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Article Text

Ten years of progress towards universal health coverage: has China achieved equitable healthcare financing?
1. Mingsheng Chen1,2,3,
2. Guoliang Zhou1,
3. Lei Si4,5
1. 1School of Health Policy and Management, Nanjing Medical University, Nanjing, China
2. 2Creative Health Policy Research Group, Nanjing, China
3. 3Center for Global Health, Nanjing Medical University, Nanjing, China
4. 4The George Institute for Global Health, UNSW Sydney, Kensington, New South Wales, Australia
5. 5UNSW Medicine, UNSW Sydney, Sydney, New South Wales, Australia
1. Correspondence to Dr Lei Si; lsi{at}georgeinstitute.org.au

## Abstract

Introduction This study aims to systematically evaluate vertical and horizontal equity in the Chinese healthcare financing system over the period 2008–2018 during the progress towards Universal Health Coverage (UHC), and to examine how both types of equity have changed during this period.

Methods Household information on healthcare payments was collected from 2398 households involving 7021 individuals in 2008, 3600 households involving 10 466 individuals in 2013 and 3660 households involving 11 550 individuals in 2018. Redistributive effects of healthcare financing system were decomposed into progressivity, pure horizontal inequity and reranking. Progressivity analysis and the Aronson-Johnson-Lambert decomposition method were adopted to measure the vertical equity and horizontal equity of healthcare financing.

Results Over the period 2008–2018, healthcare financing through indirect taxes showed a slightly prorich structure and healthcare financing through direct taxes showed a propoor structure in both urban and rural areas. Urban Employee Basic Medical Insurance experienced redistribution from the poor to the rich during the period 2008–2013, but then experienced redistribution from the rich to the poor during the period 2013–2018. Urban Resident Basic Medical Insurance (URBMI), New Rural Cooperative Medical Scheme (NRCMS), Urban and Rural Resident Basic Medical Insurance (URRBMI) and out-of-pocket payments experienced redistribution from the poor to the rich over the entire period.

Conclusion China’s healthcare financing has experienced redistribution from the poor to the rich during 10 years of progress toward the UHC. UHC improved access to and utilisation of healthcare in urban areas. The flat rate contribution mechanism should be renovated for URBMI, NRCMS and URRBMI.

• health insurance
• health policy
• health services research
• health systems

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### Key questions

• Previous studies on China’s healthcare financing mainly focused on vertical equity.

• The benefits package of the health insurance scheme in rural China needs further improvements.

• Flat rate contributions pose a challenge to health insurance premiums in China.

#### What are the new findings?

• This study is the first to systemically examine the vertical and horizontal equity of healthcare financing in China.

• Healthcare financing experienced redistribution from the poor to the rich during 10 years of progress towards Universal Health Coverage (UHC) in China.

• UHC has contributed to reducing horizontal inequity in relation to out-of-pocket payments in China’s urban areas.

#### What do the new findings imply?

• Issues of horizontal inequity arose with more people covered by health insurance programmes in China.

• UHC improved access to and utilisation of healthcare in urban China.

• The health insurance benefits package in rural China needs further improvements.

## Introduction

In 2009, the Chinese government announced a healthcare reform plan that aimed to achieve Universal Health Coverage (UHC) by expanding social health insurance to provide coverage for more than 90% of the population.1 Urban Employee Basic Medical Insurance (UEBMI), Urban Resident Basic Medical Insurance (URBMI) and the New Rural Cooperative Medical Scheme (NRCMS) are the main types of social health insurance that cover the urban and rural populations. These insurance schemes involve different contribution schedules. UEBMI is a universal health insurance scheme for workers and retirees. Contributions to the UEBMI scheme are a fixed proportion of the employee’s salary. URBMI provides cover for urban residents who are not employed, and thus ineligible for the UEBMI scheme. Target groups for the URBMI scheme were children, the elderly, the disabled, laid-off workers and other non-working urban residents. The scheme was mainly financed by government subsidies for premiums, together with flat rate household contributions. The NRCMS was implemented to meet the needs of rural populations. Since 2016, URBMI and NRCMS have been merged into a new type of resident medical insurance, Urban and Rural Resident Basic Medical Insurance (URRBMI), which covered the population in the previous URBMI and NRCMS. Similar to URBMI, the NRCMS and URRBMI featured flat rate contributions. However, although all enrollees paid the same premiums to the NRCMS, the benefit packages varied geographically.

In addition to social health insurance, taxation and out-of-pocket (OOP) payments are the two other sources of finance for healthcare in China.2 However, the impact of these sources on healthcare financing is likely to vary. Contributions towards the financing of healthcare may redistribute household income within a given population. This redistribution can be both vertical and horizontal. Vertical redistribution occurs when payments are disproportionately related to ability to pay (ATP). Horizontal redistribution occurs when people with equal ATP contribute unequally to healthcare payments because of differential treatment. Vertical and horizontal redistributions are generally defined as redistributive effects (REs). The total RE can be quantitatively decomposed into three portions: vertical equity, pure horizontal equity and reranking. Vertical equity implies that people with greater economic ability ought to pay more. Pure horizontal equity implies that people with equal economic ability should pay the same. Reranking occurs when people change rank order after payments.

With regard to UEBMI, contributions are correlated with the enrollee’s salary. Thus, the rich make a larger contribution, which contributes to vertical equity. However, the contribution is not solely linked to salary, but is also correlated with factors such as the enrollee’s age and location. Thus, the UEBMI contribution at a given income level is uncertain, and the extent of this horizontal inequity is unclear. With regard to URBMI, NRCMS and URRBMI, the flat rate contributions mean that the rich and poor pay the same premium, thereby contributing to horizontal equity and vertical inequity. Regarding finance from taxation, payments for healthcare are linked to whatever tax bases are used. For example, much higher income tax rates are imposed on high-income individuals. This tax schedule is both horizontally inequitable and vertically equitable. However, with regard to some indirect taxes, such as consumption tax, the tax burden can be transferred from one person to another, and thus the horizontal and vertical equities of this form of finance should be empirically explored.

It can be seen from the health-financing triangle presented in figure 1 that the ratio of OOP payments to total expenditure on health decreased over the period 2008–2018 in China, while social health insurance made a more substantial contribution over the same period. Thus, UHC has largely been achieved, with more than 95% of the Chinese population being covered by health insurance since 2013,3 but the overall picture of China’s healthcare financing equity has never been fully explored. First, no studies have evaluated financing equity during the 10 years of progress towards UHC in China. Second, the equity of healthcare financing depends not only on its degree of progressivity, but also on the extent of any pure horizontal inequity and reranking resulting from it.4 5 Previous studies on healthcare financing in China have only focused on vertical equity, and a comprehensive analysis of healthcare financing equity has never been undertaken. To our knowledge, only one study has examined the horizontal equity of healthcare financing in China between 2003 and 2008.6 To date, no studies have systemically examined both vertical and horizontal equity during the progress towards UHC in China. This is a serious omission. This study seeks to address this gap in the literature by evaluating the progressivity, horizontal equity and reranking associated with healthcare financing in China during the progress towards UHC. We conducted three surveys in 2008, 2013 and 2018. The aim of our analysis is to systemically evaluate equity of healthcare financing during the 10 years of progress toward UHC in China, including vertical equity, pure horizontal equity and reranking.

Figure 1

Healthcare financing triangle. Finance from out-of-pocket (OOP) payments and social health expenditure as a percentage of total expenditure on health (TEH).

## Methods

### Data sources

Data for our study were collected via three rounds of household surveys conducted in Jiangsu province in eastern China. Socioeconomic development varied geographically in Jiangsu: the overall economic status is more advanced in the south, but modest and underdeveloped in the middle and north of the province. A multistage stratified random sampling procedure was used. In 2008, the survey involved four randomly selected cities or counties. Five communities or towns were then sampled in each city or county. Two neighbourhoods or villages were then selected from each community or town. In each neighbourhood or village, 60 households were randomly selected. In 2013 and 2018, the sampling method was the same, but the sample size was expanded to include six cities or counties. These surveys contained observations from 2400 households involving 7023 individuals in 2008, 3600 households involving 10 466 individuals in 2013 and 3660 households involving 11 550 individuals in 2018. However, in 2008, two households did not report their household income and 2398 households were included in the final analysis (tables 1 and 2).

Table 1

Characteristics of the study participants in urban areas

Table 2

Characteristics of the study participants in rural areas

The surveys collected detailed information on a series of individual and family characteristics, including age, gender, marital status, highest education level attained, employment status, urban/rural classification, household expenditure, consumption of healthcare, individual and household OOP expenditure, and geographical location. Regarding household expenditure, monthly expenditure on food, clothing, traffic, communication, housing, water, electricity, fuel, education, travel, entertainment, medical care and other items were recorded. In addition, information was recorded in relation to any unexpected expenditure. Per capita household expenditure, adjusted to obtain an adult-equivalent (AE) estimate, was used as the measure of living standards.7

Regarding healthcare payments, two sources of data were used. The first involved the above-mentioned surveys, while the other source was local statistical yearbooks (which mainly provided data on tax revenues and social health insurance copayments). Indirect taxes mainly include alcohol tax, cigarette tax, electricity and gas tax, entertainment tax, excise on food, drinks, and lodging, and other consumption taxes. These indirect taxes were approximated by applying specific tax rates to the corresponding expenditures. Direct taxes for households mainly consisted of personal income tax. Personal income tax was imposed on employees, and was deducted from their pay. The personal income rates were determined by the Chinese central government and they were adjustable according to the socioeconomic development. In 2011, the government adjusted the personal income rates from 9 grades to 7 grades. Accordingly, the pre-2011 personal income tax rates were used to calculate the personal income tax in 2008 and the post-2011 rates were used for 2013 and 2018 (online supplemental table 1A,B).

### Supplemental material

As there were no taxes earmarked for healthcare in China, indirect and direct taxes were allocated on a prorata basis using information from the Chinese National Health Accounts. In 2008, tax-funded expenditure represented 91.92% of the government’s expenditure on health, which in turn accounted for 4.51% of general government expenditure. Therefore, in 2008, the proportion of household tax payments going to the health sector was 4.15%. In 2013 and 2018, the proportion of household tax payments going to the health sector was 5.29% and 6.60%, respectively.8 Regarding social health insurance, UEBMI financing contributions were measured by applying contribution rates to the earnings of the covered workers. Financing contributions on URBMI, NRCMS and URRBMI were measured by the sum of the premiums paid. Data on OOP payments included OOP expenditure on prescriptions and outpatient and inpatient services.

### Outcome variables

Vertical equity was measured using a progressivity analysis, and the Kakwani Index (KI) was adopted to evaluate the degree of vertical equity. Horizontal inequity was measured by the sum of pure horizontal inequity (H) and reranking (R), which were directly calculated using the Aronson-Johnson-Lambert (AJL) decomposition method. Vertical equity, pure horizontal inequity and reranking calculations were transformed into percentages of the total RE.

### Analyses

The unit for analysis was the household. Expenditure and healthcare payments were aggregated at the household level. Household expenditure was used as the measurement of ATP.9 Household expenditure was adjusted for household size and composition using an AE estimate. The number of AE household members was defined as

where A is the number of adults in the household, α is the cost per child, K is the number of children and β is the degree of economies of scale.9 The values of α and β were assumed to be 0.5 and 0.75, respectively.10 The population was ranked by socioeconomic status and grouped into quintiles of equal size. Household healthcare payments were also adjusted for household size and composition using an AE estimate.

#### Progressivity analysis

In this study, progressivity was measured by computing the Concentration Index (CI) and the Gini coefficient (GC), which required directly relating the covariance between variables and household fractional ranks based on people’s ATP.7 Estimates of the CI and GC can be obtained from a ‘convenient regression’ of a transformation of ATP or the healthcare payment on the fractional rank in the ATP distribution.9 Specifically,

where is the healthcare payment or ATP of household i, φ is the mean healthcare payment or ATP, is the household fractional rank based on the ATP distribution, and is the variance of the household fractional rank. The ordinary least squares regression estimate of β is an estimate of CI or GC. α is the intercept and ε is the residual.

The progressivity of the healthcare payments was then estimated using the KI, which is the difference between the CI and the GC. If the difference is positive (negative), it indicates that the payments are progressive (regressive). If the difference equals 0, it indicates that the payments are proportional. Progressivity (regressivity) means that the rich (poor) contribute a larger proportion of healthcare payments than the poor (rich) in comparison with their ATP.9

#### AJL decomposition

The horizontal inequity of the healthcare payments was measured using AJL decomposition.4 The RE of healthcare financing can be calculated as the difference in the GC caused by the healthcare payments:

and are the prepayment and postpayment GC respectively, where X denotes prepayment income, and P denotes the healthcare payment.

The AJL decomposition method shows that this difference can be written as:

where V is the vertical redistribution, H is the pure horizontal inequity and R is the degree of reranking.

V can be decomposed into a payment rate effect and a progressivity effect:

where g is the sample average healthcare payment rate (as a proportion of income) and is the KI value of the payments computed under the assumption of within-group equality.

Pure horizontal inequity was measured by the weighted sum of the group j-specific postpayment GC, with the weights being calculated as the product of the group’s population share and its postpayment income share, :

R captured the extent of reranking of households that occurred in the move from prepayment to postpayment income distributions, and was measured as the difference between and . is the postpayment GC, and is the postpayment income CI. To calculate the postpayment income CI, we ranked all households by the prepayment incomes and then, by their postpayment income within each group of prepayment ‘equals’.

### Patient and public involvement

No patients or members of the public were directly involved in this study. There are no plans to involve patients or the public in the dissemination of results.

## Results

The results regarding the progressivity, pure horizontal inequity and reranking associated with healthcare financing are presented in tables 3–5). Figures 2 and 3 show the Lorenz curve and the concentration curves associated with each healthcare financing source. Figures 4 and 5 show the degree of horizontal inequity and reranking for financing sources.

Table 3

Progressivity, pure horizontal equity and reranking associated with the Chinese healthcare financing system in 2008

Table 4

Progressivity, pure horizontal equity and reranking associated with the Chinese healthcare financing system in 2013

Table 5

Progressivity, pure horizontal equity and reranking associated with the Chinese healthcare financing system in 2018

Figure 2

Concentration curve of healthcare payments and Lorenz curve in urban areas. The figure shows actual cumulative concentration curves for healthcare payments (including those associated with indirect taxes, direct tax, UEBMI, URBMI, URRBMI and OOP payments) and the Lorenz curve in urban areas in 2008, 2013 and 2018. OOP, out of pocket; UEBMI, Urban Employee Basic Medical Insurance; URBMI, Urban Resident Basic Medical Insurance; URRBMI, Urban and Rural Resident Basic Medical Insurance.

Figure 3

Concentration curve of healthcare payments and Lorenz curve in rural areas. The figure shows actual cumulative concentration curves for healthcare payments (including those associated with indirect taxes, direct tax, the NRCMS, URRBMI and OOP payments) and the Lorenz curve in rural areas in 2008, 2013 and 2018. NRCMS, New Rural Cooperative Medical Scheme; OOP, out of pocket; URRBMI, Urban Resident Basic Medical Insurance.

Figure 4

Horizontal inequity and reranking for financing sources in urban areas in 2008, 2013 and 2018. Horizontal inequity and reranking are presented as percentages of the redistributive effect associated with each healthcare financing source in urban areas. OOP, out of pocket; UEBMI, Urban Employee Basic Medical Insurance; URBMI, Urban Resident Basic Medical Insurance; URRBMI, Urban Resident Basic Medical Insurance.

Figure 5

Horizontal inequity and reranking for financing sources in rural areas in 2008, 2013 and 2018. Horizontal inequity and reranking are presented as percentages of the redistributive effect associated with each healthcare financing source in rural areas. NRCMS, New Rural Cooperative Medical Scheme; OOP, out of pocket; URRBMI, Urban Resident Basic Medical Insurance.

### Sample characteristics

This study included 2398, 3600 and 3660 households in 2008, 2013 and 2018, respectively. Sample characteristics are shown in tables 1 and 2.

### Total healthcare financing

Total healthcare payments as a proportion of household income increased over the period 2008–2018 (urban areas: 13.38% in 2008, 15.91% in 2013, and 20.07% in 2018; rural areas: 12.55% in 2008, 11.89% in 2013 and 23.02% in 2018). The RE values were all negative over the period 2008–2018 (urban areas: –0.004377 in 2008, –0.009498 in 2013 and –0.021746 in 2018; rural areas: –0.004573 in 2008, –0.002921 in 2013 and –0.020628 in 2018). The KI values increased over the period 2008–2013 and decreased over the period 2013–2018 (urban areas: 0.0203 in 2008, 0.0409 in 2013 and –0.0216 in 2018; rural areas: 0.0574 in 2008, 0.0736 in 2013 and 0.0169 in 2018). Horizontal inequity increased over the period 2008–2013 and decreased over the period 2013–2018 (urban areas: 193.72% in 2008, 200.52% in 2013 and 82.20% in 2018; rural areas: 297.24% in 2008, 474.40% in 2013 and 140.23% in 2018).

### Indirect taxes

The RE values were all negative over the period 2008–2018 (urban areas: –0.000085 in 2008, –0.000086 in 2013 and –0.000041 in 2018; rural areas: –0.000073 in 2008, –0.000092 in 2013 and –0.000137 in 2018). The KI values remained almost unchanged, and were basically slightly regressive (it was negative but not statistically significantly so in urban areas in 2018). Horizontal inequity increased in urban areas and remained almost unchanged in rural areas over the entire period (urban areas: 27.82% in 2008, 37.78% in 2013 and 85.12% in 2018; rural areas: 26.12% in 2008, 26.16% in 2013, and 24.56% in 2018).

### Direct taxes

The RE values were all positive over the period 2008–2018 (urban areas: 0.000254 in 2008, 0.000599 in 2013 and 0.001857 in 2018; rural areas: 0.000083 in 2008, 0.000463 in 2013 and 0.002067 in 2018). The direct taxes were strongly progressive. The KI values decreased over the period 2008–2013 and increased over the period 2013–2018 (urban areas: 0.4486 in 2008, 0.2285 in 2013 and 0.2525 in 2018; rural areas: 0.5921 in 2008, 0.2179 in 2013 and 0.3558 in 2018). Horizontal inequity remained almost unchanged (urban areas: 3.32% in 2008, 20.30% in 2013 and 6.50% in 2018; rural areas: 3.05% in 2008, 10.47% in 2013 and 4.56% in 2018).

### Urban Employee Basic Medical Insurance

The RE values were negative in 2008 (–0.002698) and 2013 (–0.000710), but positive in 2018 (0.001695). UEBMI was regressive in 2008, near proportional in 2013, and progressive in 2018 (–0.0328 in 2008, 0.0037 in 2013, and 0.0438 in 2018). Horizontal inequity was small in 2008, but it increased in 2013 and 2018 (34.49% in 2008, 179.13% in 2013, and 102.07% in 2018).

### Urban Employee Basic Medical Insurance, New Rural Cooperative Medical Scheme and Urban and Rural Resident Basic Medical Insurance

We compared and discussed the URBMI, NRCMS and URRBMI over the period 2008–2018 considering that they shared the same financing mechanism and covered the same target population in urban and rural areas, respectively. In addition, given that URBMI was piloted in 2007, only a minority of the population was covered by URBMI. Consequently, URBMI in 2008 was not included in the analysis. In urban areas, the RE values of URBMI/URRBMI were negative over the period 2013–2018 (–0.001708 in 2013, and –0.003862 in 2018). The KI values were all statistically significantly negative, and were strongly regressive (–0.5441 in 2013 and –0.6361 in 2018). Horizontal inequity as a percentage of the total RE was very close to 0. In rural areas, the RE values of NRCMS/URRBMI were all negative over the period 2008–2018 (–0.000967 in 2008,–0.001953 in 2013 and –0.017468 in 2018). The KI values were all statistically significantly negative, and were strongly regressive (–0.1656 in 2008,–0.3601 in 2013 and –0.4095 in 2018). Horizontal inequity as a percentage of the total RE was very close to 0.

### OOP payments

The RE values were all negative over the period 2008–2018 (urban areas: –0.001682 in 2008, –0.007512 in 2013 and –0.019375 in 2018; rural areas: –0.003422 in 2008, –0.001204 in 2013 and –0.001880 in 2018). The KI values in urban areas increased during the period 2008–2013 but decreased during the period 2013–2018 (0.0703 in 2008, 0.0763 in 2013 and –0.0454 in 2018). Conversely, the KI values in rural areas increased during the entire period (0.0703 in 2008, 0.0969 in 2013 and 0.1038 in 2018). Horizontal inequity decreased in urban areas and increased in rural areas over the entire period (urban areas: 428.75% in 2008, 225.78% in 2013 and 73.74% in 2018; rural areas: 390.48% in 2008, 1134.41% in 2013 and 1437.78% in 2018).

## Discussion

Overall, healthcare financing in both urban and rural areas over the period 2008–2018 underwent redistribution from the poor to the rich. In other words, healthcare financing increased the wealth inequality between the rich and the poor. Vertical equity, as measured by the KI, increased over the period 2008–2013 and decreased over the period 2013–2018 in both urban and rural areas. Horizontal inequity increased over the period 2008–2013 and decreased over the period 2013–2018 in both urban and rural areas.

Healthcare financing through indirect taxes resulted in redistribution from the poor to the rich. Vertical equity remained almost unchanged, and was basically slightly regressive in both urban and rural areas (it was negative but not statistically significantly so in urban areas in 2018). Horizontal inequity increased in urban areas and remained almost unchanged in rural areas over the period 2008–2018. Indirect tax was not propoor because the rich could transfer the tax burden to the poor. In most developed countries, indirect taxes only account for a minority of general taxation. For example, in 2018, indirect taxes accounted for 9.3%, 26.5%, and 21.5% of overall taxation revenue in the USA, Australia and Canada, respectively.11 Conversely, in China, indirect tax accounts for the majority of general taxation revenue. Within the region of interest within the current study, indirect tax accounted for 61.08%, 61.18% and 54.73% of general taxation revenue in 2008, 2013 and 2018, respectively.8

Direct tax experienced redistribution from the rich to the poor, and was strongly progressive in both urban and rural areas. Horizontal inequity remained almost unchanged. For households, direct tax primarily consisted of personal income tax, which was paid based on the grade of the taxpayer’s income. As shown in online supplemental tables 1 and 2, personal income tax was a progressive mechanism that was particularly focused on the rich. Individuals with a monthly income below a specified level were either not required to pay personal income tax or paid a lower tax rate than the rich, which resulted in a propoor distribution of the tax burden in relation to healthcare financing. Specifically, in 2008, nine grades of progressive tax rates were levied on monthly incomes greater than ¥1600. From 2012, seven grades of progressive tax rates were levied on monthly incomes greater than ¥3500. This policy adjustment aimed to reduce the tax burden on the poor by decreasing the number of grades, reducing the tax rates for the lower grades, and increasing the tax rates for the higher grades. This could explain why the progressivity of direct tax decreased during the first half of the study period and increased during the second half of the study period.

UEBMI experienced redistribution from the poor to the rich during the period 2008–2013, but then experienced redistribution from the rich to the poor during the period 2013–2018, and became increasingly progressive over the entire period 2008–2018 in terms of vertical equity. The finding of UEBMI progressivity is supported by earlier WHO research showing that social health insurance tended to be regressive during progress toward UHC as the health insurance scheme began to cover the previously uninsured.12 China began to establish social health insurance at the end of 1998, and the first health insurance scheme, UEBMI, provided universal health insurance for public sector workers and retirees at the local government level. UEBMI premiums are a fixed proportion (about 2%) of the employee’s salary. 13 However, workers in non-public sectors were not covered by UEBMI. In addition, government staff and workers in important state-owned enterprises remained enrolled in the Free Medical Service programme and the Labour Medical Service programme, which were medical benefits schemes that required no employee contributions. Thus, the government has gradually expanded UEBMI to cover all types of workers, including employees working in public and private enterprises, Chinese–foreign joint ventures, social organisations and private non-enterprise units, as well as migrant workers. Moreover, the Free Medical Service programme and the Labour Medical Service programme were abolished, and individuals who were previously insured under these programmes were required to enrol in the UEBMI scheme and commence individual contributions. Because employees working in private enterprises, international joint ventures, key state-owned enterprises, and government departments generally receive higher-than-average salaries, this resulted in the progressive nature of UEBMI when China began to move toward UHC. The horizontal inequity of UEBMI was small in 2008, but increased in 2013 and 2018. Since the expansion of health insurance coverage, UEBMI has covered employees in the public, private and almost all other sectors. However, UEBMI enrollees with equivalent incomes may have different income sources. For instance, the main source of income for workers in the public sector is their salary, but migrant workers and workers in the private sector tend to have other sources of income in addition to their salaries. UEBMI premiums are based on the enrollee’s monthly salary, and thus this resulted in an increase in horizontal inequity in relation to UEBMI.

URBMI, which was established in 2007, provided cover for people in urban areas who were not covered by UEBMI, such as students, the unemployed and the aged. The NRCMS, which was established in 2003, provided cover for the general population in rural areas. URBMI premiums were ¥131.0 and ¥400.5 in 2008 and 2013, respectively.14 NRCMS premiums were ¥96.2 and ¥370.6 in 2008 and 2013, respectively.15 Since 2016, URBMI and NRCMS have been merged into URRBMI and the flat rate premium was set at ¥723.2.15 URBMI, NRCMS and URRBMI premiums were flat rate contributions, meaning that each insured individual paid the same premium, regardless of their ATP. In other words, lower-income groups contributed a greater proportion of their income than higher-income groups. This resulted in an inequitable healthcare financing distribution for URBMI, NRCMS and URRBMI, although horizontal inequity as a percentage of the total RE was 0 or very close to 0 for URBMI, NRCMS and URRBMI. Although URBMI, NRCMS and URRBMI were very horizontally equitable, they experienced redistribution from the poor to the rich because they were both very regressive. This implies that a healthcare financing system with zero horizontal inequity is not necessarily equitable because flat rate contributions result in a high level of vertical inequity. Nevertheless, the optimal range of horizontal equity for a healthcare system remains unknown at present, and thus further research should be conducted in an effort to identify the optimum balance between vertical equity and horizontal equity.

OOP payments experienced redistribution from the poor to the rich over the period 2008–2018. Progressivity in urban areas increased during the period 2008–2013 but decreased during the period 2013–2018. Conversely, progressivity in rural areas increased during the entire period 2008–2018. Owing to the nature of postpaid payments, caution is necessary when interpreting the progressivity of OOP payments. In countries where OOP payments dominate healthcare financing, the rich can afford to pay for healthcare services via OOP payments, even for high-cost medical services,16 17 while it is difficult for the poor to make OOP payments for even basic medical services.18 19 This situation is obviously inequitable, and yet the associated healthcare financing distribution has been found to be progressive. In the first half of the period covered in our study, China’s move towards UHC emphasised expanding social health insurance coverage to all of the population.20 However, during this phase, social health insurance schemes focused on providing coverage for those with a lower reimbursement level.21 22 China’s social health insurance was organised and implemented at the city or county level, and thus fragmentation limited the scope for pooling of funding and failed to provide sufficient financial protection.23 24 Local officials were forced to choose between expanding coverage to a broader population and updating benefits packages. Based on the priority underlying UHC, health insurance was provided to more people at the cost of providing appropriate benefits package.25 Moreover, the reimbursement policy was not attractive. Although deductible was not high for the insured to get reimbursed, copayment was high and ceiling was also very low, and thus most of the cost of healthcare was paid for by the patients themselves. The scope for reimbursement was very limited.26 Healthcare services were only eligible for reimbursement if they were provided by designated hospitals and pharmacies, and reimbursement was subject to the health insurance lists. However, some common and extensively used medicines and medical devices were not covered by the lists, while some expensive medicines were covered by the lists and were recommended by the hospitals. Meanwhile, retrospective payment methods, such as fee-for-service reimbursement for hospitals, might encourage overtreatment as payment to providers was mainly based on volume rather than quality of services.27 These resulted in a high level of copayments by patients. Further, URBMI and NRCMS only covered inpatient services until the end of 2011.28 Although coverage has since gradually been expanded to include outpatient services, patients remained liable for 60%–70% of their outpatient costs and more than 50% of their inpatient costs.26 Thus, OOP payments meant that the cost of medical care was prohibitive for the poor, resulting in the development of a more progressive distribution of financing in both urban and rural areas during the period 2008–2013. Subsequently, after achieving nearly universal coverage, the government has continually improved the benefits packages of health insurance schemes by expanding service coverage, reducing copayments and raising the reimbursement ceiling to provide additional financial risk protection.28 Thus, residents tended to consume more medical services during the period 2013–2018. Urban residents, especially the poor, accounted for a greater proportion of OOP payments during this period compared with the period 2008–2013, which resulted in a decrease in progressivity. However, there were significant differences among health insurance schemes in terms of benefits. For example, in 2018, reimbursements under UEBMI, URRBMI (urban) and URRBMI (rural) were ¥3379.52 (US$510.70), ¥692.36 (US$104.63) and ¥629.40 (US\$95.11) per person, respectively.3 29 Consequently, poor rural residents were less able to pay for medical services than urban residents, resulting in the continuous increase in progressivity during the period 2013–2018.

During the period 2008–2018, horizontal inequity decreased in urban areas and increased in rural areas. The extent of the horizontal inequity reflects the medical benefits provided by health insurance schemes. As part of the progress towards UHC in urban areas during the period 2008–2018, the benefits package and reimbursement levels of the main health insurance scheme, UEBMI, were continually improved. Although OOP payments increased and progressivity decreased, healthcare utilisation was improved and horizontal inequity was reduced over this period. Moreover, reranking associated with OOP payments indicated the economic rank order of individuals who funded healthcare via OOP payments. The fact that reranking decreased over this period indicates that impoverishment incurred by medical expenditure was alleviated in urban areas. Conversely, the benefits package and reimbursement levels in the main rural health insurance scheme were not improved. Thus, horizontal inequity associated with OOP payments, including reranking, continually increased during the period 2008–2018.

### Strengths and limitations

Previous studies have only focused on the vertical equity in China’s healthcare financing system, and the overall picture in relation to healthcare financing equity has never been explored. We used standardised methods to analyse data from three waves of household surveys during the 10 years of progress towards implementation of UHC, systematically evaluating the vertical and horizontal equity of the Chinese healthcare financing system. Thus, our study not only provides the most recent evidence of China’s healthcare financing distribution, but also identifies the effects of health-system financing during progress towards UHC for the first time. A limitation of our study is that we only examined one province in China, the results might not fully represent the case in other parts of China. This limitation is, however, composed of by the fact that we used percentages and indices to evaluate the implementation of national policies and programmes across the entire population. In addition, as there is a variation in economic development across geographical locations in Jiangsu which is similar to the case in China, our findings in Jiangsu might reflect the national financing distribution. Another limitation is that data collections were conducted in 2008, 2013 and 2018 cross-sectionally to evaluate the change of healthcare financing equity, therefore, the observed trend during these three time points might not fully reflect the continuous change in healthcare financing equity.

## Conclusions

China’s healthcare financing has experienced redistribution from the poor to the rich during 10 years of progress toward UHC. The flat rate contribution mechanism should be renovated for URBMI, NRCMS and URRBMI. In urban areas, the progressivity of OOP payments indicated that UHC improved access to and utilisation of healthcare, and horizontal inequity, including reranking, reflected adequate benefits package of UEBMI.

View Abstract

• ## Supplementary Data

This web only file has been produced by the BMJ Publishing Group from an electronic file supplied by the author(s) and has not been edited for content.

## Footnotes

• Handling editor Sanni Yaya

• Contributors LS led and designed the study. GZ led the data collection, analysis and interpretation. MC contributed to the study design and data analysis, and wrote the first draft of the manuscript. LS contributed to the data analysis, reviewed the manuscript and helped write the final draft manuscript. All authors reviewed the content of the final version of the manuscript.

• Funding This study is funded by the National Natural Science Foundation of China (grant number: 71503137, 71874086) and the China Medical Board (grant number: 19-346). LS is supported by an NHMRC Early Career Fellowship (grant number: GNT1139826).

• Disclaimer The funding bodies were not involved in the design of the study, or data collection, analysis, and interpretation or in writing the manuscript.

• Competing interests None declared.

• Patient consent for publication Not required.

• Provenance and peer review Not commissioned; externally peer reviewed.

• Data availability statement Data are available on request. The authors will make data available to scientists planning specified and agreed further analyses; for access, contact the corresponding author.

• Supplemental material This content has been supplied by the author(s). It has not been vetted by BMJ Publishing Group Limited (BMJ) and may not have been peer-reviewed. Any opinions or recommendations discussed are solely those of the author(s) and are not endorsed by BMJ. BMJ disclaims all liability and responsibility arising from any reliance placed on the content. Where the content includes any translated material, BMJ does not warrant the accuracy and reliability of the translations (including but not limited to local regulations, clinical guidelines, terminology, drug names and drug dosages), and is not responsible for any error and/or omissions arising from translation and adaptation or otherwise.

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# The Unapologetic Mathematician

## (Hyper-)Surface Integrals

The flip side of the line integral is the surface integral.

Given an $n$-manifold $M$ we let $S$ be an oriented $n-1$-dimensional “hypersurface”, which term we use because in the usual case of $M=\mathbb{R}^3$ a hypersurface has two dimensions — it’s a regular surface. The orientation on a hypersurface consists of $n-1$ tangent vectors which are all in the image of the derivative of the local parameterization map, which is a singular cube.

Now we want another way of viewing this orientation. Given a metric on $M$ we can use the inverse of the Hodge star from $M$ on the orientation $n-1$-form of $S$, which gives us a covector $\nu(p)\in\mathcal{T}^*_pM$ defined at each point $p$ of $S$. Roughly speaking, the orientation form of $S$ defines an $n-1$-dimensional subspace of $\mathcal{T}^*_pM$: the covectors “tangent to $S$“. The covector $\nu(p)$ is “perpendicular to $S"$, and since $S$ has only one fewer dimension than $M$ there are only two choices for the direction of such a vector. The choice of one or the other defines a distinction between the two sides of $S$.

If we flip around our covectors into vectors — again using the metric — we get something like a vector field defined on $S$. I say “something like” because it’s only defined on $S$, which is not an open subspace of $M$, and strictly speaking a vector field must be defined on an open subspace. It’s not hard to see that we could “thicken up” $S$ into an open subspace and extend our vector field smoothly there, so I’m not really going to make a big deal of it, but I want to be careful with my language; it’s also why I didn’t say we get a $1$-form from the Hodge star. Anyway, we will call this vector-valued function $dS$, for reasons that will become apparent shortly.

Now what if we have another vector-valued function $F$ defined on $S$ — for example, it could be a vector field defined on an open subspace containing $S$. We can define the “surface integral of $F$ through $S$“, which measures how much the vector field flows through the surface in the direction our orientation picks out as “positive”. And we measure this amount at any given point $p$ by taking the covector at $p$ provided by the Hodge star and evaluating it at the vector $F(p)$. This gives us a value that we can integrate over the surface. This evaluation can be flipped around into our vector field notation and language, allowing us to write down the integral as

$\displaystyle\int\limits_SF\cdot dS$

because the “dot product” $\left[F\cdot dS\right](p)$ is exactly what it means to evaluate the covector dual to $dS(p)$ at the vector $F(p)$. This should look familiar from multivariable calculus, but I’ve been saying that a lot lately, haven’t I?

We can also go the other direction and make things look more abstract. We could write our vector field as a $1$-form $\alpha$, which lets us write our surface integral as

$\displaystyle\int\limits_S\langle\alpha,\nu\rangle\omega_M=\int\limits_S\alpha\wedge*\nu$

where $\omega_M$ is the volume form on $M$. But then we know that $*\nu=\omega_S$ the volume form on $S$. That is, we can define a hypersurface integral of any $1$-form $\alpha$ across any hypersurface $S$ equipped with a volume form $\omega_S$ by

$\displaystyle\int\limits_S\alpha\wedge\omega_S$

whether or not we have a metric on $M$.

October 27, 2011

## The Fundamental Theorem of Line Integrals

At last we can state one of our special cases of Stokes’ theorem. We don’t need to prove it, since we already did in far more generality, but it should help put our feet on some familiar ground.

So, we take an oriented curve $c$ in a manifold $M$. The image of $c$ is an oriented submanifold of $M$, where the “orientation” means picking one of the two possible tangent directions at each point along the image of the curve. As a high-level view, we can characterize the orientation as the direction that we traverse the curve, from one “starting” endpoint to the other “ending” endpoint.

Given any $1$-form $\alpha$ on the image of $c$ — in particular, given an $\alpha$ defined on $M$ — we can define the line integral of $\alpha$ over $c$. We already have a way of evaluating line integrals: pull the $1$-form back to the parameter interval of $c$ and integrate there as usual. But now we want to use Stokes’ theorem to come up with another way. Let’s write down what it will look like:

$\displaystyle\int\limits_cd\omega=\int\limits_{\partial c}\omega$

where $\omega$ is some $0$-form. That is: a function. This tells us that we can only make this work for “exact” $1$-forms $\alpha$, which can be written in the form $\alpha=df$ for some function $f$.

But if this is the case, then life is beautiful. The (oriented) boundary of $c$ is easy: it consists of two $0$-faces corresponding to the two endpoints. The starting point gets a negative orientation while the ending point gets a positive orientation. And so we write

$\displaystyle\int\limits_cdf=\int\limits_{\partial c}f=\sum\limits_{j=0,1}(-1)^{1+j}f(c_{1,j})=f(c(1))-f(c(0))$

That is, we just evaluate $f$ at the two endpoints and subtract the value at the start from the value at the end!

What does this look like when we have a metric and we can rewrite the $1$-form $\alpha$ as a vector field $F$? In this case, $\alpha$ is exact if and only if $F$ is “conservative”, which just means that $F=\nabla f$ for some function $f$. Then we can write

$\displaystyle\int\limits_c\nabla f\cdot ds=f(c(1))-f(c(0))$

which should look very familiar from multivariable calculus.

We call this the fundamental theorem of line integrals by analogy with the fundamental theorem of calculus. Indeed, if we set this up in the manifold $\mathbb{R}$, we get back exactly the second part of the fundamental theorem of calculus back again.

October 24, 2011

## Line Integrals

We now define some particular kinds of integrals as special cases of our theory of integrals over manifolds. And the first such special case is that of a line integral.

Consider an oriented curve $c$ in the manifold $M$. We know that this is a singular $1$-cube, and so we can pair it off with a $1$-form $\alpha$. Specifically, we pull back $\alpha$ to $c^*\alpha$ on the interval $[0,1]$ and integrate.

More explicitly, the pullback $c^*\alpha$ is evaluated as

$\displaystyle\left[c^*\alpha\left(\frac{d}{dt}\right)\right](t_0)=\left[\alpha_{c(t_0)}\right]\left(c_{*t_0}\frac{d}{dt}\bigg\vert_{t_0}\right)$

That is, for a $t_0\in[0,1]$, we take the value $\alpha_{c(t_0)}\in\mathcal{T}^*_{c(t_0)}M$ of the $1$-form $\alpha$ at the point $c(t_0)\in M$ and the tangent vector $c'(t_0)\in\mathcal{T}_{c(t_0)}M$ and pair them off. This gives us a real-valued function which we can integrate over the interval.

So, why do we care about this particularly? In the presence of a metric, we have an equivalence between $1$-forms $\alpha$ and vector fields $F$. And specifically we know that the pairing $\alpha_{c(t)}\left(c'(t)\right)$ is equal to the inner product $\langle F(c(t)),c'(t)\rangle$ — this is how the equivalence is defined, after all. And thus the line integral looks like

$\displaystyle\int\limits_c\alpha=\int\limits_{[0,1]}\langle F(c(t)),c'(t)\rangle\,dt$

Often the inner product is written with a dot — usually called the “dot product” of vectors — in which case this takes the form

$\displaystyle\int\limits_{[0,1]}F(c(t))\cdot c'(t)\,dt$

We also often write $ds=c'(t)\,dt$ as a “vector differential-valued function”, in which case we can write

$\displaystyle\int\limits_{[0,1]}F\cdot ds$

Of course, we often parameterize a curve by a more general interval $I$ than $[0,1]$, in which case we write

$\displaystyle\int\limits_IF\cdot ds$

This expression may look familiar from multivariable calculus, where we first defined line integrals. We can now see how this definition is a special case of a much more general construction.

October 21, 2011

## The Codifferential

From our calculation of the square of the Hodge star we can tell that the star operation is invertible. Indeed, since $*^2=(-1)^{k(n-k)}\lvert g_{ij}\rvert$ — applying the star twice to a $k$-form in an $n$-manifold with metric $g$ is the same as multiplying it by $(-1)^{k(n-k)}$ and the determinant of the matrix of $g$ — we conclude that $*^{-1}=(-1)^{k(n-k)}\lvert g^{ij}\rvert*$.

With this inverse in hand, we will define the “codifferential”

$\displaystyle\delta=(-1)^k*^{-1}d*$

The first star sends a $k$-form to an $n-k$-form; the exterior derivative sends it to an $n-k+1$-form; and the inverse star sends it to a $k-1$-form. Thus the codifferential goes in the opposite direction from the differential — the exterior derivative.

Unfortunately, it’s not quite as algebraically nice. In particular, it’s not a derivation of the algebra. Indeed, we can consider $fdx$ and $gdy$ in $\mathbb{R}^3$ and calculate

\displaystyle\begin{aligned}\delta(fdx)&=-*d*(fdx)=-*d(fdy\wedge dz)=-*\frac{\partial f}{\partial x}dx\wedge dy\wedge dz=-\frac{\partial f}{\partial x}\\\delta(gdy)&=-*d*(gdy)=-*d(gdz\wedge dx)=-*\frac{\partial g}{\partial y}dy\wedge dz\wedge dx=-\frac{\partial g}{\partial y}\end{aligned}

while

\displaystyle\begin{aligned}\delta(fgdx\wedge dy)&=-*d*(fgdx\wedge dy)\\&=-*d(fgdz)\\&=-*\left(\left(\frac{\partial f}{\partial x}g+f\frac{\partial g}{\partial x}\right)dx\wedge dz+\left(\frac{\partial f}{\partial y}g+f\frac{\partial g}{\partial y}\right)dy\wedge dz\right)\\&=\left(\frac{\partial f}{\partial x}g+f\frac{\partial g}{\partial x}\right)dy-\left(\frac{\partial f}{\partial y}g+f\frac{\partial g}{\partial y}\right)dx\end{aligned}

but there is no version of the Leibniz rule that can account for the second and third terms in this latter expansion. Oh well.

On the other hand, the codifferential $\delta$ is (sort of) the adjoint to the differential. Adjointness would mean that if $\eta$ is a $k$-form and $\zeta$ is a $k+1$-form, then

$\displaystyle\langle d\eta,\zeta\rangle=\langle\eta,\delta\zeta\rangle$

where these inner products are those induced on differential forms from the metric. This doesn’t quite hold, but we can show that it does hold “up to homology”. We can calculate their difference times the canonical volume form

\displaystyle\begin{aligned}\left(\langle d\eta,\zeta\rangle-\langle\eta,\delta\zeta\rangle\right)\omega&=d\eta\wedge*\zeta-\eta\wedge*\delta\zeta\\&=d\eta\wedge*\zeta-(-1)^k\eta\wedge**^{-1}d*\zeta\\&=d\eta\wedge*\zeta-(-1)^k\eta\wedge d*\zeta\\&=d\left(\eta\wedge*\zeta\right)\end{aligned}

which is an exact $n$-form. It’s not quite as nice as equality, but if we pass to De Rham cohomology it’s just as good.

October 21, 2011

## The Hodge Star, Squared

It’s interesting to look at what happens when we apply the Hodge star twice. We just used the fact that in our special case of $\mathbb{R}^3$ we always get back exactly what we started with. That is, in this case, $*^2=1$ — the identity operation.

It’s easiest to work this out in coordinates. If $\eta=dx^{i_1}\wedge\dots\wedge dx^{i_k}$ is some $k$-fold wedge then $*\eta$ is $\pm\sqrt{\lvert g_{ij}\rvert}$ times the wedge of all the indices that don’t show up in $\eta$. But then $**\eta$ is $\pm\sqrt{\lvert g_{ij}\rvert}$ times the wedge of all the indices that don’t show up in $*\eta$, which is exactly all the indices that were in $\eta$ in the first place! That is, $**\eta=\pm\lvert g_{ij}\rvert\eta$, where the sign is still a bit of a mystery.

But some examples will quickly shed some light on this. We can even extend to the pseudo-Riemannian case and pick a coordinate system so that $\langle dx^i,dx^j\rangle=\epsilon_i\delta^{ij}$, where $\epsilon_i=\pm1$. That is, any two $dx^i$ are orthogonal, and each $dx^i$ either has “squared-length” $1$ or $-1$. The determinant $\lvert g_{ij}\rvert$ is simply the product of all these signs. In the Riemannian case this is simply $1$.

Now, let’s start with an easy example: let $\eta$ be the wedge of the first $k$ indices:

$\displaystyle\eta=dx^1\wedge\dots\wedge dx^k$

Then $*\eta$ is (basically) the wedge of the other indices:

$\displaystyle*\eta=\sqrt{\lvert g_{ij}\rvert}dx^{k+1}\wedge\dots\wedge dx^n$

The sign is positive since $\eta\wedge*\eta$ already has the right order. But now we flip this around the other way:

$\displaystyle**\eta=\pm\lvert g_{ij}\rvert dx^1\wedge\dots\wedge dx^k$

but this should obey the same rule as ever:

\displaystyle\begin{aligned}*\eta\wedge**\eta&=\pm\lvert g_{ij}\rvert\sqrt{\lvert g_{ij}\rvert}dx^{k+1}\wedge\dots\wedge dx^n\wedge dx^1\wedge\dots\wedge dx^k\\&=\pm\lvert g_{ij}\rvert(-1)^{k(n-k)}\sqrt{\lvert g_{ij}\rvert}dx^1\wedge\dots\wedge dx^n\\&=\pm\lvert g_{ij}\rvert(-1)^{k(n-k)}\omega\end{aligned}

where we pick up a factor of $-1$ each time we pull one of the last $k$ $1$-forms leftwards past the first $n-k$. We conclude that the actual sign must be $\lvert g_{ij}\rvert(-1)^{k(n-k)}$ so that this result is exactly $\omega$. Similar juggling for other selections of $\eta$ will give the same result.

In our special Riemannian case with $n=3$, then no matter what $k$ is we find the sign is always positive, as we expected. The same holds true in any odd dimension for Riemannian manifolds. In even dimensions, when $k$ is odd then so is $n-k$, and so $*^2=-1$ — the negative of the identity transformation. And the whole situation gets more complicated in the pseudo-Riemannian version depending on the number of $-1$s in the diagonalized metric tensor.

October 18, 2011

## The Divergence Operator

One fact that I didn’t mention when discussing the curl operator is that the curl of a gradient is zero: $\nabla\times(\nabla f)=0$. In our terms, this is a simple consequence of the nilpotence of the exterior derivative. Indeed, when we work in terms of $1$-forms instead of vector fields, the composition of the two operators is $*d(df)$, and $d^2$ is always zero.

So why do we bring this up now? Because one of the important things to remember from multivariable calculus is that the divergence of a curl is also automatically zero, and this will help us figure out what a divergence is in terms of differential forms. See, if we take our vector field and consider it as a $1$-form, the exterior derivative is already known to be (essentially) the curl. So what else can we do?

We use the Hodge star again to flip the $1$-form back to a $2$-form, so we can apply the exterior derivative to that. We can check that this will be automatically zero if we start with an image of the curl operator; our earlier calculations show that $*^2$ is always the identity mapping — at least on $\mathbb{R}^3$ with this metric — so if we first apply the curl $*d$ and then the steps we’ve just suggested, the result is like applying the operator $d**d=dd=0$.

There’s just one catch: as we’ve written it this gives us a $3$-form, not a function like the divergence operator should! No matter; we can break out the Hodge star once more to flip it back to a $0$-form — a function — just like we want. That is, the divergence operator on $1$-forms is $*d*$.

Let’s calculate this in our canonical basis. If we start with a $1$-form $\alpha=Pdx+Qdy+Rdz$ then we first hit it with the Hodge star:

$\displaystyle*\alpha=Pdy\wedge dz+Qdz\wedge dx+Rdx\wedge dy$

Next comes the exterior derivative:

$\displaystyle d*\alpha=\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)dx\wedge dy\wedge dz$

and then the Hodge star again:

$\displaystyle*d*\alpha=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$

which is exactly the definition (in coordinates) of the usual divergence $\nabla\cdot F$ of a vector field $F$ on $\mathbb{R}^3$.

October 13, 2011

## The Curl Operator

Let’s continue our example considering the special case of $\mathbb{R}^3$ as an oriented, Riemannian manifold, with the coordinate $1$-forms $\{dx, dy, dz\}$ forming an oriented, orthonormal basis at each point.

We’ve already see the gradient vector $\nabla f$, which has the same components as the differential $df$:

\displaystyle\begin{aligned}df&=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz\\\nabla f&=\begin{pmatrix}\displaystyle\frac{\partial f}{\partial x}\\\displaystyle\frac{\partial f}{\partial y}\\\displaystyle\frac{\partial f}{\partial z}\end{pmatrix}\end{aligned}

This is because we use the metric to convert from vector fields to $1$-forms, and with respect to our usual bases the matrix of the metric is the Kronecker delta.

We will proceed to define analogues of the other classical differential operators you may remember from multivariable calculus. We will actually be defining operators on differential forms, but we will use this same trick to identify vector fields and $1$-forms. We will thus not usually distinguish our operators from the classical ones, but in practice we will use the classical notations when acting on vector fields and our new notations when acting on $1$-forms.

Anyway, the next operator we come to is the curl of a vector field: $F\mapsto\nabla\times F$. Of course we’ll really start with a $1$-form instead of a vector field, and we already know a differential operator to use on forms. Given a $1$-form $\alpha$ we can send it to $d\alpha$.

The only hangup is that this is a $2$-form, while we want the curl of a vector field to be another vector field. But we do have a Hodge star, which we can use to flip a $2$-form back into a $1$-form, which is “really” a vector field again. That is, the curl operator corresponds to the differential operator $*d$ that takes $1$-forms back to $1$-forms.

Let’s calculate this in our canonical basis, to see that it really does look like the familiar curl. We start with a $1$-form $\alpha=Pdx+Qdy+Rdz$. The first step is to hit it with the exterior derivative, which gives

\displaystyle\begin{aligned}d\alpha=&dP\wedge dx+dQ\wedge dy + dR\wedge dz\\=&\left(\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy+\frac{\partial P}{\partial z}dz\right)\wedge dx\\&+\left(\frac{\partial Q}{\partial x}dx+\frac{\partial Q}{\partial y}dy+\frac{\partial Q}{\partial z}dz\right)\wedge dy\\&+\left(\frac{\partial R}{\partial x}dx+\frac{\partial R}{\partial y}dy+\frac{\partial R}{\partial z}dz\right)\wedge dz\\=&\frac{\partial P}{\partial y}dy\wedge dx+\frac{\partial P}{\partial z}dz\wedge dx\\&+\frac{\partial Q}{\partial x}dx\wedge dy+\frac{\partial Q}{\partial z}dz\wedge dy\\&+\frac{\partial R}{\partial x}dx\wedge dz+\frac{\partial R}{\partial y}dy\wedge dz\\=&\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dy\wedge dz\\&+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dz\wedge dx\\&+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy\end{aligned}

Next we hit this with the Hodge star. We’ve already calculated how the Hodge star affects the canonical basis of $2$-forms, so this is just a simple lookup to find:

$\displaystyle*d\alpha=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dx+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dy+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dz$

which are indeed the usual components of the curl. That is, if $\alpha$ is the $1$-form corresponding to the vector field $F$, then $*d\alpha$ is the $1$-form corresponding to the vector field $\nabla\times F$.

October 12, 2011

## A Hodge Star Example

I want to start getting into a nice, simple, concrete example of the Hodge star. We need an oriented, Riemannian manifold to work with, and for this example we take $\mathbb{R}^3$, which we cover with the usual coordinate patch with coordinates we call $\{x,y,z\}$.

To get a metric, we declare the coordinate covector basis $\{dx,dy,dz\}$ to be orthonormal, which means that we have the matrix

$\displaystyle g^{ij}=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}$

and also the inner product matrix

$\displaystyle g_{ij}=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}$

since we know that $g_{ij}$ and $g^{ij}$ are inverse matrices. And so we get the canonical volume form

$\displaystyle\omega=\sqrt{\det\left(g_{ij}\right)}dx\wedge dy\wedge dz=dx\wedge dy\wedge dz$

We declare our orientation of $\mathbb{R}^3$ to be the one corresponding to this top form.

Okay, so now we can write down the Hodge star in its entirety. And in fact we’ve basically done this way back when we were talking about the Hodge star on a single vector space:

\displaystyle\begin{aligned}*1&=dx\wedge dy\wedge dz\\ *dx&=dy\wedge dz\\ *dy&=-dx\wedge dz=dz\wedge dx\\ *dz&=dx\wedge dy\\ *(dx\wedge dy)&=dz\\ *(dx\wedge dz)&=-dy\\ *(dy\wedge dz)&=dx\\ *(dx\wedge dy\wedge dz)&=1\end{aligned}

So, what does this buy us? Something else that we’ve seen before in the context of a single vector space. Let’s say that $v$ and $w$ are two vector fields defined on an open subset $U\subseteq\mathbb{R}^3$. We can write these out in our coordinate basis:

\displaystyle\begin{aligned}v&=v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z}\\w&=w_x\frac{\partial}{\partial x}+w_y\frac{\partial}{\partial y}+w_z\frac{\partial}{\partial z}\end{aligned}

Now, we can use our metric to convert these vectors to covectors — vector fields to $1$-forms. We use the matrix $g_{ij}$ to get

\displaystyle\begin{aligned}g(v,\underline{\hphantom{X}})&=v_xdx+v_ydy+v_zdz\\g(w,\underline{\hphantom{X}})&=w_xdx+w_ydy+w_zdz\end{aligned}

Next we can wedge these together

\displaystyle\begin{aligned}g(v,\underline{\hphantom{X}})\wedge g(w,\underline{\hphantom{X}})=&(v_yw_z-v_zw_y)dy\wedge dz\\&+(v_zw_x-v_xw_z)dz\wedge dx\\&+(v_xw_y-v_yw_x)dx\wedge dy\end{aligned}

Now we come to the Hodge star!

\displaystyle\begin{aligned}*(g(v,\underline{\hphantom{X}})\wedge g(w,\underline{\hphantom{X}}))=&(v_yw_z-v_zw_y)dx\\&+(v_zw_x-v_xw_z)dy\\&+(v_xw_y-v_yw_x)dz\end{aligned}

and now we’re back to a $1$-form, so we can use the metric to flip it back to a vector field:

\displaystyle\begin{aligned}g\left(*(g(v,\underline{\hphantom{X}})\wedge g(w,\underline{\hphantom{X}})),\underline{\hphantom{X}}\right)=&(v_yw_z-v_zw_y)\frac{\partial}{\partial x}\\&+(v_zw_x-v_xw_z)\frac{\partial}{\partial y}\\&+(v_xw_y-v_yw_x)\frac{\partial}{\partial z}\end{aligned}

Here, the outermost $g(\underline{\hphantom{X}},\underline{\hphantom{X}})$ is the inner product on $1$-forms, while the inner ones are the inner product on vector fields. This is exactly the cross product of vector fields on $\mathbb{R}^3$.

October 11, 2011

## The Hodge Star in Coordinates

It will be useful to be able to write down the Hodge star in a local coordinate system. So let’s say that we’re in an oriented coordinate patch $(U,x)$ of an oriented Riemannian manifold $M$, which means that we have a canonical volume form that locally looks like

$\displaystyle\omega=\sqrt{\lvert g_{ij}\rvert}dx^1\wedge\dots\wedge dx^n$

Now, we know that any $k$-form on $U$ can be written out as a sum of functions times $k$-fold wedges:

$\displaystyle\eta=\sum\limits_{1\leq i_1<\dots

Since the star operation is linear, we just need to figure out what its value is on the $k$-fold wedges. And for these the key condition is that for every $k$-form $\zeta$ we have

$\displaystyle\zeta\wedge*(dx^{i_1}\wedge\dots\wedge dx^{i_k})=\langle\zeta,dx^{i_1}\wedge\dots\wedge dx^{i_k}\rangle\omega$

Since both sides of this condition are linear in $\zeta$, we also only need to consider values of $\zeta$ which are $k$-fold wedges. If $\zeta$ is not the same wedge as $\eta$, then the inner product is zero, while if $\zeta=\eta$ then

\displaystyle\begin{aligned}(dx^{i_1}\wedge\dots\wedge dx^{i_k})\wedge*(dx^{i_1}\wedge\dots\wedge dx^{i_k})&=\langle dx^{i_1}\wedge\dots\wedge dx^{i_k},dx^{i_1}\wedge\dots\wedge dx^{i_k}\rangle\omega\\&=\det\left(\langle dx^{i_j},dx^{i_k}\rangle\right)\omega\\&=\det\left(\delta^{jk}\right)\omega\\&=\sqrt{\lvert g_{ij}\rvert}dx^1\wedge\dots\wedge dx^n\end{aligned}

And so $*(dx^{i_1}\wedge\dots\wedge dx^{i_k})$ must be $\pm\sqrt{\lvert g_{ij}\rvert}$ times the $n-k$-fold wedge made up of all the $dx^i$ that do not show up in $\eta$. The positive or negative sign is decided by which order gives us an even permutation of all the $dx^i$ on the left-hand side of the above equation.

October 8, 2011

## The Hodge Star on Differential Forms

Let’s say that $M$ is an orientable Riemannian manifold. We know that this lets us define a (non-degenerate) inner product on differential forms, and of course we have a wedge product of differential forms. We have almost everything we need to define an analogue of the Hodge star on differential forms; we just need a particular top — or “volume” — form at each point.

To this end, pick one or the other orientation, and let $(U,x)$ be a coordinate patch such that the form $dx^1\wedge\dots\wedge dx^n$ is compatible with the chosen orientation. We’d like to use this form as our top form, but it’s heavily dependent on our choice of coordinates, so it’s very much not a geometric object — our ideal choice of a volume form will be independent of particular coordinates.

So let’s see how this form changes; if $(V,y)$ is another coordinate patch, we can assume that $U=V$ by restricting each patch to their common intersection. We’ve already determined that the forms differ by a factor of the Jacobian determinant:

$\displaystyle dx^1\wedge\dots\wedge dx^n=\det\left(\frac{\partial x^i}{\partial x^j}\right)dy^1\wedge\dots\wedge dy^n$

What we want to do is multiply our form by some function that transforms the other way, so that when we put them together the product will be invariant.

Now, we already have something else floating around in our discussion: the metric tensor $g$. When we pick coordinates $x^i$ we get a matrix-valued function:

$\displaystyle g^x_{ij}=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)$

and similarly with respect to the alternative coordinates $y^i$:

$\displaystyle g^y_{ij}=g\left(\frac{\partial}{\partial y^i},\frac{\partial}{\partial y^j}\right)$

So, what’s the difference between these two matrix-valued functions? We can calculate two ways:

\displaystyle\begin{aligned}g^x_{ij}&=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)\\&=g\left(\sum\limits_{k=1}^n\frac{\partial y^k}{\partial x^i}\frac{\partial}{\partial y^k},\sum\limits_{l=1}^n\frac{\partial y^l}{\partial x^j}\frac{\partial}{\partial y^l}\right)\\&=\sum\limits_{k,l=1}^n\frac{\partial y^k}{\partial x^i}\frac{\partial y^l}{\partial x^j}g\left(\frac{\partial}{\partial y^k},\frac{\partial}{\partial y^l}\right)\\&=\sum\limits_{k,l=1}^n\frac{\partial y^k}{\partial x^i}\frac{\partial y^l}{\partial x^j}g^y_{kl}\end{aligned}

That is, we transform the metric tensor with two copies of the inverse Jacobian matrix. Indeed, we could have come up with this on general principles, since $g$ has type $(0,2)$ — a tensor of type $(m,n)$ transforms with $m$ copies of the Jacobian and $n$ copies of the inverse Jacobian.

Anyway, now we can take the determinant of each side:

$\displaystyle\lvert g^x_{ij}\rvert=\left\lvert\frac{\partial y^i}{x^j}\right\rvert^2\lvert g^y_{ij}\rvert$

and taking square roots we find:

$\displaystyle\sqrt{\lvert g^x_{ij}\rvert}=\left\lvert\frac{\partial y^i}{x^j}\right\rvert\sqrt{\lvert g^y_{ij}\rvert}$

Thus the square root of the metric determinant is a function that transforms from one coordinate patch to the other by the inverse Jacobian determinant. And so we can define:

$\displaystyle\omega_U=\sqrt{\lvert g^x_{ij}\rvert}dx^1\wedge\dots\wedge dx^n\in\Omega^n_M(U)$

which does depend on the coordinate system to write down, but which is actually invariant under a change of coordinates! That is, $\omega_U=\omega_V$ on the intersection $U\cap V$. Since the algebras of differential forms form a sheaf $\Omega^n_M$, we know that we can patch these $\omega_U$ together into a unique $\omega\in\Omega^n_M(M)$, and this is our volume form.

And now we can form the Hodge star, point by point. Given any $k$-form $\eta$ we define the dual form $*\eta$ to be the unique $n-k$-form such that

$\displaystyle\zeta\wedge*\eta=\langle\zeta,\eta\rangle\omega$

for all $k$-forms $\zeta\in\Omega^k(M)$. Since at every point $p\in M$ we have an inner product and a wedge $\omega(p)\in A^n(\mathcal{T}^*_pM)$, we can find a $*\eta(p)\in A^{n-k}(\mathcal{T}^*_pM)$. Some general handwaving will suffice to show that $*\eta$ varies smoothly from point to point.

October 6, 2011

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Browse Questions

# A company makes 3 model of calculators:A,B and c at factory I and factory II.The company has orders for at least 6400 calculators of model A;4000 calculator of model B and 4800 calculator of model C.At factory I ,50 calculators of model A,50 of model B and 30 of model C are made every day;at factory II,40 calculators of model A,20 of model B and 40 of model C are made every day.It costs Rs 12000 and Rs 15000 each day to operate factory I and II respectively.Find the number of days each factory should operate to minimize the operating costs and still meet the demand.

Toolbox:
• Let $R$ be the feasible region for a linear programming problem and let $z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
• If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
Let $x$ calculators be produced by factory I and $y$ calculators be produced by factory II
Models$\qquad$Factory I$\qquad$Factory II$\qquad$Maximum limit
A $\qquad\qquad$50 $\qquad\qquad$40 $\qquad\qquad\;\;$6400
B $\qquad\qquad$50 $\qquad\qquad$20 $\qquad\qquad\;\;$4000
C $\qquad\qquad$30 $\qquad\qquad$40 $\qquad\qquad\;\;$4800
Hence the total cost for $x$ calculators by factory I and $y$ calculators by factory II is $12000x+15000y$
It is given that factory I can produce 50 calculators and factory II can produce 40 calculators of type A and the two factories can produce at 6400 calculators.
$\therefore 50x+40y\leq 6400$
Step 2:
Similarly factory I produces 50 calculators of type B and factory II produces 20 calculators of type B and the two factories can produce at least 4000 calculators.
$\therefore 50x+20y\geq 4000$
Finally factory I produces 30 calculators of type C and factory II produces 40 calculators of type C and the two factories can produce at least 4800 calculators.
$\therefore 30x+40y\geq 4800$
Hence the objective function $Z=12000x+15000y$
Subject to constraints,
$50x+40y\geq 6400$------(1)
$5x+4y\geq 640$
$50x+20y\geq 4000$------(2)
$30x+40y\geq 4800$
$3x+4y\geq 480$------(3)
and $x,y\geq 0$
Step 3:
The graph can be plotted for the three lines to find the feasible region bounded by threes three lines.
Consider the line$AB :5x+4y\geq 640$ when $x=0,y=0\Rightarrow 5\times 0+6\times 0\geq 640$
Which is not true.
Hence $O(0,0)$ does not belong to the region $5x+4y\geq 640$
It is clear that the region above $5x+4y=640$ represents $5x+4y\geq 640$
Similarly consider the line $EF :3x+4y=480$
Put $x=0,y=0\rightarrow 0\geq 480$
Which is not true.
Hence the region $3x+4y \geq 480$ lies above the line $3x+4y\geq 480$
Finally consider the line $CD :5x+2y=400$
Put $x=0,y=0\rightarrow 0\geq 400$
Which is not true.
Hence the region $5x+2y \geq 400$ lies above the line $5x+2y= 400$
Step 4:
On solving the equations $5x+4y=640$ and $3x+4y=480$ we obtain the point of intersection to be $Q(80,60)$
On solving the equations $5x+4y=640$ and $5x+2y=400$ we obtain the point of intersection to be $P(32,120)$
The point C is point of intersection is C(0,200) and the coordinates of F is (160,0) .Let us obtain the values of the objective function as follows :
Let us obtain the value of the objective function as follows :
At the point $(x,y)$ the value of the objective function $Z=12,000x+15000y$
At $C(0,200)$ the value of the objective function $Z=12000\times 0+15000\times 200=30,00,000$
At $C(32,120)$ the value of the objective function $Z=12000\times 32+15000\times 120=2184000$
At $Q(80,60)$ the value of the objective function $Z=12000\times 80+15000\times 60=1860000$
At $F(160,0)$ the value of the objective function $Z=12000\times 160+15000\times 0=1920000$
It is clear that $Z$ is minimum at $Q(80,60)$
Hence the number of days each factory should operate to minimize the operating cost is 80days by factory I and 60 days by factory II


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IIT JAM Follow
May 21, 2020 7:11 pm 50 pts
36 (vil) The Integral of A Binomial Differential are constants not equal to zero. The integral x (a+bx"P dx A binomial differential is a differential of the form xm (a + bxn)P dx where m, n, peQand a, b is expressible in terms of elementary functions in the following three cases Case 1: (i) If p e N then expand by the formula of Newton Binomial. (i) If p < 0, then we put x = t where k is the common denominator of the fractions m and n. Case 2: If = Integer then we put a + bx = t where a is the denominator of the fraction p. Case 2: If+1 n
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Problem 11-46 Current and potential change

In a particular lightning bolt, $24 \;C$ of charge are transferred in $0.21\; s$ through a potential difference of $5.5 \times10^8\; V.$

(a) What is the average current?
(b) How much electric potential energy is dissipated?

The relationship between charge $(q)$, time $(t)$ and current $(I)$ is:

(A)   $I = qt$

(B)   $I = t/q$

(C)   $I = q/t$


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# Tag Info

0

I offer the following based on Monte Carlo integration, but trying to make use of the facilities Mathematica provides: weights = {2/3, 1/3}; means = {-3, 2}; sd = Sqrt[2]; distributions = NormalDistribution[#, sd] & /@ means; fdist = MixtureDistribution[weights, distributions]; qdist1 = TransformedDistribution[Log[PDF[fdist, f]], Distributed[f, ...

0

Here is a solution via the LibraryLink technique: First, let us make a comparison between BezierNonzeroBasis[n, u] and BernsteinBasis[n, Range[0, n], u] Do[BezierNonzeroBasis[10, 0.1], {10000}] // AbsoluteTiming Do[BernsteinBasis[10, Range[0, 10], 0.1], {10000}] // AbsoluteTiming BezierNonzeroBasis[10000, 0.1]; // AbsoluteTiming BernsteinBasis[10000, ...

1

Association/<||> objects are Atomic and thus unmatchable before 10.4 AtomQ@Association[] yields True. This is confusing because it is not stated anywhere in the manual. For example tutorial/BasicObjects#15871 claims that only numbers (including complex ones), Strings and Symbols are atomic objects. guide/AtomicElementsOfExpressions does not mention ...

2

In general one should not expect to obtain a general symbolic solution (a function x[a,b]) to the given equation since there are two independent variables, see e.g. Solve symbolically a transcendental trigonometric equation and plot its solutions for certain aspects regarding transcendental equations. To get an idea how the solution depends on parameters a ...

2

Perhaps this! It works if you only need the plot and not the values, or the function. That would require more work. ContourPlot3D[ Evaluate[-x + (Log[f2[x]] - Log[f1[x]] + Log[b] - Log[a])/((1 - a) + (1 - 1/f1[x]) - (1 - b) + (1 - 1/f2[x]))] , {a, 0, 1}, {b, 0, 1}, {x, 0, 1} , AxesLabel -> {"a", "b", "x"} , Contours -> {0} , Mesh -> None ...

23

Short answer The local variables of the form varname$... are used by the system, and it is unwise to use symbols with such names as local variables. With, like many other lexical scoping constructs, performs excessive renamings, often even in cases where it isn't strictly necessary. This probably has to do with efficiency - full analysis may be more costly.... 10 There is also a newer package, HolonomicFunctions, that has an implementation of Chyzak's generalization of Zeilberger's algorithm. To perform the desired task, use the following commands: smnd = Simplify[ G /. HoldPattern[HypergeometricPFQ[pl_List, ql_List, x_]] :> (Times @@ (Pochhammer[#, k] & /@ pl)) / (Times @@ (Pochhammer[#, k] & /@ ql)... 3 Perhaps you need to define MTensor for variable left and right. The modified code is: #include "WolframLibrary.h" DLLEXPORT int NonzeroBasis(WolframLibraryData libData, mint Argc, MArgument *Args, MArgument Res) { /*define the argument-varible*/ mreal saved; mint i, p; mreal u; MTensor tensor_U; mreal *U; /*-------------... 2 Is this a "standard" Manipulate which will not grow larger? Ok, use one of already posted answers. Here is an alternative in case where Manipulate is supposed to be generated from a package function or is a part of a bigger code. ClearAll["Test*"]; BeginPackage["Test"] myManipulate::usage = "myManipulate[] generates a demo Manipulate"; Begin["... 12 I appear to be in the minority but I never write big blocks of Mathematica code, it's just too difficult to read. The way I look at it you have to consider how a reader will understand your app. So I make the main block very small, like this: Manipulate[ Row[{ vectorPlotAndTrajectory[y0, b], Show[fy, position[y0, b]] }] , {b, -7.99, 8} , {{y0, 2}... 7 Here is the way it would look if it were my code.My formatting style is much the same as Szabolcs' (including replacing your grid with a column), but I don't like his dangling ], and you won't see those. Also, I always start control specifications on a new line, and I mostly start option specifications on a new line, too. Note: I have redefined ϕ to ... 13 Personally, I use lots of newlines and let the Front End indent. There doesn't seem to be anything special in your code, other than a lot of nesting that is in fact necessary in this case. You were using Grid incorrectly. Grid[a,b] is wrong. Grid[{{a,b}}] or Grid[{{a},{b}}] are correct. I guess you wanted Column[{a,b}], so I changed that. Using Text ... 17 Description In software engineering, it is a good practice to comment your code. I would advise you to utilise comments to partition your code in a following way. Additionally, if you have ever worked with any other programming languages, you could employ indentation. Alternatively, you could modularize your application as to develop it in smaller, more ... 16 Update: The very likely reason for the garbled error messages is that you have a Chinese version of Visual Studio printing errors in Chinese, and there is a mismatch in the character encoding of these messages and how Mathematica tried to interpret them. CreateLibrary has two very useful options: "ShellCommandFunction" and "ShellOutputFunction". Set ... 8 For me (Windows 10) I get a readable error message like So I add mint saved;. Compiling again I get which leads me to change your MArgument_setReal(Res,tensor_N); to MArgument_setMTensor(Res,tensor_N); and then things work. BTW: I use Needs["CCompilerDriver"]; srccode=" #include \"WolframLibrary.h\" DLLEXPORT int NonzeroBasis(WolframLibraryData ... 5 This is a form of memoization by dynamically defining a Pattern with the results of the executed Expression as clarified by @ciao. The standard form is documented in Functions that remember values they have found and looks like: f[x_]:=f[x]=rhs So as an example: f[x_] := f[x] = f[x - 1] + f[x - 2] behaves the same as i:f[x_] := i = f[x - 1] + f[x - ... 11 Well, facinating question! Code first: Clear["*"]; n = 5; steps = 1000; dir = {{0, 0}, {1, 0}, {0, 1}}; start = FromDigits[#, 2] & /@ Transpose@Append[RandomChoice[dir, n - 1], RandomInteger[1, 2]]; move[pt_] :=With[{c = RandomChoice@Pick[-dir, BitAnd @@ (# + pt) & /@ -dir, 0]}, pt + c +RandomChoice@DeleteCases[dir, -c]]; move[{2^n, 0}] := {2^n -... 2 I'm a beginner, and I find it interesting to explore the documentation. Some of the commands I've read about have given me interesting ideas for programs. Also, I'm not sure if this has been mentioned, but the Wolfram Demonstrations Project has a bunch of amazing programs you can play with and download the code from. 4 Use memoization. See here for a description of the memoization in general, and here for its Mathematica implementation. This will avoid having to recalculate all the previous values to determine the next one: Clear[x, T, a, b] T[x_] := T[x] = Piecewise[{{1 - x, 0 <= x < 1/7}, {(x + 6)/7, 1/7 <= x <= 1}}] a[n_] := a[n] = n/(n + 5)^5 b[n_] := b[n] ... 3 The following code will help: cc = ConnectedComponents@g; pts = Map[{#, PropertyValue[{g, #}, VertexCoordinates]} &, cc, {2}]; pre = First@MinimalBy[#, Last] & /@ Apply[{#1[[1]] <-> #2[[1]], Norm[#1[[2]] - #2[[2]]]} &, Tuples /@ Subsets[pts, {2}], {2}]; spt = EdgeList@ FindSpanningTree[ CompleteGraph[Length@cc, EdgeWeight -... 5 Fourier transforms do not return the expected result Many users know a particular definition of the Fourier transform. When Mathematica returns an unexpected result and users cannot comprehend what is happening, often it is because Mathematica is using a different definition. In particular, this is the general definition of a transform pair:$\$ F(\omega) = ...

1

Here is yet another version fl[x0_] := Cases[Range[x0], y_ /; Mod[x0, y] == 0] On my computer my version is slightly faster than everything else offered so far. In particular, I get the following AbsoluteTiming results on fl[10000000]: "Cases": 9.4 sec "Select": 9.8 sec "Table": 10.8 sec "AppendTo": 15.5 sec "Sow": 15.6 sec Added: Using Divisible[x0,...

2

Another approach is to use Table combined with Nothing which is only available in version 10.2 or later. f[x0_] := Table[If[Mod[x0, x] == 0, x, Nothing], {x, 1, x0}] and then fl[10] (* {1, 2, 5, 10} *) I found this to be slightly faster than using Select or the Sow and Reap approach.

2

Use of "Sow" and "Reap" allow a simple modification of the existing code. E.g. fl[x0_] := Module[{x = 1}, While[x <= x0, If[Mod[x0, x] == 0, Sow[x]]; x++]] and Reap[fl[12]] returns {Null, {{1, 2, 3, 4, 6, 12}}}

7

When you're processing an element at a time using procedural constructs like Module, While, If, and Print in Mathematica, it's usually a sign that you're fighting the language rather than using it. It's much easier and more concise to use functional constructs that work on arrays: fl[x0_] := Select[Range[x0], Mod[x0, #] == 0 &]

1

I'm sure there are a lot of ways to generate random functions. Here's one: a := RandomChoice[{Times, Plus}] @@ (RandomChoice[{Sqrt, Identity, Power}] /@ (RandomSample[{x, y, z}, 3] RandomInteger[{1, 5}, 3])^RandomInteger[{-3, 3}, 3]); To make a matrix of such functions: Table[a, 3, 3] // MatrixForm Of course the output is different each time. ...

5

Preamble Leonid's method is new to me and quite interesting. I expect that as with most of his methods it is well reasoned and has advantages that are not immediately apparent. Nevertheless I also find value in alternative methods, so here is one of mine. I shall use his example code so that these methods may be compared directly. Boilerplate General::...

12

Implementation This is indeed an important problem. It is usually best to have a separate function testing various options. Here is the solution I propose: a wrapper that would factor out the testing functionality from the main function. Here is the code: ClearAll[OptionCheck]; OptionCheck::invldopt = "Option 1 for function 2 received invalid value 3`"...

2

It was clarified in comments that expressions are supposed to be read into Mathematica from a text file. To achieve what OP wants one can read the file as list of held expressions, perform appropriate replacements, then release hold. If we have a test.txt file with following contents: a * b * c * d a^dag We can read it in following way: ReadList["test....

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# Monkey Monkey

The monkey B is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkey A and B are 5Kg and 2Kg respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry the monkey B with it. Let the force be F such that a<F<b , Find (a+b) $$**Assumptions**$$ Take g = $$10m/s^{2}$$

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In class I was taking notes about equivalence relations defined as:

Given a generic relation $$R$$ on $$X$$, $$xIy$$ if both $$xRy$$ and $$yRx$$

Now, I don't really understand the following proposition:

If $$R$$ (the original relation) is reflexive and transitive, then $$I$$ (the derived relation) is reflexive, symmetric and transitive.

I understand this is an instance of $$\succsim$$ where the derived relations are $$\sim$$ and $$\succ$$, but the proposition holds only for the former (the strict preference obviously is not symmetric). Why is that so?

Also, how can I prove that $$R$$ having the two properties leads to $$I$$ having the same two properties + symmetry?

Q1. Why is $$\succ$$ not symmetric?

Q2. Why is $$\sim$$ (a) symmetric; (b) reflexive; and (c) transitive?

A1. Suppose $$x \succ y$$. Then by Definition 4 (refer to basic setup and definitions below), $$x \succsim y$$ and $$y \not\succsim x$$. Therefore, by the same definition, $$y \not\succ x$$. Thus, $$\succ$$ is not symmetric.

A2. Let $$x,y,z\in X$$.

A2(a)$$\hspace{10pt}$$ $$x\sim y$$ $$\hspace{10pt}$$ $$\iff$$ $$\hspace{10pt}x\succsim y$$ and $$y\succsim x$$ (Definition 3) $$\hspace{10pt}$$ $$\iff$$ $$\hspace{10pt}y\sim x$$.

A2(b) By Definition 2, $$x \succsim x$$. And so, by Definition 3, $$x \sim x$$.

A2(c) Suppose $$x\sim y$$ and $$y\sim z$$. (We will show that $$x\sim z$$.)

By Definition 2, we have $$x\succsim y$$ and $$y\succsim z$$. And now by the transitivity of $$\succsim$$, we have $$x\overset{1}{\succsim} z$$.

Similarly, we also have $$y\succsim x$$ and $$z\succsim y$$. And so again by the transitivity of $$\succsim$$, we have $$z\overset{2}{\succsim} x$$.

Given $$\overset{1}{\succsim}$$ and $$\overset{2}{\succsim}$$, by Definition 3, we have $$x\sim z$$.

Basic setup and definitions.

Definition 1. A (binary) relation $$R$$ (on a set $$X$$) is:

• Symmetric if for all $$x,y\in X$$, $$xRy \iff yRx$$;

• Reflexive if for all $$x\in X$$, $$xRx$$;

• Transitive if for all $$x,y,z\in X$$, $$xRy$$ and $$yRz$$ implies $$xRz$$.

Let $$X$$ be our set of alternatives.

Definition 2. Let $$\succsim$$ be a reflexive and transitive relation on $$X$$.

We interpret $$\succsim$$ as the weakly-preferred-to relation.

Definition 3. $$x \sim y$$ if $$x \succsim y$$ and $$y \succsim x$$.

We interpret $$\sim$$ as the indifference relation.

Definition 4. $$x \succ y$$ if $$x \succsim y$$ and $$y \not\succsim x$$.

We interpret $$\succ$$ as the strictly-preferred-to relation.

• The defining characteristics of the weak preference relation $\succsim$ is its transitivity and completeness. The reflexivity of $\succsim$ is a consequence. – Kenneth Rios Sep 30 '18 at 1:48
• @KennethRios: I wanted to directly address the question(s), without also having to prove that reflexivity is a consequence. The question(s) included, in particular, the following: "Now, I don't really understand the following proposition: If $R$ (the original relation) is reflexive and transitive, then $I$ (the derived relation) is reflexive, symmetric and transitive." – user18 Sep 30 '18 at 2:03
• Sure. Just preempting confusion in case OP wrongly thinks that all relations that are both reflexive and transitive are preference relations. – Kenneth Rios Sep 30 '18 at 3:19

If both $$xRy$$ and $$yRx$$ $$\forall x,y \in X$$, then the indifference relation $$I$$ is symmetric by definition. Not all symmetric relations are equivalence relations. However, you are given that $$R$$ is also reflexive and transitive. So we have that $$I$$ is symmetric, reflexive, and transitive. Therefore $$I$$, or $$\sim$$, is an equivalence relation.

You can verify that $$\sim$$ is an equivalence relation by reading off the definitions of the three properties. $$R$$, or $$\succsim$$, is a (weak) preference relation, also known as a total preorder. Total preorders are transitive, complete, and therefore reflexive.

Keep in mind that reflexivity and transitivity does not necessarily imply symmetry: $$R = \{(x,x), (y,y), (x,y)\}$$ given $$X = \{x,y\}$$ is clearly reflexive and transitive but not symmetric because $$xRy$$ but not $$yRx$$.


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If you find any mistakes, please make a comment! Thank you.

## The group of units of a field acts on the field by left multiplication

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.1

Solution: Let $a \in F$. We have $1 \cdot a = 1a = a$ by the definition of identity. Now let $g_1, g_2 \in F^\times$. Then $$g_1 \cdot (g_2 \cdot a) = g_1 \cdot g_2 a = g_1(g_2a) = (g_1g_2)a = (g_1g_2) \cdot a,$$ by the associativity of multiplication in $F$. So left multiplication by nonzero elements is in fact a group action.

#### Linearity

1. No. There is no need to consider $a=0$ as $a\notin F^\times$.


text:
# Give an example of each or state that the request is impossible. For a

Give an example of each or state that the request is impossible. For any that are impossible, give a compelling argument for why that is the case.
a) A sequence with an infinite number of ones that does not converge to one.
b) A sequence with an infinite number of ones that converges to a limit not equal to one.
c) A divergent sequence such that for every $$\displaystyle{n}\in{N}$$ it is possible to find nn consecutive ones somewhere in the sequence.

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Kayla Kline

Step 1
Take sequence $$\displaystyle{x}_{{{n}}}={\left(-{1}\right)}^{{{n}}}$$ where 1 occurs infinitely many times but does not converge to 1.
This is because if we choose $$\displaystyle\epsilon\le{2}$$, given any $$\displaystyle{N}\in{\mathbb{{{N}}}}$$, where exists $$\displaystyle{n}={2}{N}+{1}$$ such that
$$\displaystyle{\left|{\left({1}\right)}^{{{n}}}-{1}\right|}={2}\geq\epsilon$$
Step 2
b) Suppose $$\displaystyle{x}_{{{n}}}$$ is a convergent sequence with infinitely many 1 but converges to $$\displaystyle{x}\ne{1}$$. Let $$\displaystyle\epsilon={\left|{1}-{x}\right|},$$ then for any $$\displaystyle{N}\in{\mathbb{{{N}}}}$$, there exist $$\displaystyle{n}\geq{N}$$ such that $$\displaystyle{x}_{{{n}}}={1}$$. So
$$\displaystyle{\left|{x}_{{{n}}}-{x}\right|}={\left|{1}-{x}\right|}\geq\epsilon$$
So this is not possible
Step 3
c) Consider a sequence $$\displaystyle{\left({x}_{{{n}}}\right)}$$ given by
$$x_{n}=\begin{cases}r & if\ n=\frac{r(r+1)}{2}-1\ \text{for some}\ r\in\mathbb{N}\\ 1 & \text{otherwise}\end{cases}$$
(or put any thing in place of r other than 1)
There n consecutive 1's occur in the sequence. Some of the terms in sequence are
$$\displaystyle{\left({x}_{{{n}}}\right)}_{{{n}\in{\mathbb{{{N}}}}}}={\left({1},\ {2},\ {1},\ {1},\ {5},\ {1},\ {1},\ {1},\ {9},\ {1},\ {1},\ {1},\ {1},\ {14},\ \cdots\right)}$$
This is constructed this way
$$\begin{matrix} 1 & 2 \\ 1 & 1 & 5 \\ 1 & 1 & 1 & 9 \\ \cdots\end{matrix}$$

###### Not exactly what you’re looking for?
scoollato7o

Step 1
a) Consider the sequence $$\displaystyle{x}_{{{n}}}={\left(-{1}\right)}^{{{n}}}$$
Here the occurrence of 1 is infinite times and it does not converge to 1.
That is, on choosing $$\displaystyle\epsilon\le{2}$$, given any $$\displaystyle{N}\in{\mathbb{{{N}}}}$$, there exist $$\displaystyle{n}={2}{N}+{1}$$ such that $$\displaystyle{\left|{\left({1}\right)}^{{{n}}}-{1}\right|}={2}\geq\epsilon$$
Step 2
b) Consider a convergent sequence $$\displaystyle{\left({x}_{{{n}}}\right)}$$ with infinitely many 1 but it converges to $$\displaystyle{x}\ne{1}$$.
Let $$\displaystyle\epsilon={\left|{1}-{x}\right|}$$ then for any $$\displaystyle{N}\in{\mathbb{{{N}}}}$$, there exist $$\displaystyle{n}\geq{N}$$ such that $$\displaystyle{x}_{{{n}}}={1}$$. So,
$$\displaystyle{\left|{x}_{{{n}}}-{1}\right|}={\left|{1}-{x}\right|}\geq\epsilon$$
Thus, this is not possible.
Step 3
c) Consider a sequence $$\displaystyle{\left({x}_{{{n}}}\right)}$$ given by
$x_{n}=\begin{cases}r & if\ n=\frac{r(r+1)}{2}-1\ \text{for some} r\in\mathbb{N}\\ 1 & \text{otherwise}\end{cases}$
There exist n consecutive 1's occur in the sequence. Some of the terms in the sequence will be
$(x_{n})_{n\in\mathbb{N}}=(1,\ 2,\ 1,\ 1,\ 5,\ 1,\ 1,\ 1,\ 9,\ 1,\ 1,\ 1,\ 1,\ 14,\ \cdots)$
That is,
$$\begin{matrix} 1 & 2 \\ 1 & 1 & 5 \\ 1 & 1 & 1 & 9 \\ \cdots \end{matrix}$$
Thus, it is impossible.

content_user

Step 1
a) Consider the sequence
$$\{a_{n}\}=\{(-1)^{n}\}=\{-1,\ 1,\ -1,\ 1,\ \cdots\}$$
$$a_{n}$$ has an infinite number of ones. Indeed, if n is even, $$(-1)^{n}=1$$, and there are infinite even numbers, hence infinite ones. However, $$\{a_{n}\}$$ does not converge to one, because $$\{a_{n}\}$$ is divergent (it oscillates between 1 and -1).
b) This is impossible. We will use the definition of limit to see why.
Let $$\epsilon>0$$. Suppose that $$\{a_{n}\}$$ has an infinite number of ones, and suppose that $$\{a_{n}\}$$ converges to L. Then, there exists some M>0 such that if $$m\geq M$$ then $$|a_{n}-L|<\epsilon$$. Now, there exist infinite natural numbers $$n\geq M$$ such that $$a_{n}=1$$ by hypotheses. Then $$|1-L|<\epsilon$$
Hence L and 1 get arbitrarily close, therefore L=1
c) It is possible to construct such a sequence:
$$\{a_{n}\}=\{1,\ 0,\ 1,\ 1,\ 0,\ 1,\ 1,\ 1,\ 0,\ 1,\ 1,\ 1,\ 1,\ 0,\ 1,\ 1,\ 1,\ 1,\ 1,\ 0,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 0,\ \cdots\}$$
even if there is not a simple explicit formula. The sequence is divergent because it has infinite zeroes and infinite ones (thus L should be either 0 or 1, but it can't be both).


text:
# Finding the Satellite Orbital Radius

1. Oct 25, 2014

### logan3

1. The problem statement, all variables and given/known data
If NASA wants to put a satellite in a circular orbit around the sun so it will make 2.0 orbits per year, at what distance (in astronomical units, AU) from the sun should that satellite orbit? The earth's orbit is 1.0 AU from the sun.

$T_E = \frac {1.0orbit}{yr}$
$T_s = \frac {2.0orbit}{yr}$
$r_E = 1.0AU$

2. Relevant equations
$\frac {T_E^2} {r_E^3} = \frac {T_s^2} {r_s^3} => r_s = \sqrt[3]{\frac {T_s^2 r_E^3}{T_E^2}}$

3. The attempt at a solution
Using Kepler’s 3rd law of planetary motion, we know that the orbital period squared is proportional to the radius of the orbit cubed. Since the proportion is approximately the same for all objects orbiting relative to the Sun, then: $\frac {T_E^2} {r_E^3} = \frac {T_s^2} {r_s^3}$

Solving for the radius of the satellite and plugging in the values, we get:
$r_s = \sqrt[3]{\frac {T_s^2 r_E^3}{T_E^2}} = \sqrt[3]{\frac {(\frac {2.0orbit}{yr})^2 (1.0AU)^3}{(\frac {1.0orbit}{yr})^2}} = \sqrt[3]{4.0AU^3} = 1.5874AU \sim 1.6AU$

Thank-you

Last edited: Oct 25, 2014
2. Oct 25, 2014

### ehild

The satellite makes 2 orbits per year, what is its orbital period?

3. Oct 25, 2014

### SteamKing

Staff Emeritus
The further away from the sun, the longer the orbital period. That's Kepler's (unpublished) Fourth Law.

4. Oct 25, 2014

### logan3

$T_s = \frac {2 orbits} {1yr} x \frac {2\pi} {1 orbit} x \frac {1yr} {365days} x \frac {1day} {24hrs} x \frac {1hr} {3600s} = \frac {\pi} {7.884x10^6s}$
$G = 6.67x10^{-11} Nm^2/kg^2$
$M_\odot = 1.9885x10^{30} kg$
$1 AU = 1.4959x10^{11} m$

$T_s^2 = \frac {4\pi^2 r^3}{GM_\odot} \Rightarrow r = \sqrt[3]{\frac {T_s^2 GM_\odot}{4\pi^2}} = 81.102m x \frac {1 AU}{1.4959x10^{11} m} = 5.4216x10^{-10}AU$

5. Oct 25, 2014

### Staff: Mentor

You don't need meters. The approach in post 1 was fine, you just plugged in the wrong orbital period.
And no, an orbital radius of a few meters is certainly wrong (this would be inside the sun!!).

T is the duration of an orbit, not the number of orbits per time (that is the inverse value).

6. Oct 25, 2014

### ehild

2 total orbits last one year, how long time is one orbit? (in years)

7. Oct 25, 2014

### SteamKing

Staff Emeritus
5.4216*10-10AU works out to be about 81 meters. Does this sound like a reasonable answer?

If you make 2 orbits per year, the orbital period is going to be a half year for each orbit.

You know that the earth's orbital period = 1 year (it's kind of a definition) and you have Kepler's third law which relates orbital distance to orbital period.

Try again, and put a little thought into your calculations.

8. Oct 25, 2014

### logan3

For the satellite, one orbit (in years) is half a year. Thus,

$T_E = 1yr$
$T_s = 0.5yr$
$r_E = 1.0AU$

$r_s = \sqrt[3]{\frac {T_s^2 r_E^3}{T_E^2}} = \sqrt[3]{\frac {(0.5 yr)^2 (1.0 AU)^3}{(1 yr)^2}} = \sqrt[3]{0.25 AU^3} = 0.62996 AU \sim 0.63 AU$

In order for the orbital period to decrease, the object must be closer to the Sun. 0.63AU is closer than the Earth and has a shorter orbital period (0.5 yr), therefore, numerically the result corresponds to the aforementioned reasoning.

Using the second method:
$T_s = 0.5 yr x \frac {365 days} {1 yr} x \frac {24 hrs} {1day} x \frac {3600 s} {1 hr} = 1.5768x10^7 s$
$G = 6.67x10^{-11} Nm^2/kg^2$
$M_\odot = 1.9885x10^{30} kg$
$1 AU = 1.4959x10^{11} m$

$T_s^2 = \frac {4\pi^2 r^3}{GM_\odot} \Rightarrow r = \sqrt[3]{\frac {T_s^2 GM_\odot}{4\pi^2}} = 9.4177x10^{11} m x \frac {1 AU}{1.4959x10^{11} m} = 0.62957 AU \sim 0.63 AU$

The answers from both methods match.

Thank-you