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# Stone–Wales defects in hexagonal boron nitride as ultraviolet emitters

## Abstract

Many quantum emitters have been measured close or near the grain boundaries of the two-dimensional hexagonal boron nitride where various Stone–Wales defects appear. We show by means of first principles density functional theory calculations that the pentagon–heptagon Stone–Wales defect is an ultraviolet emitter and its optical properties closely follow the characteristics of a 4.08-eV quantum emitter, often observed in polycrystalline hexagonal boron nitride. We also show that the square–octagon Stone–Wales line defects are optically active in the ultraviolet region with varying gaps depending on their density in hexagonal boron nitride. Our results may introduce a paradigm shift in the identification of fluorescent centres in this material.

## Introduction

Two-dimensional (2D) materials are rich in novel phenomena in phyics. Hexagonal boron nitride (h-BN) is one of the key 2D materials in the field, which consists of boron–nitrogen bonds in a honeycomb lattice. Because of the strong polarization of the covalent bond between boron and nitrogen, it exhibits a large gap of about 6 eV1 unlike the semimetallic graphite or graphene. Recently, room temperature quantum emitters have been found in h-BN2,3. These findings have attracted large interest as they may serve as a basis to realize room temperature photon quantum blockade or very sensitive quantum sensors4. In particular, the first quantum emitters were found to emit in the visible2 and one in the ultraviolet (UV) region3. The origin of these emitters is unknown but assumed to come from point defects with states in the fundamental band gap of h-BN2. It can be envisioned in 2D materials that these quantum emitters may be created in a well-controlled fashion because the composition of the top layer can be directly manipulated with different techniques (e.g., ref. 5), but first, the nature of these quantum emitters should be identified, in order to further develop quantum optics measurements on these quantum emitters and their deterministic creation.

We focus our attention to the UV quantum emitter3 that has a zero-phonon line (ZPL) energy at around 4.08 eV with prominent phonon sideband peaks3,6 (see Fig. 1). We note that Huang–Rhys factor (S) of 1.3 was deduced in ref. 6 but S = 1 − 2 was estimated in ref. 7 for the PL spectrum. By integrating the area of ZPL emission and the total emission as plotted in Fig. 1c in ref. 6, its ratio, i.e. the Debye–Waller factor (DW), is ≈0.14 which corresponds to S ≈ 2. The optical lifetime of the emitter was observed at 1.15 ± 0.05 ns6. We note that this colour centre is distinct from similar UV emitters8,9 that have ZPL emission at around 4.1 eV but less pronounced phonon sideband with S ≈ 1 (see “Results and discussion” and Supplementary Note 1 for further discussion).

Here, we computed the optical properties of the pentagon–heptagon Stone–Wales defect by Kohn–Sham hybrid density functional theory (DFT) in h-BN. We find that the calculated ZPL energy, the phonon modes participating in the fluorescence spectrum (see Fig. 1), and the optical lifetime agree with the observed data on the 4.08-eV quantum emitter. The metastable triplet state has characteristic zero-field splitting due to the low symmetry of the defect where the corresponding spin states may be selectively addressed under illumination. Our study implies that abundant emitters may exist with relatively high formation energies near or inside grain boundaries of h-BN, and extended defects could be the origin of other colour centres in this material.

## Results and discussion

### Electronic structure and formation energy

We apply first principles plane-wave DFT calculations on the pentagon–heptagon rings Stone–Wales defect in monolayer h-BN (see “Methods”). The pentagon–heptagon structure in h-BN can be created by rotating one boron–nitrogen pair by 90° in the hexagonal lattice about the axis perpendicular to the h-BN sheet which automatically introduces a nitrogen antisite and a boron antisite with creating a nitrogen–nitrogen bond and a boron–boron bond, respectively [see Fig. 1a]. We find that the nitrogen-nitrogen bond creates a level at Ev + 0.44 eV whereas the boron–boron bond creates a level at Ec − 0.40 eV in the fundamental band gap, where Ev and Ec are the valence band maximum and conduction band minimum, respectively (see Fig. 2). This structural defect is isovalent with the perfect lattice, thus the lower energy level is fully occupied whereas the upper level is empty with constituting a closed shell singlet electronic configuration. This is an electrically and optically active defect as the defect may be ionized and optical transition can occur between the occupied and empty defect levels in the gap. Indeed, previous scanning electron microscope measurements associated a 2.5-eV energy gap with the pentagon–heptagon Stone–Wales defect in h-BN10, but the calculated levels do not confirm this interpretation because the electronic gap rather remains in the UV region (≈5.2 eV).

The formation energy (Eform(SWq)) and adiabatic ionization energies of the defect can be calculated by the thermodynamic equation developed by Zhang and Northrup11 with using a charge correction energy (Δq) with a charge q of the defect from refs 12,13 as follows:

$${E}^{{\rm{form}}}({{\rm{SW}}}^{q})={E}_{{\rm{tot}}}({{\rm{SW}}}^{q})-{E}_{{\rm{tot}}}({\rm{BN}})+q({E}_{{\rm{v}}}+{E}_{{\rm{F}}})+{\Delta }^{q},$$
(1)

where Etot(SWq) and Etot(BN) are the total energy of the defect in the charge state q and the perfect supercell, respectively, whereas EF is the Fermi level between Ev and Ec. The calculated formation energy of the neutral defect is 7.2 eV, which is relatively high and it is basically in accord with a previous DFT PBE study14. The calculated donor and acceptor levels in the single sheet of h-BN are resonant with Ev and Ec, respectively.

### Optical properties

In the neutral defect, the optical transition may occur between the occupied and unoccupied defect states in the gap in the UV region. In the Franck–Condon approximation, the strength of optical transition does not change as a function of the coordinate of the ions and the participation of the phonons in the optical transition, i.e. the phonon sideband in the luminescence spectrum, can be calculated as the overlap between the phonon modes in the adiabatic potential energy surface (APES) of the electronic ground state and of the electronic excited state. We apply this theory to characterize the optical transition of this defect. We further simplify this picture by assuming that the APES in the electronic ground state and excited state is very similar, and thus the corresponding phonons are equivalent, i.e., we apply Huang–Rhys approximation. In the Huang–Rhys approximation, the calculated Huang–Rhys factor (S) has a direct relation to the observed Debye–Waller factor, DW as $${\mathrm {DW}}=\exp (-S)$$.

As a consequence, the calculation of the luminescence spectrum requires several steps: (i) computing the electronic excited state with geometry optimization, (ii) computing the phonons in the electronic ground state, (iii) computing the overlap between the phonon modes in the electronic ground and excited states. We go through these steps one-by-one, with providing analysis on the results.

### Electronic excited state

Before computing the electronic excited state, it is intriguing to analyse the electronic structure by group theory. The pentagon–heptagon Stone–Wales defect has a single mirror plane symmetry, C1h. Both the lower energy and upper energy defect states show a symmetry because they are basically pz orbitals localized on the nitrogen and boron atoms around the nitrogen–nitrogen and boron–boron bonds, respectively. The closed shell singlet state is an $${\,}^{1}{A}^{\prime}$$(g) state. Excited states can be constructed by promoting an electron from the lower defect level to the upper defect level in the gap. This can produce a $${\,}^{3}{A}^{\prime}$$ triplet and a $${\,}^{1}{A}^{\prime}$$(e) singlet excited state. The $${\,}^{3}{A}^{\prime}$$ state is dark, and it is lower in energy than the $${\,}^{1}{A}^{\prime}$$(e) because of the exchange interaction of the two electrons. The many-body $${\,}^{1}{A}^{\prime}$$(e) is a correlated state that can be only described by two Slater-determinants. The spinpolarized hybrid DFT method might not be able to produce an accurate charge density of this interacting many-electron system; therefore, the ΔSCF method has larger inaccuracy than anticipated (e.g., ≈0.1 eV in energy, see ref. 15) for less severe excited states. This can affect the optimized geometry in the excited state and the total energy of the $${\,}^{1}{A}^{\prime}$$(e) state. The latter can be corrected by estimating the exchange energy by the calculated DFT total energies of the singlet and triplet excited states similarly to a previous work16. At the ground state geometry, the corrected excitation energy is 4.53 eV, which is reduced by 0.47 eV in the geometry optimization procedure. The final calculated ZPL energy is 4.06 eV, which perfectly agrees with the ZPL energy of the 4.08-eV emitter.

### Phonon density of states

The calculated phonon densities of states for the perfect and defective supercells are shown in Fig. 3a. As can be seen, the defect introduces new phonon modes at about 206 meV slightly above the phonon bands, and two other ones at 149 and 54 meV, respectively. The two highest energy phonon modes asymmetrically stretch the boron–boron and nitrogen–nitrogen bonds, as reported in Fig. 3b, c. In the optical excitation, these bonds are indeed stretched [see Fig. 1a]; therefore, they should appear in the corresponding optical excitation spectrum.

### Computed photoluminescence spectrum and optical lifetime

Next, we turn to the contribution of phonons to the optical transition. The computed luminescence spectrum is shown in Fig. 1b. The calculated phonon prominent peaks in the phonon sideband occurs exactly at those energies that were observed in the experiment, and their corresponding replica can be also well recognized. The first intense peak near the ZPL line consists of predominantly two broad phonon modes at 149 and 206 meV that were analysed above. On the other hand, the present calculational approach seems to overestimate the intensity of the phonon sideband and the calculated S ≈ 3 is larger than the observed one at ~2 for the 4.08-eV emitter6. Since the optimized geometry of the $${\,}^{1}{A}^{\prime}$$(e) state directly enters the calculation of the luminescence phonon sideband and the Huang–Rhys factor that suffers from an error due to the spin contamination it can cause a larger inaccuracy in the calculated intensity of the phonon sideband than anticipated for other defects (e.g., refs 17,18). We note that our PBE (Perdew, Burke, and Ernzerhof) calculations predict S ≈ 2 for this defect, which may be due to a fortuituous cancellation of errors in the PBE functional for this particular state that finally provides a good optimized geometry for the electronic excited state. As a consequence, the calculated phonon sideband based on the PBE geometries provide excellent agreement with the observed PL phonon sideband. The main point of these findings is that the peak positions in the phonon sideband of the 4.1-eV emitter can be well-explained by the (quasi)localized phonon modes associated by the Stone–Wales defect. Our analysis emphasizes the importance of the local or quasilocal phonon modes in understanding the PL spectrum. The prominent phonon modes either coincide or are very close to the corresponding bulk phonon modes in energy, e.g., 200-meV Raman mode of bulk h-BN, but they are rather localized on the defect with producing relatively sharp phonon replicas in the luminescence spectrum.

Finally, we address the optical lifetime of the defect which depends on the radiative and non-radiative rates. The non-radiative decay from the $${\,}^{1}{A}^{\prime}$$(e) may occur via the intersystem crossing (ISC) towards the $${\,}^{3}{A}^{\prime}$$ state. The $${\,}^{3}{A}^{\prime}$$ triplet state splits via dipolar electron spin–spin interaction where the calculated zero-field splitting falls in the GHz region (see Fig. 4). Group theory analysis implies that the ISC towards the ms = 0 state is faster than that towards ms = {x, y} states. The ISC rate can be calculated as (refs 18,19,20),

$${\Gamma }_{{\rm{nr}}}=4\pi \hslash {\lambda }_{z}^{2}F(\Delta ),$$
(2)

where λz = 0.31 GHz is the calculated spin–orbit coupling between $${\,}^{1}{A}^{\prime}$$(e) and $${\,}^{3}{A}^{\prime}$$, whereas F is the phonon overlap spectral function between $${\,}^{1}{A}^{\prime}$$(e) and $${\,}^{3}{A}^{\prime}$$ separated by Δ = 0.3 eV, which yields very small value because S ≈ 0.1 between the two states. As a consequence, Γnr 0.1 kHz. Thus, the non-radiative decay rate can be neglected in the optical lifetime, which can be then calculated from the radiative lifetime as

$$\tau =\frac{3\pi {\epsilon }_{0}\hslash {c}^{3}}{n{\omega }^{3}| \mu {| }^{2}},$$
(3)

where n = 2.5 is the refractive index of h-BN at $$\hbar$$ω = 4.08 eV ZPL energy (see ref. 21), μ is the optical transition dipole moment as derived from the calculated imaginary part of the dielectric function between the corresponding Kohn–Sham wave functions representing the ground and excited states at the ground state geometry, ϵ0 is the vacuum permittivity, c is the speed of light. We obtain τ = 2.4 ns, which broadly agrees to the observed one. We again assumed Franck–Condon theory in applying the ZPL energy in Eq. (3). We note that the radiative lifetime of another UV emitter in h-BN, the carbon-dimer defect (see below), was derived similarly in ref. 16.

We conclude from the calculated ZPL energy, characteristic phonon replica in the phonon sideband, and the optical lifetime that the 4.08-eV quantum emitter can be associated with the pentagon–heptagon Stone–Wales defect in h-BN.

### Comparison to other models

UV emission with the ZPL energy at around 4.1 eV was often found in polycrystalline h-BN; therefore, these emitters were erroneously associated with the free exciton of h-BN in the past22. Other early experimental and theoretical studies assumed23,24 that the UV emitters are associated with the presence of carbon defects because carbon is a common impurity during the growth of h-BN. Later, it was proposed based on some analogue with the donor–acceptor pair emission in GaN and related materials, that the UV emission may include two types of defects with donor and acceptor nature involving a carbon-related defect25,26. However, the observed fast optical lifetime goes strongly against this argument and rather indicates a defect with localized electronic ground and excited states as alluded in ref. 16. Recently, a carbon-dimer defect has been proposed as the origin for the 4.1-eV emitters based on first principles calculations16.

The confusion about the origin of the UV emitters may be caused by the fact that these UV emitters could have very similar ZPL energies at around 4.1 eV but they belong to different colour centres. In a recent work9, a series of UV emitters with ZPL energies at around 4.1 eV has been detected, including emitters with S ≈ 1 phonon sideband8. They prepared 13C doped h-BN too and observed the same series of UV emitters, but no isotope shift was found in the corresponding PL spectra9 (see Supplementary Note 1). Furthermore, one UV emitter with ZPL energy at 4.08 eV always appeared in the sample whereas the PL intensity of the other UV emitters with blueshift compared to the 4.08-eV emitter exhibited some correlation with the carbon content of h-BN. Our study focuses on the single UV emitter3 with ZPL emission at 4.08 eV, which has a broad prominent peak shifted from the ZPL energy between 150 and 200 meV, and it has a Huang–Rhys factor at S ≈ 2. This type of UV emitter was previously measured as ensembles in polycrystalline h-BN6.

We realized that the 4.08-eV emitter may be situated near or inside the grain boundaries of polycrystalline h-BN. In the grain boundary regions, pentagon–heptagon structural defects have been observed in the hexagonal lattice27, a very well known type of defects in carbon nanotubes or graphene sheets called Stone–Wales defect. The Stone–Wales defect contains boron and nitrogen antisites in h-BN [see Fig. 1a]; therefore, it has relatively high formation energy14,28. However, the large strain at the grain boundaries mediates the formation of Stone–Wales defects29. We note that the series of pentagon–heptagon structural defects were often observed in lines in the core of grain boundaries27. Simulations showed30 that those arrangements can decrease the strain in the h-BN sheet. The Stone–Wales defect is a pair of pentagon–heptagon defects which introduces a small strain to the lattice and is commensurate with the periodicity of the host h-BN lattice. Previous studies already confirmed10,14,31 that pentagon–heptagon Stone–Wales defects may introduce levels into the band gap but the feasible optical signature has not yet been explored. We find here that they produce the commonly observed UV emission.

### 4.08-eV optical centre as a quantum bit

We note that the calculated very slow ISC rate implies that the high-spin triplet state cannot be effectively pumped optically from the ground state. Continuous wave photo-excitation might drive the system very slowly to the triplet state but absorption of the UV photon in the triplet state would immediately photo-ionize the pentagon–heptagon Stone–Wales defect into a dark state, thus the lifetime of the triplet state under continuous UV illumination would be short. On the other hand, injection of holes and electrons into h-BN can result in trapping the exciton into the triplet state by pentagon–heptagon Stone–Wales defects which should have a long lifetime in dark. We tentatively propose that the triplet state plays a role in the ~100 ns antibunching signal in the cathodoluminescence measurements of the 4.08-eV emitter32. The relatively long lifetime of the triplet state would make possible to induce quantum bit operation with alternating magnetic fields in the microwave region when the mobile Wannier–Mott exciton is trapped into this state of the pentagon–heptagon Stone–Wales defect.

### Structural defects and quantum emitters

Beside pentagon–heptagon Stone–Wales defect, square-octagon Stone–Wales defects can occur in h-BN at the grain boundaries31. Among the different types of square-octagon Stone–Wales defects10,31,33, scanning tunneling microscope studies clearly identified those structures that have N–N and B–B bonds (see Fig. 5) which often forms line defects33. Here we only consider this type of square–octagon Stone–Wales defect.

The square-octagon Stone–Wales line defect with N–N and B–B bonds create bands: the N–N occupied bands (blue curves) appear above and near the valence band maximum of the pristine h-BN whereas the B–B empty band (green curve) appears below the conduction band minimum of the pristine h-BN (see Fig. 5). The line defect related band structure is direct at the Γ-point in which the band gap strongly varies by the distance between the line defects because the strain field induced by them in a rigid environment forced by the grain boundaries (see Fig. 5c). We note that the modelling of strain field acting on the line defects caused by the grain boundaries is oversimplified by the picture of self-interaction of line defects but the physical trends of strain can be well recognized by this model. The band gap of isolated square-octagon Stone–Wales line defect approaches 4.6 eV. These line defects could also reside relatively close to each other33, and then the band gap can reduce substantially in the order of 100 meV in a rigid environment. Thanks to the direct gap nature, the line defect is optically active and may also contribute to the ultraviolet emission in h-BN.

In summary, we have reported the results of hybrid functional calculations for the pentagon–hexagon Stone–Wales defect in h-BN. The calculated optical properties of this Stone–Wales defect, in particular, the phonon modes participating in the optical transitions, reproduce all the known features of a 4.08-eV quantum emitter. Our study shows that polycrystalline h-BN may contain such defects that have relatively high formation energy but still abundant and optically active. Our findings complement previous plausible models based on phenomenological approaches34 and recent first principles calculations16 on the origin of 4.1-eV emitters which are all based on extrinsic point defects.

Our results imply that intrinsic structural defects may affect the overall optical properties of h-BN. The extended Stone–Wales line defects introduces occupied and empty bands in the fundamental band gap of pristine h-BN which effectively reduces the band gap for proximate point defects. As a consequence, the ionization energies of these point defects are reduced that can affect their photo-stability as found for divacancy defects inside stacking faults in 4H SiC35. Moreover, some common defects in h-BN, like carbon substitutional defects28, have optical transition between the in-gap defect level and band edges. Since the effective band edges shift closer to the in-gap defect level for defects near the Stone–Wales line defect the corresponding optical signal will change with showing generally longer wavelength emission. Stone–Wales line defects also generate a strain field that can act on the optical transition of point defects, in which the optical transition occurs between in-gap defects states but sensitive to the strain field, e.g., vacancy complexes36.

As many optical centres and quantum emitters were found near or inside the grain boundaries or cracks of h-BN, this study may turn the direction of research towards exploring the magneto-optical properties of structural defects and the interaction between point defects and structural defects for identification of the observed colour centres in h-BN.

## Methods

### First principles methodology

First principles calculations are performed using Kohn–Sham spinpolarized DFT with the VASP package37. We use the screened hybrid functional of Heyd, Scuseria, and Ernzerhof (HSE)38. In this approach, the short-range exchange potential is calculated by mixing a fraction of nonlocal Hartree-Fock exchange with the generalized gradient approximation of PBE39. The screening parameter is set to 0.2 Å−1 and the mixing parameter to α = 0.32. These parameters closely reproduce the experimental band gap and structural parameters of h-BN, similarly to GaN and AlN materials40. The valence electrons are separated from the core electrons by projector augmented wave potentials41. An energy cutoff of 500 eV is used for the plane-wave basis set. We model the pentagon–heptagon Stone–Wales (SW) defect in a 162-atom h-BN supercell in which Γ-point calculation for mapping the Brillouin-zone suffices. The vacuum size is set to 25 Å. In the geometry optimization procedures, the force criterion is set to 0.001 eV/Å. The phonons in the Γ-point of the applied supercell are determined by calculating the numerical derivatives of the forces for creating the Hessian matrix in the ground state electronic configuration within PBE. The spin–orbit coupling is calculated in non-collinear approximation with the spin axis set perpendicular to the h-BN sheet within HSE. The square-octagon line defect was calculated by the same parameters but modelled in a rectangular unit cell of h-BN using 4 × 1 × 1 k-point mesh. We note that previous calculations on well-localized defect states showed that monolayer and bulk models of h-BN produce very similar ZPL energies for the same defect, also as a function of strain36. We expect here the same behaviour for the considered defects.

The magneto-optical properties are calculated with employing the ab initio toolkit developed in our group18. Briefly, the excited state is calculated by the ΔSCF method15 where the correction for the total energy of the open-shell singlet excited state is applied, similarly to a previous work16. The phonon sideband in the calculated luminescence spectrum is calculated within the Huang–Rhys approximation using our home-built implementation17 in the spirit of ref. 42. The dipolar electron spin–spin interaction part of the zero-field splitting is calculated as implemented by Martijn Marsman (see also ref. 43).

## Data availability

The data that support the findings of this study are available from the corresponding author upon reasonable request.

## Code availability

The codes that were used in this study are available upon request to the corresponding author.

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## Acknowledgements

A.G. acknowledges the support from the National Research, Development and Innovation Office in Hungary for Quantum Technology Program (Grant No. 2017-1.2.1-NKP-2017-00001), National Excellence Program (Grant No. KKP129866), the Ministry of Innovation and Technology and the National Research, Development and Innovation Office within the Quantum Information National Laboratory of Hungary, and the European Commission for the EU H2020 Asteriqs project (Grant No. 820394). V.I. acknowledges the MTA Premium Postdoctoral Research Program and the support from the Knut and Alice Wallenberg Foundation through WBSQD2 project (Grant No. 2018.0071). G.T. was supported by the János Bolyai Research Scholarship of the Hungarian Academy of Sciences and the ÚNKP-20-5 New National Excellence Program for Ministry Innovation and Technology from the source of the National Research, Development and Innovation Fund. We acknowledge the fruitful discussions from Guillaume Cassabois and technical help from Joel Davidsson and Bálint Somogyi.

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Hamdi, H., Thiering, G., Bodrog, Z. et al. Stone–Wales defects in hexagonal boron nitride as ultraviolet emitters. npj Comput Mater 6, 178 (2020). https://doi.org/10.1038/s41524-020-00451-y

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• ### Vibrational signatures for the identification of single-photon emitters in hexagonal boron nitride

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# Add & subtract polynomials: find the error

## Problem

Javier tried to subtract two polynomials. His work is shown below:
\begin{aligned} &\phantom{=}(x^4-3x^2y^2+y^4)-(-11x^4+4x^2y^2-3y^4) \\\\ &=x^4-3x^2y^2+y^4+11x^4-4x^2y^2+3y^4&\text{Step }1 \\\\ &=(1-3)x^6+(-3+1)y^6 +(11-4)x^6+(-4+3)y^6&\text{Step }2 \\\\ &=-2x^6-2y^6+7x^6-y^6&\text{Step }3 \end{aligned}
At which step did Javier make an error, if at all?

format: plain


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 Zh. Mat. Fiz. Anal. Geom.: Year: Volume: Issue: Page: Find

 Mat. Fiz. Anal. Geom., 1996, Volume 3, Number 3/4, Pages 231–260 (Mi jmag494)

Gårding domains for unitary representations of countable inductive limits of locally compact groups

A. I. Danilenko

Department of Mechanics and Mathematics, Kharkov State University, 4, Svobody Sq., 310077, Kharkov, Ukraine

Abstract: Let $G$ be the inductive limit of an increasing sequence of locally compact second countable groups $G_1\subset G_2\subset\cdots$. Given a strongly continuous unitary representation $U$ of $G$ in a separable Hilbert space $\mathcal H$, we construct an $U$-invariant, separable, nuclear, Montel $(\mathrm{DF})$-space $\mathcal F$ which is densely (topologically) embedded in $\mathcal H$ and such that the restriction of $U$ to $\mathcal F$ is a weakly continuous representation of $G$ by continuous linear operators in $\mathcal F$. Moreover, $\mathcal F$ is a domain of essential self-adjointness for the generator of each one-parameter subgroup of $G$, and all such generators keep $\mathcal F$ invariant.

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Citation: A. I. Danilenko, “Gårding domains for unitary representations of countable inductive limits of locally compact groups”, Mat. Fiz. Anal. Geom., 3:3/4 (1996), 231–260

Citation in format AMSBIB
\Bibitem{Dan96} \by A.~I.~Danilenko \paper G{\aa}rding domains for unitary representations of countable inductive limits of locally compact groups \jour Mat. Fiz. Anal. Geom. \yr 1996 \vol 3 \issue 3/4 \pages 231--260 \mathnet{http://mi.mathnet.ru/jmag494}

• http://mi.mathnet.ru/eng/jmag494
• http://mi.mathnet.ru/eng/jmag/v3/i3/p231

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This publication is cited in the following articles:
1. Beltita D., Neeb K.-H., “A Nonsmooth Continuous Unitary Representation of a Banach-Lie Group”, J. Lie Theory, 18:4 (2008), 933–936
2. Neeb K.-H., “On Differentiable Vectors for Representations of Infinite Dimensional Lie Groups”, J. Funct. Anal., 259:11 (2010), 2814–2855
3. Dawson M., Olafsson G., “Conical Representations For Direct Limits of Symmetric Spaces”, Math. Z., 286:3-4 (2017), 1375–1419


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# Crystal structure of NaCl

Chemistry Level 1

In mineralogy and crystallography, a crystal structure is a unique arrangement of atoms or molecules in a crystalline liquid or solid. A crystal structure describes a highly ordered structure, occurring due to the intrinsic nature of molecules to form symmetric patterns. A crystal structure can be thought of as an infinitely repeating array of $$3$$D 'boxes', known as unit-cells. The above shows the crystal structure of sodium chloride (table salt). How many particles are there in an unit cell of sodium chloride?

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text:
Double Doors Regional P1 - Tudor Gets a Goat

View as PDF

Points: 3
Time limit: 1.0s
Memory limit: 64M

Problem type

Tudor likes goats!

One day, Tudor finds himself inside a mall, the Honty Mall. He is participating in a game show where he has the chance to either win a goat or a car. In the rewards segment of the game show, Tudor gets to select one of three rooms, each of which is blocked from view by a set of double doors. Behind one set of double doors is a car, which Tudor doesn't want. Behind each of the other two sets of double doors is a goat, which Tudor will take home and nurture. Tudor will be happy with either goat.

Because Tudor often has trouble distinguishing among sets of double doors, the rooms are conveniently numbered 1, 2, and 3. Tudor merely has to say the number of the room he wants to inspect, after which he gets to open the double doors and claim his prize.

Jason, the host of the game, really wants Tudor to get a goat. To help him in his quest, Jason chooses to open the double doors of a room that Tudor didn't select which contains a goat. However, Tudor can't actually take that goat. That would be unethical.

With this information, Tudor suddenly starts to panic. Should he switch his room selection to the other room to get the goat?

Input Specification

The input consists of exactly three lines.

The first line of the input contains a single integer, . The car is behind room number .

The second line of the input contains a single integer, . Tudor has selected room number .

The third line of the input contains a single integer, . Jason has opened room number .

Output Specification

If Tudor should switch rooms, output Switch. Otherwise, output Stay.

Sample Input

1
2
3

Sample Output

Stay

Sample Input

1
1
3

Sample Output

Switch


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Home > Standard Error > Standard Error Of Estimate Formula

# Standard Error Of Estimate Formula

## Contents

When the margin of error is small, the confidence level can low or high or anything in between. Regressions differing in accuracy of prediction. All Rights Reserved. Generated Sat, 15 Oct 2016 19:19:45 GMT by s_ac15 (squid/3.5.20) Source

Point Estimate vs. A confidence interval is a type of interval estimate, not a type of point estimate. TI-84/83 Making Predictions using Linear Regression Equation - Dauer: 4:16 gregthemathguy 13.503 Aufrufe 4:16 Correlation and Linear Regression using a TI 84 - Dauer: 6:19 Jackson Fox 107.403 Aufrufe 6:19 Finding Estimate the sample standard deviation for the given data.
3. http://onlinestatbook.com/2/regression/accuracy.html

## Standard Error Of Estimate Formula

I'm not quite sure what to do. Because we are examining two simple random samples from less than 10% of the population, each sample contains at least 30 observations, and neither distribution is strongly skewed, we can safely I found my estimate for $\sigma$ to be 0.3385 and my estimate for $\mu$ to be 1.348, using sample variance and sample mean. The system returned: (22) Invalid argument The remote host or network may be down.

An estimate of a population parameter may be expressed in two ways: Point estimate. Standard Error Of The Regression Interval Estimate Statisticians use sample statistics to estimate population parameters. Transkript Das interaktive Transkript konnte nicht geladen werden. The estimation with lower SE indicates that it has more precise measurement.

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## Standard Error Of The Regression

Figure 1. Confidence Intervals Statisticians use a confidence interval to express the precision and uncertainty associated with a particular sampling method. Standard Error Of Estimate Formula Barron's AP Statistics with CD-ROM (Barron's AP Statistics (W/CD))Martin Sternstein Ph.D.List Price: $29.99Buy Used:$0.01Buy New: $3.50Texas Instruments TI-Nspire CX Graphing CalculatorList Price:$165.00Buy Used: $84.99Buy New:$126.99Approved for AP Statistics Standard Error Of Estimate Excel Wird verarbeitet...

Table 1. this contact form This standard error calculator alongside provides the complete step by step calculation for the given inputs.

Example Problem:
Estimate the standard error for the sample data 78.53, 79.62, 80.25, 81.05, 83.21, How to cope with too slow Wi-Fi at hotel? Standard Error of Sample Estimates Sadly, the values of population parameters are often unknown, making it impossible to compute the standard deviation of a statistic. Standard Error Of Estimate Interpretation

For example, suppose the local newspaper conducts an election survey and reports that the independent candidate will receive 30% of the vote. Notation The following notation is helpful, when we talk about the standard deviation and the standard error. Margin of Error In a confidence interval, the range of values above and below the sample statistic is called the margin of error. have a peek here Why would a password requirement prohibit a number in the last character?

the data are not paired), we can conclude that the diﬀerence in sample means can be modeled using a normal distribution.Conditions for normality of $\bar{x}_1-\bar{x}_2$If the sample means, $\bar{x}_1$ and $\bar{x}_2$, each Linear Regression Standard Error Isn't that more expensive than an elevated system? Browse other questions tagged point-estimation or ask your own question.

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# Market price of risk on two assets

Under the assumptions of the Black--Scholes model, I read that the market price of risk of two assets $$S_1$$ and $$S_2$$ are the same, if they both follow Geometric Brownian motion driven by the same Brownian motion.

The claim is that if \begin{align*} dS_1(t)&=\mu_1S_1(t)dt+\sigma_1S_1(t)dW(t),\qquad\text{and} \\ dS_2(t)&=\mu_2S_2(t)dt+\sigma_2S_2(t)dW(t) \end{align*} then $$\frac{\mu_1-r}{\sigma_1}=\frac{\mu_2-r}{\sigma_2}$$ where $$r$$ is the risk-free rate. The 'proof' of this relies on constructing a portfolio of $$\sigma_2S_2$$ units of $$S_1$$ and $$-\sigma_1S_1$$ units of $$S_2$$ and assuming that this portfolio is self-financing, then using Ito's formula on the value of this portfolio to show that it only has a drift term. I don't believe the assumption that this portfolio is self-financing holds.

Does the claim hold, and if so is there a proof of this result?

EDIT:

Thought about this a bit more and realised it falls out of the Second Fundamental Theorem of Asset Pricing where the risk-neutral measure is unique if and only the market is arbitrage-free and complete.

Assuming that the market is arbitrage-free and complete, we can construct measures $$\mathbb{Q}_1$$ and $$\mathbb{Q}_2$$ such that $$W_1(t)=W(t)+\frac{\mu_1-r}{\sigma_1}t,\qquad W_2(t)=W(t)+\frac{\mu_2-r}{\sigma_2}t$$ are $$\mathbb{Q}_1$$ and $$\mathbb{Q}_2$$ Brownian motions respectively. Both these measures give rise to a measure such that discounted asset prices are martingales. By uniqueness, $$\mathbb{Q}_1=\mathbb{Q}_2$$ and so $$\frac{\mu_1-r}{\sigma_1}=\frac{\mu_2-r}{\sigma_2}.$$

Here is a simple solution using the equivalence of no arbitrage and the existence of a stochastic discount factor. Let the SDF be $$\Lambda(t)$$. This evolves as

$$\frac{d\Lambda(t)}{\Lambda(t)}=-rdt-\varphi(t) dW(t),$$

where we used the fact that the drift of the SDF is the risk-free rate and that there is only one source of uncertainty. The standard pricing conditions for the stocks are

$$(\mu_1-r)dt=-\frac{dS_1(t)}{S_1(t)}\frac{d\Lambda(t)}{\Lambda(t)}=\sigma_1\varphi(t)dt$$

$$(\mu_2-r)dt=-\frac{dS_2(t)}{S_2(t)}\frac{d\Lambda(t)}{\Lambda(t)}=\sigma_2\varphi(t)dt.$$

That is the market price of risk $$\varphi(t)$$ is given by

$$\varphi(t)=\frac{\mu_1-r}{\sigma_1}=\frac{\mu_2-r}{\sigma_2}$$

• If $P=\sigma_2S_1S_2 -\sigma_1 S_2S_1$, then $dP =( \sigma_2-\sigma_1)d(S_1S_2)$, which is $dP = ( \sigma_2-\sigma_1) (S_1dS_2 +S_2dS_1+dS_1dS_2$) – ir7 Jul 28 '20 at 17:22
• In your first equation you are stating the portfolio is self-financing (for bank account rate set to 0). – ir7 Jul 28 '20 at 17:41
• @ir7 Now that I think of it you seem correct. How would you correct for that? – fesman Jul 28 '20 at 19:04
• I'm struggling too.I think I did it, but with a different portfolio (see my updated answer). – ir7 Jul 28 '20 at 20:12
• I think same technique (if valid) works for the given weights too (my second update). – ir7 Jul 28 '20 at 21:37

Another way to look at it, is that we have a one-dimensional Brownian motion process driving the market but two risky assets. The market price of risk process (giving the equivalent martingale measure), $$\lambda$$, must then respect two conditions:

$$\lambda \sigma_1 =\mu_1 -r$$ $$\lambda \sigma_2 =\mu_2 -r$$

which implies

$$\frac{\mu_1-r}{\sigma_1}=\frac{\mu_2-r}{\sigma_2}.$$

Update: One other way (same strategy as in the question, but different portfolio).

For a self-financing portfolio $$(\gamma^1, \gamma^2,\beta)$$, we have:

$$P_t = \gamma^1_tS_t^1 + \gamma^2_tS_t^2 + \beta_tB_t$$

and

$$dP_t = \gamma^1_t dS_t^1 + \gamma^2_t dS_t^1 +\beta_tdB_t$$

which is the same as

$$dP_t = \gamma^1_t dS_t^1 + \gamma^2_t dS_t^1 +r(P_t - \gamma^1_tS_t^1 - \gamma^2_tS_t^2) dt$$

(used $$dB_t = rB_t dt$$ in the last step)

It turns out that $$\beta_t$$ needs to be risky, function of assets. We take:

$$\gamma_t^1 = (\sigma_1 S_t^1)^{-1}$$

$$\gamma_t^2 = (\sigma_2 S_t^2)^{-1}$$

and $$\beta$$ defined by equation:

$$d\beta_t = B_t^{-1}(\gamma_t^1 dS_t^1 + \gamma_t^2 dS_t^2 )$$

This is self-financing because:

$$dP_t = d(\gamma^1_tS_t^1 + \gamma^2_tS_t^2 + \beta_tB_t)$$ $$= d(\sigma_1^{-1} + \sigma_2^{-1} + \beta_tB_t)$$ $$= B_t d\beta_t + \beta_tdB_t$$ $$= \gamma_t^1 dS_t^1 + \gamma_t^2 dS_t^2 + \beta_tdB_t.$$

(we used the fact that quadratic covariation between $$\beta_t$$ and $$B_t$$ is $$0$$)

Finally, some straightforward calculations take us now to:

$$dP_t= \gamma^1_t dS_t^1 + \gamma^2_t dS_t^1 +r(P_t - \gamma^1_tS_t^1 - \gamma^2_tS_t^2) dt$$

$$= \left(rP_t + \frac{\mu_1-r}{\sigma_1} -\frac{\mu_2-r}{\sigma_2} \right)dt$$

Update 2: For the weights in the question, we can choose $$\beta$$ such that

$$d \beta = - B^{-1}(\sigma_2 S^1 dS^2 - \sigma_1 S^2 dS^1 + (\sigma_2 -\sigma_1)dS^1dS^2)$$

For $$P = \sigma_2 S^2S^1 - \sigma_1 S^1S^2 + \beta B$$

we then have:

$$dP = (\sigma_2 -\sigma_1)d(S^1S^2) + Bd\beta + \beta dB$$ $$= (\sigma_2 -\sigma_1)(S^1dS^2 + S^2dS^1 + dS^1dS^2) + Bd\beta + \beta dB$$ $$= \sigma_2S_2 dS^1 -\sigma_1 S^1 dS^2 + \beta dB$$

So, the final portfolio dynamics is:

$$dP= \sigma_2 S^2dS^1 - \sigma_1 S^1dS^2 +r(P_t - \sigma_2 S^2S^1 + \sigma_1 S^1S^2 ) dt$$

$$= \left(rP + \sigma_2(\mu_1-r)S^1S^2 - \sigma_1(\mu_2-r)S^1S^2\right) dt$$

• Nice argument adding the bond to make the portfolio self-financing – StackG Jul 31 '20 at 8:18
• @StackG Thank you. – ir7 Jul 31 '20 at 14:13


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# Optimized portfolios with variational quantum algorithms

June 17, 2020

Blog post written by: Oscar Javier Hernandez

###### Quantum computers have the potential to show large speed-ups over their classical counterparts and disrupt industries such as finance in the future. Following the work of M. Hodson et. al [arxiv:1911.05296] in this blog post we implement variational quantum algorithms to solve the financial portfolio optimization problem. Investment constraints are imposed in two different ways, the first method uses a penalization term while the second method constrains the feasible space of solutions.  This work was carried out with assistance from Guoming Wang from Zapata Computing and a mentor from the quantum open source foundation (QOSF). The accompanying Jupyter notebook and source code are available as an open-source project for anyone else to use.

link to the Quantum open source foundation

### Introduction

Given the set of all possible stocks, what is the best way to choose the subset that optimizes financial objectives such as maximizing returns or minimizing the investment risks for an asset manager or individual? This problem is known as the portfolio-optimization problem and is rooted in the work of Harry Markowitz from 1952. In the original formulation, the optimal portfolio allocation is the one that produces the maximum ratio between the portfolio return and its risk. This problem can be stated as an optimization problem where the objective is to minimize the quadratic objective function

$$C({\bf w})= \lambda{\bf w}^T \Sigma {\bf w} -(1-\lambda ){\bf u} \cdot {\bf w},$$

where $\Sigma$ is the $N \times N$ covariance matrix of the $N$-selected stocks , ${\bf w}$ represents the percentage % of invested capital in a given asset (with $\sum_i w_i =1$), ${\bf u}$ is the vector representing the expected returns of the stocks and $0 \leq \lambda \leq 1$ is the asset manager control parameter. The first term ${\bf w}^T \Sigma {\bf w}$ in this expression represents the volatility of the chosen portfolio, while the second term ${\bf u} \cdot {\bf w}$ represents the expected returns. Therefore, a low value of $\lambda$ favors the expected returns of the portfolio while a value approaching $1$ favors low volatility. This simple cost function can be extended to include other effects, such as rebalancing costs and market impact.

In this post, we will focus on solving the discrete version of this problem. In this formulation, instead of the continuous holding vector ${\bf w}$ which represent percentage allocations, we have the discrete holding vector ${\bf z}$ that consists of integer entries where $+1$ is a hold position, $-1$ a short position and $0$ indicates no position. For a set of $N$ stocks, there are $3^N$ feasible portfolio holding combinations to compute and therefore solving the discrete portfolio optimization problem becomes exponentially more difficult with larger $N$. Furthermore, in the discrete version of this problem, we are interested in imposing the constraint

$$\sum_i z_i =D,$$

that represents our desire to have exactly $D$ equally valued positions to be invested in each stock.

Due to the exponentially increasing problem size, this problem can be very slow to solve on a classical computer when $N$ becomes large. For an $N$ -qubit Quantum computer, $2^N$-possible bitstrings can be stored as a quantum state, allowing quantum computers the potential to sample these exponentially growing model spaces and determine the optimal values more efficiently than a classical computer. Currently available quantum hardware exists as noisy intermediate-scale quantum computers (NISQ) available through cloud-based APIS. Quantum algorithms such as the Quantum approximate optimization algorithm (QAOA) are available that can be executed on these NISQ devices as well as on simulators.

The improved speed and solution quality that can be provided by quantum algorithms is very attractive for solving mathematical finance problems. In this post we implement the algorithm presented in arxiv:1911.05296 for solving the discrete portfolio optimization with investment constraints on a universal gate-based quantum computer using two different variational quantum algorithms to handle constraints, run it on a simulator and compare their results.

In the next section we will provide an overview of the two specific algorithms that were used to solve this problem.

### The Quantum approximate optimization algorithm and the Quantum alternating operator ansatz

The quantum approximate optimization algorithm (QAOA) is a variational algorithm developed to approximately solve difficult combinatorial optimization problems. These problems are computationally difficult due to the large model space sizes involved. It has been demonstrated that with high probability, the QAOA algorithm can determine good approximate solutions to these class of problems. In fact, for some very specific tasks such as MAXCUT, there exist theoretical guarantees about the performance of the QAOA. However, this is still not the case for all problems that the algorithm has been applied to and is under active investigation.

The key ingredients of the QAOA algorithm is the reference Hamiltonian $H_0$ whose ground state $|\psi_0\rangle$ is known and a cost function $H_C$ encoded in the computational basis that we want to minimize. Once these ingredients are chosen the variational quantum state $|\psi (\vec{\gamma}, \vec{\beta}) \rangle$ is constructed where

$$|\psi (\vec{\gamma}, \vec{\beta}) \rangle = \prod_{k=1}^{p} U(B,\beta_k)U(C,\gamma_k)|\psi_0\rangle,$$
with the unitary operators given by
$$U(B,\beta_k) = \text{Exp}\left( -i \beta_k H_0 \right),$$

and
$$U(C,\gamma_k) = \text{Exp}\left( -i \gamma_k H_C \right),$$

for arbitrary angles $\beta_k$ and $\gamma_k$ for $k=1,\cdots, p$ that are the components of the vectors $\vec{\gamma},\vec{\beta}$. The parameter $p$ in the above expression is the depth of the desired circuit. The angle vectors $\vec{\gamma}$, ${\vec{\beta}}$ must be determined classically from the minimization of the expectation value of the cost Hamiltonian that encodes our cost function
$$E = \langle\psi \left(\vec{\gamma}, \vec{\beta}\right) | H_C | \psi \left(\vec{\gamma}, \vec{\beta}\right) \rangle.$$

In the limit that the circuit depth $p$ goes to infinity, the expectation value $E$ will approach the optimal value of the cost function $H_c$.

$$\lim_{p \rightarrow \infty } E \rightarrow \text{min}_{{\bf z}}\lbrace H_C({\bf z})\rbrace.$$

One limitation in this approach is that hard constraints, such as our desired condition where $\sum z_i =D$ is not straightforward to enforce. Therefore, following the work of M. Hodson et. al [arxiv:1911.05296] we consider two methods to introduce the constraints. The first and more common approach is to introduce a quadratic penalization function $P_A$ to the cost function where

$$P_A = A\left(\sum_i z_i-D\right)^2,$$

and $A$ is the penalty scaling parameter that serves as an additional hyperparameter for the problem. With this term, the cost function that we are to minimize is

$$C(Z) = C_0(Z)+ A\left(\sum_i z_i-D\right)^2,$$

where $C_0$ is the original cost function for the problem. Due to the quadratic nature of $P_A$, it is clear that any deviation from the constraint $\sum_i z_i =D$ will increase the energy of the new cost function. Therefore the minimum of $C(Z)$ must have $P_A=0$. The addition of this penalty function with a careful choice of $A$ to the cost function helps push the search space of the QAOA algorithm to the subset where the constraints are satisfied. The QAOA algorithm with soft constraints will be denoted as QAOA-soft.

An alternative way to include constraints into the quantum circuit is to apply the Quantum Alternating Operator Ansatz method  (QAOA-hard for our purposes) which encodes the constraints into a carefully chosen initial state $|\psi_0 \rangle$ along with a special unitary operator $U(B,\beta_k)$ (known as the mixer) that preserves quantum states that satisfy the desired constraints. More details about the specific forms used for the portfolio problem are found in the notebook accompanying the post.

In this work, we used the QAOA-soft along with the QAOA-hard to solve the portfolio optimization problem. The notebook containing the results can be found in the following repositiory.

### The $N=4$ stock portfolio optimization problem

To illustrate how these quantum algorithms solve the portfolio optimization problem. In this section we solve the case where the number of stocks in our portfolio is $N=4$ and we want the total amount of investments to be $D=2$. In this case the number of qubits required is $8$. Following the work of arxiv:1911.05296 we use the expected returns for the share prices of the Australian ASX.20 market in 2017 using data for 252 trading days. The subset of stock chosen for our portfolio where AMP, ANZ, BHP and BXB. The expected returns of these four stocks are given by

$${\bf u}\cdot 10^4 = [4.01, 0.61, 9.16,-6.19],$$

while the covariance matrix between these stocks that was used was

$$\Sigma\cdot 10^6 = \begin{bmatrix} 99.8 & 42.5 & 37.2 & 40.3 \\ 42.5 &100.5 & 41.1 & 15.2 \\ 37.2 & 41.1 &181.3 &17.9 \\ 40.3 & 15.2 & 17.9 & 253.1 \end{bmatrix}.$$

With such few stocks, it is possible to determine the optimal portfolio that minimizes the cost function using a brute force search. In this case the optimal solution is

$$\text{Brute-Force optimal state: } [1, 0, 1, 0]$$

while the value of this optimal cost function is
$$\text{Brute-Force optimal energy: } 0.00018825 \ .$$

With the known solution in hand we will construct two different quantum circuits to carry out the optimization. The QAOA with soft constraints (QAOA-soft) and the QAOA which imposes hard constraints (QAOA-hard). The asset manager control parameter used was $\lambda=0.9$ (favoring risk reduction) and the penalty scaling parameter for the QAOA-soft is $A=0.03$ as was used in the reference article. For simplicity, we will set the transaction costs to zero although the associated notebook can also handle non-zero transaction costs.

#### QAOA with soft constraints

A small segment of the 8-qubit QAOA-soft circuit is shown below for illustration. The full diagram is available on the accompanying notebook.

Although the entire circuit cannot be shown very concisely in such an image. The above figure begins with all of the qubits in the $|0\rangle$ state which is subsequently operated upon by the Hadamard gate $H$ that places each qubit into an eigenstate of the $X$ operator that serves as the reference Hamiltonian $H_0$ that we discussed earlier. After all of the qubits are in the initial reference state $|\psi_0\rangle$ the next operators encode the cost function as a series of rotation operators acting on the reference state. We will leave the discussion of the details of the these operators for the accompanying notebook.

Next, having constructed the circuit we conduct a grid search of the parameters $\beta$ and $\gamma$ for $p=1$. The heatmap of this grid search is shown below.

The grid-search allows us to determine the optimal parameters $\beta^*$ and $\gamma^*$. Once these are found (in this case $\beta^* = 0.63 \pi$ and $\gamma^* = 1.89 \pi$), we carry out a simulation of 100 trials of this quantum circuit with these parameters and obtain the following histogram of results.

As demonstrated in the above figure, the optimal feasible solution ${\bf x}^*=[1,0,1,0]$ exists as a possible result. The probability of the optimal result, which we denote at $p({\bf x}^*)$ is $7\%$ and there are many other possible results. There are also many solutions that do not satisfy our $D=2$ constraint. However, the advantage provided by the QAOA method is that by applying this quantum circuit to our problem we have restricted the space of possible solutions to one that is smaller containing the optimal solution ${\bf x}^*$ and that is easier to search than a brute force approach.

For circuits with depth $p \geq 2$ it is computationally expensive to carry out a grid search of the $\vec{\beta}$ and $\vec{\gamma}$ parameters, therefore we apply optimization algorithms to minimize the energy expectation value $E = \langle\psi \left(\vec{\gamma}, \vec{\beta}\right) | H_C | \psi \left(\vec{\gamma}, \vec{\beta}\right) \rangle$. In the notebook, we used several approaches such as gradient-descent, the nelder-mead optimizer built into the scipy package and the cross-entropy method. It was found in our experiments that the cross-entropy method was better at finding the optimal parameters for $\vec{\beta}$ and $\vec{\gamma}$ for these circuits. The results of these numerical experiments are plotted in the figure below.

In the above figure, we have used the cross-entropy optimization for determining the optimal angles for the circuit. The expectation value $\langle \psi | C(Z)|\psi \rangle$ is the expectation value of the cost function which decreases sharply when $p \geq 2$. The expectation value $\langle x^*|C(Z)|x^*\rangle$ is the value of the cost function for the optimal state ${\bf x}^*$ that was determined by the quantum circuit. This value also decreases sharply and does not improve for circuits with depth $p \geq 2$ in our quantum circuits. Finally, the probability of obtaining this optimal state $x^*$ is denoted by $p(x^*)$ and remains in the range $0.05 \leq p(x^*) \leq 0.12$ for $p=1,2,3,4$. Based on our numerical experiments, the QAOA-soft circuit where $p=2$ has the largest probability of finding the optimal state with $p(x^*)=0.12$.

Next we look at the approximation ratio $\alpha$, which is the ratio between the true optimal energy and the lowest energy solution obtained from the QAOA-soft

$$\alpha = \frac{E_{\rm QAOA}}{E_{\rm minimum}} \leq 1,$$

where $\alpha=1$ indicates that the optimal solution was found. In the plot below, we see the approximation ratio as a function of circuit depth for three different optimization methods.

In the above figure we see observe that in our numerical experiments, the gradient decent method oscillated the most, followed by the nelder-mead optimizer. In contrast, the cross-entropy optimizer had $\alpha=1$ for all quantum circuits $p \geq 2$. Therefore, we find that in our numerical experiments, the for this portfolio optimization problem the cross-entropy method consistently found the optimal solution.

#### QAOA with hard constraints

Next, we construct the QAOA circuit that imposes the hard-constraints directly into the quantum circuit. The following figure shows the first few operations on this circuit which demonstrate some differences to the QAOA-soft form of this circuit.

Here we see that in contrast to the QAOA-soft circuit, the initial state preparation of $|\psi_0 \rangle$ does not apply the Hamard gate $H$ to all qubits. In this case, the combination of $H$, $X$ and controlled-$X$ gates initializes a state $|\psi_0 \rangle$ that satisfies the constraint $D=2$. However, this initial state does not minimize the cost function in this initial part of the circuit.

As before, we construct the $p=1$ circuit and conduct a grid search of the parameters $\beta$ and $\gamma$. The heatmap of this grid search is shown below.

There appear to several optimal solutions with the same energy in this case. To illustrate what a random solution in this space looks like, we choose $\beta= 0.5 \pi$ and $\gamma = 0.5 \pi$ and simulate the $p=1$ QAOA-hard circuit 100 times. The figure below shows the solutions obtained from these measurements.

We observe that indeed the QAOA-hard algorithm is able to constrain the feasible subspace of the solutions to the states where $D=2$. As for the QAOA-soft circuit, the convergence plot for the cross-entropy method is below

We observe that the probabilities of the optimal solutions $p(x^*)=1$ using this QAOA-hard algorithm as opposed to the QAOA-soft except where $p=3,4$ where the optimization may not have converged. In contrast, for the QAOA-soft method the highest probability reached was only $p(x^*)=0.12$.  We then determine the approximation ratio as a function of circuit depth below.  For both methods the approximation ratio reached $\alpha=1$ at higher circuit depths indicating that the optimal solution was found.

From the above figure we can see that for the QAOA-hard circuit the approximation ratio remained in the range $0.9 \leq\alpha \leq 1.0$. This is similar to the approximation ratio determined by the QAOA-soft circuit using only the cross-entropy and nelder-mead optimization methods which had the range of approximately $0.85 \leq\alpha \leq 1.0$.

### Summary

In this post, we have introduced the concepts behind the quantum approximate optimization algorithm (QAOA) in the context of the financial portfolio optimization problem. Furthermore, we have described how it is possible to modify this algorithm to incorporate investment constraints using two different variational quantum algorithms. The first method introduces a penalty function (QAOA-soft), while the second approach uses an different formulation of the QAOA known as the Quantum alternating operator ansatz that embeds the hard constraints directly into the feasible subspace of solutions (QAOA-hard). We applied these hybrid quantum algorithms to a $N=4$ stock portfolio using data from the work of M. Hodson et. al [arxiv:1911.05296] and compared different optimization routines to determine the optimal angles for the quantum circuits. From our numerical experiments we found that the cross-entropy optimization method yielded consistently better results for determining the best approximation ratio $\alpha$ which were consistently $\alpha \geq 0.9$. Furthermore, the QAOA-hard was the algorithm that reached the highest optimal solution probabilities $p(x^*)=1$ for circuit depths $p=1,2$, while the highest optimal solution probability reached by the QAOA-soft circuit was $p(x^*) \approx 0.12$ indicating that the QAOA-hard circuit at low circuit depths is the preferable algorithm to use for the portfolio optimization problem.

### Further work

While we have carried out an interesting numerical investigation of the circuits and methods presented in M. Hodson et. al [arxiv:1911.05296], more careful analysis is required to rigorously replicate the results of that study. For example, further numerical experiments and simulations are required to estimate the distribution $P(\alpha)$ of the approximation constants, along with the properties of the optimal-solution probability $p(x^*)$ for both the QAOA-soft and QAOA-hard algorithms.

In the future, it would also be interesting to run this algorithm on existing NISQ hardware, which may require tailoring the quantum circuits to available hardware and to check the robustness of the optimization against noise.


text:
# Normalizing boundary conditions

1. Nov 21, 2014

### garcijon

Hi All,

This is my first post on these forums. I am not looking for a solution to this problem but more interested in seeing if someone can point me to a resource that can explain the following. Thanks in advance for any help.

I'm trying to solve a pde which gives a temperature profile.

We end up changing over to spherical coordinates where our boundary conditions are not completely clear although our initial condition is.

The professor writes something like this. basically it normalizes the boundary conditions to (0,1). I can't wrap my head around this and googling for a while hasn't given me any results that clarify this.

2. Nov 26, 2014

### Greg Bernhardt

This is a straightforward linear rescaling of the form $$f : [a,b] \to [0,1] : x \mapsto \frac{x - a}{b - a}.$$ Observe that $f$ is strictly increasing with $f(a) = 0$ and $f(b) = 1$.
If what you are struggling with is "why does he bother to do this?" then the concept you are looking for is nondimensionalization. Also it's easier to work with $G(t = 0) = G(r \to \infty) = 0$ rather than $G(t = 0) = rT_0$ and $G(r \to \infty) = \infty$.


text:
By Sunil Bhardwaj

3768 Views

It is observed that when the projectile is moving with the energy of 20MeV then the mass of moving projectile (m) and the mass in the rest $$({ m }_{ 0 })$$ are equal. $$\therefore m = { m }_{ 0 }\qquad \qquad \qquad ...(below 20 MeV)$$ But in 1946, Lorenz found that when the projectile is moving with energy more than 20MeV, in that case the mass of the projectile is given by, $$m = \frac { { m }_{ 0 } }{ \sqrt { 1-\frac { { v }^{ 2 } }{ { c }^{ 2 } } } } \qquad \qquad \qquad ...(1)$$ where v = Velocity of projectile and c = Velocity of light.
As v < c $$\therefore \frac { v }{ c } < 1$$ or $$\frac { { v }^{ 2 } }{ { c }^{ 2 } } < 1$$ $$\therefore \sqrt { 1-\frac { { v }^{ 2 } }{ { c }^{ 2 } } } < 1\qquad \qquad \qquad ...(2)$$ So from eq (1) and (2) we can say,$$\boxed { m > { m }_{ 0 } } .$$For a particle moving with 50% of the velocity of light, the mass m is 1.155mo. and when the particle is moving with 99% of the velocity of light then m = 7mo. It means as the velocity of projectile increases in the cyclotron, the mass increases and time taken by projectile is not constant. As a result the particle are not reaching at the constant time in the gap. It means the particle arrive at the gap either a little before or little after the instant at which the potential of the dees changes. The particles therefore acquire little or no additional energy. The Cyctotron therefore cannot impart energy to the projectile beyond a certain limit. In 1945, V.Veksler in the Russia and Monillan in US showed that relativistic increase in the mass can be compensated by adjusting the magnetic field. This is called the principle of phasestability. Such type of cyclotron in which this adjustment is made are known as Syachro Cyctotron In 1945, a cyclotron based on this principle was built at Berkeley, California. It has magnets with pole faces 184 inches in diameter and weighs about 4000 tons. Protons with 730 MeV energy have been produced with this synchro-cyctotron. At Batavia, Illinois in the United States is a synchrocyctotron which can impart 400 Billion electron volt (BeV).

MCQ on Nuclear Chemistry from Physical Chemistry
###### Prof. Gianfranco Coletti

Shared publicly - 2019-08-23 00:00:00

Don’t want your columns to simply stack in some grid tiers? Use a combination of different classes for each tier as needed. See the example below for a better idea of how it all works.

###### Prof. Maheshwar Sharon

Shared publicly - 2019-08-24 00:00:00

For grids that are the same from the smallest of devices to the largest, use the .col and .col-* classes. Specify a numbered class when you need a particularly sized column; otherwise, feel free to stick to

###### sunil

Shared publicly - 2023-02-28 11:09:52

this is

###### ss

Shared publicly - 2023-02-28 10:48:10

gsgsg

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text:
# Heat Flow – PDE problem

1. Apr 15, 2007

### mrjeffy321

I have got a heat flow partial differential equation problem that is giving me a little problem due to the direction the temperature is changing.

I have a bar (which lies along the X axis) which is initially at a uniform temperature which (for simplicity sake) we will call zero degrees.
At time, t, = 0 and onward one side of the bar is heated so that it is kept at a constant temperature of, say, 100 degrees.
I am asked to find the temperature of the bar as a function of the position, x, along the bar, and time t.

Setting up the heat flow equation, I have,
Del square u = 1/alpha * the first partial derivative of u with respect to time.
I can rearrange this and solve for a product solution of u(x, t) as X(x) * T(t). then separate out thr variables into two ordinary differential equations for X(x) and T(t) and get,
X’’ + k^2 * X = 0,
And,
T’ + (k*alpha)^2 * T = 0

The X equation is going to give me some sine and/or cosine solution, and the T equation should give me a exponential solution.

Solving for T,
T = A * e^-(K * alpha)^2 * t
Where A is just some constant.
I am ignoring the + exponent solution since this is non-physical, temperature cannot go to infinite as t goes to infinity.

But what worries me is that the equation does not satisfy the initial boundary conditions.

At t = 0, then entire rod should be at zero. And at t = infinite, the temperature of the rod should be some constant function of x (steady state solution when the two ends of the rod are held at different temperatures).
But my equation says that the temperature drops to zero as t goes to infinite and has some maximum value (A) when t = 0.

Clearly I need to modify my solution for T(t) so get it to show an increasing temperature which will approach some non-zero value.

2. Apr 16, 2007

### HallsofIvy

Staff Emeritus
Do you mean that one end, say x= 0, is kept at 0 degrees and the other at 100 degrees? And what is the length of the bar? I'll call it "L".
What happens is that your initial temperature distribution is not continuous. Since you'll be integrating to find the temperature function, that's not a problem.

Calling the temperature $\phi(x,t)$, you are solving
$$\frac{\partial^2 \phi}{\partial x^2} = \frac{1}{\alpha^2}\frac{\partial \phi}{\partial t}$$
with boundary conditions $\phi (0,t)= 0$, $\phi(L,t)= 100$ and initial condition $\phi(x,0)= 0$ if x is not L, $\phi(L,0)= 100$.

This may not be the way you were taught to do this but I would recommend subtracting of the function 100x/L for all t, which satisfies the boundary conditions. Since
[tex]\frac{\partial^2 (100x/L)}{\partial x^2}= 0[/itex]
$u(x,t)= \phi(x,t)- 100x/L$ satisfies the same differential equation as $\phi(x,t)$, u(0,t)= u(L,t)= 0, and u(x,0)= -100x/L for x not equal to L, u(L,0)= 0. Since in integrating, the value at a single point is unimportant, you can ignore that last value. Now, it should be easy to solve the equation for X, determine the correct values for k and use that in the T equation.

3. Apr 16, 2007

### mrjeffy321

Yes.

Since the x = 0 side of the bar will always be 0°, I could eliminate the possible solution to the X(x) function involving Cosine, since cos (0) = 1. I figured out that if I said that the length of the bar, L, was only 1/4 of the period of the Sine function solution, this would keep the x = L side from being zero.

I also found the solution for T(t) as being e^-(kα)^2 * t. But the problem I was having was that as time goes to infinite, my T(t) function goes to zero. Since T(t) is multiplied by X(x) to get the temperature function u(x, t), this would make the entire function zero.
What should happen is that some linear temperature distribution should develop in the bar to make u(x, infinity) = 100 * x / L.

I don’t know about subtracting….seems like adding this value (100x/L) would work though. T(t) goes to zero as time goes to infinite, u(x, infinity) will just be = +100x/L, which does give the correct distribution. Although this would work at t = infinity, wont it interfere at other times….for example t = 0? When t = 0, T(t) = 1, so u(x, t) would be = sin (n * pi * x / (2L)) * 1 + 100x/L. Which is not true.


text:
# Basic Special Relativity Question

Discussion in 'Physics & Math' started by Fednis48, Apr 22, 2013.

1. ### OnlyMeValued Senior Member

Messages:
3,914
I too am interested, but I agree that getting into that side discussion now would be distracting. I do hope you have something new in mind, like something in the last couple of years that I have not heard about.., but let it wait a bit.

3. ### TachBannedBanned

Messages:
5,265
I fully understand that you don't have a clue how to run this experiment. Your first answer, where you acknowledged that you don't know how to run the experiment was the only honest one, all the subsequent ones where your attempts at covering up that you do not have a clue. You do not know how to do this measurement.
Let's try to help you out a little: you have a very long set of cameras strung along the $x"$ axis in $S"$ . The cameras are snapping pictures of the passing train , with the rod inside, at a very fast shutter setting. The cameras time stamp each snapshot and the snapshot snapping is synchronized in $S"$. Each camera is a one pixel motion detector, each camera can only capture only one pixel at a time: on if the rod is lined up with the camera objective and off if it isn't. What data do you collect from these cameras and what does the data tell you?

So, they are wrong. So are you. Do you still want the references that show that length contraction is physical or are you going to dig your heels in your new-found lie?

Old enough to prove that you don't know what you are talking about.

Last edited: May 7, 2013

5. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

Messages:
3,390
Sure.

Otherwise this thread (90% of which I haven't read) may be too much semantics. Since relativity is counter intuitive, experiments are worth a thousand words.

7. ### TachBannedBanned

Messages:
5,265
deleted

Last edited: May 7, 2013
8. ### TachBannedBanned

Messages:
5,265
One proof that the Lorentz contraction is physical: here and here. It explains the physics of heavy ion collisions I was talking about. I will give one proof at a time since Pete is intent on dragging this charade indefinitely, so I am certain that he'll argue about the validity of each proof. This will make it painful but any interaction with Pete is painful, especially when science disagrees with his prejudices.

9. ### OnlyMeValued Senior Member

Messages:
3,914
While both of these experiments can be classified as "physical", neither one represents direct evidence. The measurements in experiments like these are interpreted from the data. They support the theoretical basis, but they do not represent direct measurement of length contraction... So in a strict sense continuing to refer to this kind of evidence, as "physical" evidense, while technically accurate, is misleading, as it implies that length contract has been "proven".

I did not retain a copy but I remember reading an analysis of the heavy ion collision reference that did raise alternate explanations for the bunching. There is a lot going on in the process of accelerating the ions that has to be accounted for and ruled out, which may contribute to the bunching.

We now have a great deal of evidence, physical observations.., that supports the existence of dark matter, still dark matter remains theoretical.., waiting for some "direct" observation/evidense.

BTW Tach neither of the links mentions length contraction, though it has been discussed elsewhere.., in the at least heavy ion case.

10. ### TachBannedBanned

Messages:
5,265
All measurements in experiments are interpreted from the (collected) data. Did you ever run an experiment?

You are wrong on both accounts, both articles specifically mention length contraction, you need to read carefully.

11. ### OnlyMeValued Senior Member

Messages:
3,914
Yes, Lorentz Contraction is mentioned, but there is nothing that explains how that conclusion is reached... In these references.

Where the ion pancaking is concerned, it is not a measured phenomena, it is suggested by a theoretical interpretaion of how the data from the impact, works out. That is actually the best supporting "evidence", as far as I am concerned, but it only supports rather than proves the contraction, because it is the theoretical interpretation that leads to that conclusion. What they are actually saying (not in these references) is that the way the scattering "looks" after impact, is what would be expected if the ions were length contracted. The actual data does not say that length contraction is the only possibility. It says the Lorentz-contraction is an explanation...

I don't mean to suggest there is nothing that supports the theory, just that we as yet have no direct measurements that proves it.

12. ### TachBannedBanned

Messages:
5,265
Good, you changed your tune. It means that found the relevant sections but you still do not understand the explanation, which is not unexpected knowing you.

You are the same crackpot you've always been, when presented with scientific evidence, you come up with some cockamammie "explanation" of your own.

13. ### OnlyMeValued Senior Member

Messages:
3,914
Reference an autaul research paper on the "pancaking" of heavy ions, and my "incoherent babbling" will be supported by the research, in intent, if not in the wording. I believe I have even seen some reference to similar conclusions involving individual high energy proton/neutron collisions. No where I have ever, in either lay publication or research paper, seen anyone doing the research suggest that they were measuring the shape of the heavy ions or protons/neutrons directly.

Indirect measurement "supports" the theory. Direct measurement would prove the theory. We are unable to carry out experiments that would prove Lorentz contract direct at classical scales of length and velocity.., and we have no rods and clocks useful in carrying out direct measurement at the scales of subatomic particles or heavy ions... That leaves the evidence referenced as "supporting" Lorentz contraction, not as "proving" Lorentz contraction.

To the extent that your posts degrade to the point of name calling, your credibility is also degraded... Since that is where we seem to be now, it would be best to leave this discussion until the discussion of the OP has been resolved.

P.S. here are the two mentions of Lorentz contractions in the articles you referenced,

The ions appear as flat disks because of the relativistic effect called "Lorentz contraction".

and
One can see an example of a Au+Au collisionat a centre-of-mass energy of 200 GeV per nucleon pair and at an impact parameter of 5 fm as calculated in RQMD in the animation on the right. You can see the Au nucleus as a Lorentz-contracted pancake-shaped object moving in one direction (at about 99.995% of the speed of light, spheres look like pancakes, they are contracted 100 times in their direction of motion).

In the first note, "The ions appear...", not a proof of measurement...

In the second, the reference is to an animation, but the lead in describing the experimental context, references the impact not a measurement of the ions, and also includes, "You can see the Au nucleus as a Lorentz-contracted pancake-shaped object.., spheres look like pancakes...".

In one the ions "appear" to be pancakes and the other they "look" like pancakes.., in an animation of the impact.

Last edited: May 7, 2013
14. ### TachBannedBanned

Messages:
5,265
Knowing you , it doesn't really matter what you have "seen" and what you "haven't seen". BTW: what is an "autaul"?

1. You cannot "prove a theory". Ever.
2. For 100 points: what can you actually do?
3. For additional 20 points, what is "indirect measurement"? How do you perform it?

15. ### OnlyMeValued Senior Member

Messages:
3,914
Since all you have left is spelling and semantic errors, in statements which are not proof read... There really is no discussion here.

16. ### TachBannedBanned

Messages:
5,265
No, items 1 and 3 point at your crackpot statements, as does this one:

17. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
I'm glad we agree that it can be done. I knew you'd figure it out.
But as I said, we actually have detectors not just on the x''-axis, but covering the x''-y'' plane in the area of interest.
As I said before, the data collected is the $(t'',x'',y'')$ coordinates on the rod worldline, and possibily the rod element's k-values as well.
It tells us this:

Are say saying length contraction is not frame dependent?
Or are you disagreeing with the semantics of "physical effect"?

18. ### TachBannedBanned

Messages:
5,265
Well, since you were clearly unable to figure it out by yourself, somebody needed to help you. So I helped you. <shrug>

OK, go ahead, let's see if you can finish the setup, now that you have received appropriate tutoring.

So, how would you do that?

What gives you this bright idea?

I have provided you with the proof that, contrary to your fringe misconceptions, length contraction is a physical effect. have you read the references? I gave you two.

19. ### UndefinedBannedBanned

Messages:
1,695
You have the nerve to call other people "cranks" while you come up with rubbish "proofs" of "length contraction" like this reference?

How many times have you told a "crank" that animations and simulations only show the assumptions and interpretations that went into them, and that they are not proof of anything except your theory interpretations of the inputs and not the actual physical phenomena itself? No one "sees" anything. No one "proofs" anything by such simulation and animation computer code "illustrations" of their theoretical interpretations rather than actual known facts of what is happening physically.

You are a sham "expert" who depends on others to do the work while you pretend to "know" and "correct" using others work references which you yourself do not even understand in any way that is relevant to the issue. You are the worst sort of "crank", one that pretends to "authority" he doesn't have by any measure of competency in physics understandings. You are a troll and a crank that science doesn't need in any way shape or form. Game playing nincompoop and stupid troll to boot. Where are the moderators? Don't they care that you lie and muddy the threads like this, pretending to knowledge and authority you don't have? The site is getting a bad reputation I think from your silly games. Moderators should do something about it by telling you to put up "proof" of your "correct solution" or shut up and get lost.

20. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
Previously, you said it can't be done.

How would we collect the k-values of the rod elements? Write them on the rod so the cameras can see them. Embed RFID chips. Paint the rod as a rainbow along its length. Like I said before, the details aren't important, unless you're actually planning to set up a relativistic train.

Semantics. 'Physical effect' is an ambiguous term. You mean something different to what Gron and Johanessen mean.
All I'm saying is that length contraction is frame dependent, the bending of the rod is frame dependent, and that they are exactly as physical as each other.

They mentioned length contraction of individual ions, but didn't mention the density of the ion beam, and didn't describe how the effect was measured.
I'm interested in this:

21. ### TachBannedBanned

Messages:
5,265
I dared you to come up with the setup, I still say that and I still dare you. You aren't able to complete the setup, even after I gave you a running start.

The cameras capture only one pixel in the setup I helped you with, so they cannot capture any numbers. If you have a different setup in mind, please post it, my challenge to you is on.

Quite the opposite, the details are important, I challenge you to come up with a complete setup.

You are being evasive: length contraction is measurable, the rod "bending" is not. You haven't managed to come up with any experimental setup that can do it.

I'll show you, it is a different setup, first do the honest thing and admit that length contraction is "physical".

22. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
I'm not interested in your dares, Tach. I'm satisfied that the data can be collected in theory. The Lorentz transformation from $S'$ tells us what events would be measured in $S''$. The practical details aren't going to change that.
If you want to talk about how to set up a practical experiment, open another thread.

Yes, it is. Recording rod events in $S''$ reveals this:

Why are you playing games?
Look, this is as clear as I can be, because I don't know precisely what you mean by 'physical':

Length contraction is physical, in the sense that it is measurable with clocks and rulers moving relative to an object.
Length contraction is not physical, in the sense that it doesn't affect the physical structure of the object.

The rod bending is physical, in the sense that it is measurable with $S''$clocks and rulers.
The rod bending is not physical, in the sense that it doesn't affect the physical structure of the rod.

23. ### TachBannedBanned

Messages:
5,265
So, you are unable to construct a valid setup.

You keep repeating the same mistake, Lorentz transforms aren't measuring devices.

I challenged you to produce a setup in this thread, you made a laughable attempt and you failed. The fact is, you are unable to show a valid way as to how the "rod bending" effect would be measured experimentally. Digging your heels or trying to evade doesn't work, you already know that the effect is not measurable because it isn't physical.

Posting animations doesn't prove your case.

True.

False, the links I provided show exactly the opposite, ions are length contracted. I can provide even more proof that you are wrong. You managed to score higher on the crackpot scale than even Undefined and OnlyMe have done in this thread.


text:
# Revision history [back]

While your computations for y and z are correctly done (I think) in Z/(p-1), your computations with K need to be done in Z/(p) instead. So add another ring:

R = IntegerModRing(8885569518)
S = IntegerModRing(8885569519)
K = 263785120
a = 11
b = 17

Ka = S(K)^a
(etc.)

y = 1/R(a)
(etc.)


text:
This function finds a problem set by its name, unpacks it into your current working directory, and provides some hints to get going.

get_pset(pname, newname = NULL)

## Arguments

pname Name of a problem set Your preferred name for the unpacked problem set. Default: NULL (use pname as the folder name)

## Details

By default the problem set will unpack into a folder of the same name. If you'd prefer to unpack the problem set under a different name, perhaps because you've decided to start fresh, or because for whatever reason there's already a folder with that name where you want to unpack it, use the newname argument to asign a new one. (Note that this only renames the top folder. The contents are keep their original names.)

preview_pset to see problem set questions without unpacking anything, and list_psets to see the complete set of problem sets.


text:
Count the number of regions made by Lissajous curve

How many disconnected regions does Lissajous curve divide the plane into?

For example: Let

g = ParametricPlot[{Sin[2 t], Sin[t]}, {t, 0, 2 π}]

The number of disconnected regions made by curve g is 3.

• And what have you tried? – Kuba May 12 '15 at 11:20
• Seems like a mathematics question. Isn't it just 2 plus the number of crossings, provided it's a closed curve with simple crossings? – Michael E2 May 12 '15 at 11:56
• Not sure how general this is: perhaps GraphicsMeshMeshInit[];1 + Length@ GraphicsMeshFindIntersections[ ParametricPlot[{Sin[# t], Sin[ (# - 1) t]}, {t, 0, 2 Pi }]] & /@ {2, 3, 4, 5}? – kglr May 12 '15 at 12:07
• @kguler Oh,guy,why the answer of Length@GraphicsMeshFindIntersections[ ParametricPlot[{Sin[2 t], Sin[(2 - 1) t]}, {t, 0, 2 Pi}]] is 2?But it should be 1 .Somthing wrong? – WateSoyan May 12 '15 at 12:15
• @WateSoyan, for a closed curve the starting point is also an intersection (it intersects with the ending point). – kglr May 12 '15 at 12:45

I don't like to think too much :P

Manipulate[
{#, Composition[
# - 1 &,
Length,
Union,
Flatten,
MorphologicalComponents,
Binarize,
Rasterize
]@#} &@
ParametricPlot[{Sin[ n t], Sin[m t]}, {t, 0, 2 Pi}, Axes -> False,
PlotStyle -> Thick]
, {n, 2, 10, 1}, {m, 1, 9, 1}]

• A similar method was used to count the self-intersections of Lissajous curve. – WateSoyan May 12 '15 at 12:10
• @WateSoyan it depends on what do you mean by similar, but it doesn't seem so. – Kuba May 12 '15 at 12:18
• Your code is so applicable that I benefit a lot from it.:) – WateSoyan May 12 '15 at 12:45
• You might want to add imagesize, which is a function of n and m – Alexey Bobrick May 15 '15 at 14:52
GraphicsMeshMeshInit[];
eps = 1/100000000;
Manipulate[Labeled[plt = ParametricPlot[{Sin[n t], Sin[(n - 1) t]}, {t, eps, 2 Pi},
Axes -> False, PlotStyle -> Thick]; plt /. Line -> Polygon,
Grid[{{"n", "p"}, {n, 2 + Length@GraphicsMeshFindIntersections[plt]}},
ItemStyle -> Directive[16, "Panel"]], Top], {n, 2, 10, 1}]

• Your method is nice.Thank you.But could you explain why your code should return a true result? – WateSoyan May 12 '15 at 12:47
• @WateSoyan, rough/raw intuition: every time a simple curve intersects itself it creates a new region/polygon (try it with pen and paper). – kglr May 12 '15 at 12:56
• Yes,in this cases, it's true.But I think it's not always be true in other cases. – WateSoyan May 12 '15 at 13:05
• By any chance, @Wate, are you in any way familiar with Euler's $v-e+f=2$? – J. M.'s ennui May 12 '15 at 14:34
• However, @Wate, you only asked about Lissajous curves in the question; these curves will only exhibit double nodes, so kguler's counting strategy works here. If you want to ask about rosettes instead, ask a new question. – J. M.'s ennui May 13 '15 at 2:16