

Gets the term at the given index.

Parameters
 index is the index of the term
Returns
the activated term at the given index

## ◆ highestActivatedTerm()

 virtual const Activated* fl::Aggregated::highestActivatedTerm ( ) const
virtual

Iterates over the Activated terms to find the term with the maximum activation degree.

Returns
the term with the maximum activation degree

## ◆ isEmpty()

 virtual bool fl::Aggregated::isEmpty ( ) const
virtual

Indicates whether the vector of activated terms is empty.

Returns
whether the vector of activated terms is empty

## ◆ membership()

 virtual scalar fl::Aggregated::membership ( scalar x ) const
virtual

Aggregates the membership function values of $$x$$ utilizing the aggregation operator.

Parameters
 x is a value
Returns
$$\sum_i{\mu_i(x)}, i \in \mbox{terms}$$

Implements fl::Term.

## ◆ numberOfTerms()

 virtual std::size_t fl::Aggregated::numberOfTerms ( ) const
virtual

Returns the number of activated terms.

Returns
the number of activated terms

## ◆ operator=()

 Aggregated& fl::Aggregated::operator= ( const Aggregated & other )

## ◆ parameters()

 virtual std::string fl::Aggregated::parameters ( ) const
virtual

Returns the parameters of the term.

Returns
"aggregation minimum maximum terms"

Implements fl::Term.

## ◆ range()

 virtual scalar fl::Aggregated::range ( ) const
virtual

Returns the magnitude of the range of the fuzzy set,.

Returns
the magnitude of the range of the fuzzy set, i.e., maximum - minimum

## ◆ removeTerm()

 virtual const Activated& fl::Aggregated::removeTerm ( std::size_t index )
virtual

Removes the term at the given index without deleting the term.

Parameters
 index is the index of the term
Returns
the removed term

## ◆ setAggregation()

 virtual void fl::Aggregated::setAggregation ( SNorm * aggregation )
virtual

Sets the aggregation operator.

Parameters
 aggregation is the aggregation operator

## ◆ setMaximum()

 virtual void fl::Aggregated::setMaximum ( scalar maximum )
virtual

Sets the maximum of the range of the fuzzy set.

Parameters
 maximum is the maximum of the range of the fuzzy set

## ◆ setMinimum()

 virtual void fl::Aggregated::setMinimum ( scalar minimum )
virtual

Sets the minimum of the range of the fuzzy set.

Parameters
 minimum is the minimum of the range of the fuzzy set

## ◆ setRange()

 virtual void fl::Aggregated::setRange ( scalar minimum, scalar maximum )
virtual

Sets the range of the fuzzy set to [minimum, maximum]

Parameters
 minimum is the minimum of the range of the fuzzy set maximum is the maximum of the range of the fuzzy set

## ◆ setTerms()

 virtual void fl::Aggregated::setTerms ( const std::vector< Activated > & terms )
virtual

Sets the activated terms.

Parameters
 terms is the activated terms

## ◆ terms() [1/2]

 virtual const std::vector& fl::Aggregated::terms ( ) const
virtual

Returns an immutable vector of activated terms.

Returns
an immutable vector of activated terms

## ◆ terms() [2/2]

 virtual std::vector& fl::Aggregated::terms ( )
virtual

Returns a mutable vector of activated terms.

Returns
a mutable vector of activated terms

## ◆ toString()

 virtual std::string fl::Aggregated::toString ( ) const
virtual

Returns the representation of the term in the FuzzyLite Language.

Returns
the representation of the term in FuzzyLite Language
FllExporter

Reimplemented from fl::Term.

The documentation for this class was generated from the following file:


text: Example questions: What is 18% of 2,800? 2. Video Tutorial--Another empirical formula problem--Khan Academy. Percent Composition by Mass Worksheet Experimental Percent Composition by Mass 1. Cu: _____ Br: _____ 2)NaOH 104.1g/mol Si 1x28.1=28.1 F 4x19=76 104.1g/mol 5. Example 2: Find the percent composition of copper (Cu) in CuBr2. … All Grade Worksheets Percent position And Molecular Formula from Percent Composition And Molecular Formula … Percent Composition Worksheet 1. Percent position Worksheet from Percent Composition Worksheet Answer Key With Work, source: homeschooldressage.com. Honors Chemistry Worksheet on Percentage Composition Page 2. Title: Molar Mass Percent Composition Subject: Chemistry Author: Jeff Christopherson Last modified by: UNIT5 Created … Fe%=(m Fe /m total)*100 (1.26/(1.26+0.54))*100= 70.0% O%=(m O /m total)*100 (0.54/(1.26+0.54))*100= 30.% 2. Analytical Chemistry. 200 total. C6H6O2. Total mass in the sample. Zn3(PO4) 2. PercentCompWkANS. Percent composition tells you by mass what percent of each element is present in a compound. 104.1 g 1 mol ** How many atoms of fluorine are in 15.0 g of silicon tetrafluoride? 3. … This set of … Percent Composition and Molecular Formula Worksheet Key. Ni 1x58.7= 58.7 O 2X16=32 H 2X1=2 92.7g/mol %Ni = 58.7/92.7 x 100 = 63.3% % O = 32/92.7 x 100 = 34.5% % H = 2/92.7 x 100 = 2.25 6. Displaying top 8 worksheets found for - Percentage Composition. To find the percent composition of red M&M’s in a bag of 200 M&M’s, if there are 26 red, you would use this formula. Percentage Composition. Find the percent composition (by mass) of nitric acid. ID: 1253749 Language: English School subject: Chemistry Grade/level: 10 Age: 15-18 Main content: Percent Composition Other contents: Add to my workbooks (0) Download file pdf Embed in my website or blog Add to Google Classroom What is the percentage of sodium (by mass) in sodium phosphate? by: TemplateFans. Chapter 1 Worksheet 1 and KEY Significant Figures, Scientific Notation, and Rounding 1 Determine the number of si nificantfi ures in the following values: Value 140.74 0.0041 31.00 1300 847.040 2) Round the followin 3.76411 —+3.76 3.76811 —3.77 3.76511 —3.77 11.048176—11.0 8.75510 —8.76 # of sig. 7D Percentage position WS Google Docs from Percent Composition Worksheet Answer Key With Work, source: docs.google.com. percentage composition worksheet. What is the percent composition of each element in the new compound? Examples of … A chemical compound is the combination of two or more elements. Percent position And Molecular Formula Worksheet from Percent Composition And Molecular Formula Worksheet, source:guillermotull.com. Tes Classic Free Licence. Other resources by this author. 6. Via advice on presentation composing, to developing e-book wrinkles, or figuring out what sort of content to use for a person’s structure, most of us ensure that all of our … 2. CaSO4. Some of the worksheets for this concept are Work 8 empirical formulas h o n o 4i, Formula work, , Percent composition work ii, Percent composition and molecular formula work, Percent composition and molecular formula work, Empirical formula work, Empirical formulas work 1. 2013-08-26. other Category. percentage composition worksheet. How can I re-use this? \[\% \: \text{by mass} = \frac{\text{mass of element in} \: 1 \: \text{mol}}{\text{molar mass of … Quiz & Worksheet - How to Calculate Percent Composition and Determine Empirical Formulas Quiz; Course; Try it risk-free for 30 days Instructions: Choose an answer and hit 'next'. Mass Percentage Composition, Definition, Examples, Calculations, Problems, How to Calculate Percentage Composition, water, sodium chloride, carbon dioxide. 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Percent Composition Worksheet - Solutions Find the percent compositions of all of the elements in the following compounds: 1) CuBr2 Cu: 28.4% Br: 71.6% 2) NaOH Na: 57.5% O: 40.0% H: 2.5% 3) (NH4)2S N: 41.1% H: 11.8% S: 47.1% 4) N2S2 N: 30.4% S: 69.6% Percent Composition Worksheet - Courses Percent Composition Worksheet W 342 Everett Community College Tutoring Center Student Support Services … Some of the worksheets for this concept are The percent composition work answers, Percentage composition work, Percent composition work, Percent by mass work with answers, Percent composition work, Percent composition work ii, Instructors guide percent composition, Manhasset union school district home. Steps for Solving Empirical Formula and Molecular Formula. Percent Composition Practice. Find the percent compositions of all of the elements in the following compounds: 1)CuBr. Percent Composition Worksheet Chemistry Percent Composition Worksheet Percent Composition Worksheet Percent Composition Empirical Formula Worksheet and Answers Empirical Formula Worksheet Answer Key . How many molecules are in 15.0 g of SiF4? Calculating the value of a percentage of a number worksheets with a variety of percent amounts and with various complexities of the numbers used. Video Tutorial--Empirical Formula by Ms. E. Video Tutorial--Empirical Formulas 2 by Ms. E. Video Tutorial--How to determine the empirical formula. (NH4) 2SO4. Molecular Mass And Percent Composition Worksheet Answer Key Author: hiring.gumiviet.com-2020-11-04T00:00:00+00:01 Subject: Molecular Mass And Percent Composition Worksheet Answer Key Keywords: molecular, mass, and, percent, composition, worksheet, answer, key Created Date: 11/4/2020 12:02:52 PM home / other. SiF4 = 28.1 + (4 x 19) = 104.1 g/mol. Show your work and always include units. 4. (504) Calculating the Value of a Percent (All percents from 1% to 99%) (Whole Numbers) Calculating the Value of a Percent (Select percents) (Whole Numbers) Calculating … Na 2S: 2(23.0) + 32.1 = 78.1 g/mol 9H 2O: 9(18.0) = 162.0 g/mol Na 2S•9H 2O (total): 240.1 g/mol %H 2O = 162.0/240.1*100 = 67.5% 2. N20 NH4N03 4. Calculate the percent composition of carbon for the following compounds. C3H3O mass = 55 g/mole . So, in the M&M’s problem: 26 red x 100 = 13%. 0 Comments. # of sig. Chemistry; Chemistry / Compounds and mixtures; 14-16; View more. Created: Apr 11, 2013. Thus, it helps in chemical analysis of the given compound. Calculate the percent composition for each element in C6Hp06 Glucose. 7. 15.0 g x 1 mol x 6.02 x 1023 molecules = 8.67 x 1022 molecules. Ca3P2. 8. 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Percent Composition Worksheet. Info. Video tutorial from Khan Academy--Empirical & Molecular Formulas. Students will solve word problems involving percentages. Hi, In this gallery we will show … 8 Images of Percent Composition Worksheet Answer Key. 5. What percentage of water found in Na 2S•9H 2O? They will solve word problems involving percentage increments and decrements. 1. Gallery Type. Calcium chloride can exist as the ANHYDROUS compound CaCl2 or as three different hydrated forms which are mono, di, and hexahydrates. What’s the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen? Percent Composition , Percent yield worksheet & 4 Percent Yield with answer keycx""sc from Percent Composition Worksheet Answers While we talk about Percent Composition Empirical Formula Worksheet, below we can see several variation of photos to give you more ideas. A compound is a pure substance made up of different elements. 8. KCl. 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Calculating the Value of a Percentage of a Number. 25th Aug 2019 @ 10 min read. 100 Molecular Formula Worksheet Answers from Percent Composition And Molecular Formula Worksheet, source:rtvcity.com. … Updated: Apr 16, 2013. pub, 136 KB. For the hydrate sodium sulfate decahydrate, calculate the following: (a) the percent of sodium (by mass) in the hydrate (14.3%) (b) percent of TOTAL oxygen … Students may require extra paper to write and solve their equations. mole conversion worksheet – streamcleanfo from Percent Composition Worksheet Answers, source: streamclean.info. Calculate the molar mass for the following compounds SiF4 KOH 2. Calculate the molar masses and percentage composition of each of the following compounds. _____ 4 F atoms per molecule of SiF4, therefore 4 x 8.67 x 1022 = 3.47 x 1023. CaC03 CH4 3. So, it consists of atoms of different elements. A 5.0 g sample of a hydrate of BaCl 2 was heated, and only 4.3 g of the anhydrous salt remained. Percent Composition Worksheet - Solutions Find the percent compositions of all of the elements in the following compounds: 1) CuBr2 Cu: 28.4% Br: 71.6% 2) NaOH Na: 57.5% O: 40.0% H: 2.5% 3) (NH4)2S N: 41.1% H: 11.8% S: 47.1% 4) N2S2 N: 30.4% S: 69.6%. The percentage composition of a given element is expressed using the following formula, $$\%C_{E}=\frac{g^{E}}{g^{T}}\times 100$$ Find the percent composition (by mass) of nickel (II) hydroxide. Mg(NO2) 2. % Ni = _____ Simply because we want to give solutions a single authentic and also dependable reference, most of us provide helpful information about many subjects and also topics. Kindergarten Maths Worksheets Printable . COMPOSITION OF HYDRATES - KEY 1. 5. figures values to 3 si Value 14 3.70 X 10 1.05 x 101 7.0400 x 10 2495 ificant figures. Percent Composition Worksheet by using Beneficial Themes. 3 additional worksheets are provided at the basic and intermediate skills levels. What’s the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen? HNO3 H 1x1=1 N 1x14=14 O 3x16=48 63g/mol % N = 14/63 x 100 … The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound. Empirical Formulas Percent Composition Answer Sheet - Displaying top 8 worksheets found for this concept.. molecular and empirical formula worksheet, empirical formula worksheet answer key and empirical formula worksheet answer key are three of main things we will show you based on the post title. Categories & Ages. The equation for percent composition is (mass of element/molecular mass) x … Report a problem. (42.1%) 3. About this resource. Continue with more related ideas as follows lewis dot structure … Is 110 grams/mole, what ’ s the Empirical Formula of a molecule 18.7... Chemical analysis of the compound and multiplied by \ ( 100\ % )! 1.26 g of SiF4, 2013. pub, 136 KB ( II ) hydroxide for - Composition... 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text:
Model predictive control reframes the task of following a vehicle into an optimization problem. The solution to this optimization is a time ordered set of optimal actuator inpus for steering $\delta$ and acceleration $a$, which can be positive or negative. When these inputs are integrated using an initial state and the underlying vehicle model, the result is an optimal trajectory. A trajectory is a time ordered set of vehicle states. To find this optimal trajectory, a cost function is utilized, which also accounts for the vehicle constraints. The problem of following a reference trajectory can be solved by minimizing the distance to this reference. Therefore, the actuator inputs are optimized during each step in time in order to minimize the cost of the predicted trajectory. Once the optimal trajectory is found, only the first set of actuation commands is applied to the vehicle. The rest of the optimized trajectory is ignored because the sourrounding environment could have been changed already and the underlying model is only approximate. In the next time step the process, to calculate a new optimal trajectory over a finite horizon is repeated with the new initial state. That is why this method is also known as receding horizon control.

## Cost function

To design a cost function we think about the reference which provides the desired position an orientation of the vehicle. Provided that we have a given reference path. A high cost will occur when the vehicle is far away from the center lane and has a different heading. Therefore, the cross track error (cte) and the orientation error are used in the cost function:

One problem with this cost function is that the vehicle might just stop to avoid increasing costs. The mentioned cost function also does not incorporate the vehicle constraints, which can lead to drastically changing actuator inputs and may damage the actuators.

## Avoid Stopping

To deal with stopping along the way, a desired reference velocity can help to overcome this issue. The cost function will penalize the vehilce for not maintaining that reference velocity.

Another option is to measure the eclidean distance between the vehicle’s current and the goal position and then add that distance to the cost.

To account for the mentioned actuator limits the cost function can also be used to incorporate the inputs. This can be used to penalize the magnitude of the inputs as well as the change rates. If we want to change lanes we would have a large cross-track error, due to an offset to the reference. However, we want to penalize turning the steering wheel sharply, which will yield a smoother lane change.

The control input magnitude can be added to the cost in the following way:

To capture the change rate of the control input in order to add some termporal smoothness, we add the following to the cost function:

This term captures the difference between the next control input and the current one. This reduces the change rate and gives us further control over the inputs.

## Hyperparameters

Model predictive control allows to choose the number of optimization steps $N$. Another tuning parameter is the sampling rate or the time in between the timesteps $dt$, which defines how much time elapsed between actuations. The product of these two parameters results in the prediction horizon $T$, which is the duration over which future predictions are made. In the case of a self driving vehicle the duration of this horizon should be a few seconds because the environment keeps changing. Long horizons increase the computation costs and do not provide more safety because the environment won’t stay the same in the near future.

### Horizon Length

The number of optimization steps determins the size of the control input vector $[\delta_1, a_1, \delta_2, a_2, \dots, \delta_{N-1}, a_{N-1}]$. Integrating this vector using the vehicle model results in an optimal trajectory, found by minimizing a cost function. Increasing the horizon results in higher computational costs.

### Timestep Duration

For a computers continuous trajectories need to be discretized. The step size of these discretizations is $dt$. Large values of $dt$ result in a high discretization errors. To accurately represent a continous trajectory $dt$ should be small. On the other hand, low values increase the computation and are not always necessary. A lower limit depends on the rate new measurements are received.

To find parameters for $N$ and $dt$ is to define a resonable $T$ and think about the consequences of the two parameters.

## MPC Algorithm

To setup the mpc algorithm it is necessary to define the previously mentioned time horizon $T$, which specifies the duration of the trajectory. This is done by choosing $N$ and $dt$. Since, mpc is based on a vehicle model, a dynamic or kinematic model has to be chosen. Model predictive control, tries to reduce the cross-track error to a reference path. Therefore the cte and orientation errors are also included in the model:

The constraints of an optimization problem are used to limit the actuator values.

To obtain an optimal trajectory a cost function like the following can be minimized:

After setting up the requirements for an optimization solver, the feedback loop is executed in the following way:

• Pass initial state $[x_1, y_1, \psi_1, v_1, cte_1, e\psi_1]$ to the model predictive controller
• Call the optimization solver
• Solver finds optimal control inputs \ref{inputs} that minimizes the cost function \ref{cost}

## Latency

The time it takes for a command to adjust the actuator and until the actuator reaches its final position is known as latency. This time needs to be taken into consideration while solving for the optimal actuator inputs.


text:
# Citi Bike Ridership Mini-Research: Young People Are More Likely to Use Citi Bikes on Weekends

Abstract

This study aims to find out whether or not young people ride bikes on weekends more often than that of middle-aged people. The analysis performs a hypothesis test (Z-test) to compare the ratio of the number of young people using citi bikes on weekends over weekdays to that of mid-age people. The result shows that under 5% significance level, the ratio of the number of young people biking on weekends over week days(7 days) is greater than the counterpart middle-aged people.

Keywords: Citi Bike, Hypothesis Test, Z-test , Age

## Introduction

In a fast pace modern city like New York, citi bike has become not only one of the most popular alternatives for commuting, but also a crucial component of a city’s gradually formed network system of both transportation and social activities. As a source from which quite comprehensive data sets can be acquired, citi bike is a great subject for researchers to study citizen’s behavior through patterns in its ridership. This study aims to find out whether young people ride bikes on weekends more often than that of middle-aged people, with the assumption that the bikers’ usage of citi bikes fully reflects their personal preferences– biking only for general use rather than heavily commuting purpose.

## Data Availability and Processing

All processed data used to perform the statistical test is from:

https://s3.amazonaws.com/tripdata which is documented on a monthly basis. The data wrangling process follows the idea of reproducibility and includes the following stages:

1. 1.

Enable checking and downloading data to a pointed directory each time when searching for data of a specific month, so the existed data becomes retrievable. We choose February 2015 citi bike data for our research.

2. 2.

Read the data with Pandas Dataframe; select and modify the attributes as needed(i.e. create a binary ”age group” by calculating the ages using ”birth year”). Label each row with $$18\leqslant age<40$$ and $$40\leqslant age<60$$ as young and middle-aged respectively.

3. 3.

Plot histograms to visualize the normalized fractions of young and middle-aged bikers’ average biking trip counts as well as each individual group on each day of the week.