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# Large potential reduction in economic damages under UN mitigation targets

## Abstract

International climate change agreements typically specify global warming thresholds as policy targets1, but the relative economic benefits of achieving these temperature targets remain poorly understood2,3. Uncertainties include the spatial pattern of temperature change, how global and regional economic output will respond to these changes in temperature, and the willingness of societies to trade present for future consumption. Here we combine historical evidence4 with national-level climate5 and socioeconomic6 projections to quantify the economic damages associated with the United Nations (UN) targets of 1.5 °C and 2 °C global warming, and those associated with current UN national-level mitigation commitments (which together approach 3 °C warming7). We find that by the end of this century, there is a more than 75% chance that limiting warming to 1.5 °C would reduce economic damages relative to 2 °C, and a more than 60% chance that the accumulated global benefits will exceed US$20 trillion under a 3% discount rate (2010 US dollars). We also estimate that 71% of countries—representing 90% of the global population—have a more than 75% chance of experiencing reduced economic damages at 1.5 °C, with poorer countries benefiting most. Our results could understate the benefits of limiting warming to 1.5 °C if unprecedented extreme outcomes, such as large-scale sea level rise8, occur for warming of 2 °C but not for warming of 1.5 °C. Inclusion of other unquantified sources of uncertainty, such as uncertainty in secular growth rates beyond that contained in existing socioeconomic scenarios, could also result in less precise impact estimates. We find considerably greater reductions in global economic output beyond 2 °C. Relative to a world that did not warm beyond 2000–2010 levels, we project 15%–25% reductions in per capita output by 2100 for the 2.5–3 °C of global warming implied by current national commitments7, and reductions of more than 30% for 4 °C warming. Our results therefore suggest that achieving the 1.5 °C target is likely to reduce aggregate damages and lessen global inequality, and that failing to meet the 2 °C target is likely to increase economic damages substantially. Your institute does not have access to this article ## Relevant articles Open Access articles citing this article. • ### Informing environmental health and risk priorities through local outreach and extension Environment Systems and Decisions Open Access 02 June 2022 • ### National attribution of historical climate damages Climatic Change Open Access 01 June 2022 • ### Contrasting influences of biogeophysical and biogeochemical impacts of historical land use on global economic inequality Nature Communications Open Access 05 May 2022 ## Access options Buy article Get time limited or full article access on ReadCube.$32.00

All prices are NET prices.

## References

1. UNFCCC Adoption of the Paris Agreement. I: Proposal by the President. Draft Decision CP.21 https://unfccc.int/resource/docs/2015/cop21/eng/l09r01.pdf (United Nations Office, Geneva, 2015).

2. Nature Editorial Board. The maximum climate ambition needs a firm research backing. Nature 537, 585–586 (2016).

3. Hallegatte, S. et al. Mapping the climate change challenge. Nat. Clim. Chang. 6, 663–668 (2016).

4. Burke, M., Hsiang, S. M. & Miguel, E. Global non-linear effect of temperature on economic production. Nature 527, 235–239 (2015).

5. Collins, M. et al. in Climate Change 2013: the Physical Science Basis: Working Group I Contribution to the Fifth Assessment Report of the Intergovernmental Panel on Climate Change (eds Stocker, T. et al.) Ch. 12 (Cambridge Univ. Press, Cambridge, 2014).

6. Riahi, K., van Vuuren, D. P. & Kriegler, E. The shared socioeconomic pathways and their energy, land use, and greenhouse gas emissions implications: an overview. Glob. Environ. Change 42, 153–168 (2017).

7. Rogelj, J. et al. Paris agreement climate proposals need a boost to keep warming well below 2 °C. Nature 534, 631–639 (2016).

8. Schleussner, C.-F. et al. Science and policy characteristics of the Paris Agreement temperature goal. Nat. Clim. Chang. 6, 827–835 (2016).

9. Rogelj, J., McCollum, D. L., Reisinger, A., Meinshausen, M. & Riahi, K. Probabilistic cost estimates for climate change mitigation. Nature 493, 79–83 (2013).

10. Hallegatte, S., Fay, M., Bangalore, M., Kane, T. & Bonzanigo, L. Shock Waves: Managing the Impacts of Climate Change on Poverty (World Bank Publications, Washington, 2015).

11. Dell, M., Jones, B. F. & Olken, B. A. Temperature shocks and economic growth: evidence from the last half century. Am. Econ. J. Macroecon. 4, 66–95 (2012).

12. Schlenker, W. & Roberts, M. J. Nonlinear temperature effects indicate severe damages to US crop yields under climate change. Proc. Natl Acad. Sci. USA 106, 15594–15598 (2009).

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14. Carleton, T. A. & Hsiang, S. M. Social and economic impacts of climate. Science 353, aad9837 (2016).

15. Deryugina, T. & Hsiang, S. M. Does the Environment Still Matter? Daily Temperature and Income in the United States. NBER working paper 20750 (National Bureau of Economic Research, Cambridge, 2014); http://www.nber.org/papers/w20750.

16. Colacito, R., Hoffmann, B. & Phan, T. Temperatures and Growth: A Panel Analysis of the United States. IDB working paper IDB-WP-676 (Inter-American Development Bank, Washington, 2016); https://publications.iadb.org/handle/11319/7654.

17. Zhang, P., Deschenes, O., Meng, K. & Zhang, J. Temperature effects on productivity and factor reallocation: evidence from a half million Chinese manufacturing plants. J. Environ. Econ. Manage. 88, 1–17 (2018).

18. Taylor, K. E., Stouffer, R. J. & Meehl, G. A. An overview of CMIP5 and the experiment design. Bull. Am. Meteorol. Soc. 93, 485–498 (2012).

19. O’Neill, B. C., Kriegler, E. & Ebi, K. L. The roads ahead: narratives for shared socioeconomic pathways describing world futures in the 21st century. Glob. Environ. Change 42, 169–180 (2017).

20. Burke, M. et al. Opportunities for advances in climate change economics. Science 352, 292–293 (2016).

21. Seneviratne, S. I., Donat, M. G., Pitman, A. J., Knutti, R. & Wilby, R. L. Allowable CO2 emissions based on regional and impact-related climate targets. Nature 529, 477–483 (2016).

22. Cai, Y., Lenton, T. M. & Lontzek, T. S. Risk of multiple interacting tipping points should encourage rapid CO2 emission reduction. Nat. Clim. Chang. 6, 520–525 (2016).

23. Hsiang, S. et al. Estimating economic damage from climate change in the United States. Science 356, 1362–1369 (2017).

24. Hof, A. F. et al. Global and regional abatement costs of nationally determined contributions (NDCs) and of enhanced action to levels well below 2 °C and 1.5 °C. Environ. Sci. Policy 71, 30–40 (2017).

25. Mastrandrea, M. D. et al. The IPCC AR5 guidance note on consistent treatment of uncertainties: a common approach across the working groups. Clim. Change 108, 675 (2011).

26. Matsuura, K. & Willmott, C. J. Terrestrial Air Temperature: 1900–2010 Gridded Monthly Time Series (version 3.01) (Center for Climatic Research, Univ. Delaware, Newark, 2012); http://climate.geog.udel.edu/~climate/.

27. Tebaldi, C. & Knutti, R. The use of the multi-model ensemble in probabilistic climate projections. Phil. Trans. R. Soc. Lond. A 365, 2053–2075 (2007).

28. Balk, D. et al. Determining global population distribution: methods, applications and data. Adv. Parasitol. 62, 119–156 (2006).

29. Hartmann, D. L. et al. Observations: atmosphere and surface. In Climate Change 2013: the Physical Science Basis: Working Group I Contribution to the Fifth Assessment Report of the Intergovernmental Panel on Climate Change (eds Stocker, T. et al.) (Cambridge Univ. Press, Cambridge, 2013).

30. Stern, N. The Economics of Climate Change: The Stern Review. (Cambridge Univ. Press, Cambridge, 2006).

31. Weitzman, M. A review of the Stern review on the economics of climate change. J. Econ. Lit. 45, 703–724 (2007).

32. Nordhaus, W. D. A Question of Balance: Weighing the Options on Global Warming Policies. (Yale Univ. Press, Yale, 2008).

33. Goulder, L. H. & Williams, R. C. III The choice of discount rate for climate change policy evaluation. Clim. Change Econ. 3, 1250024 (2012).

34. Arrow, K. J. et al. Should governments use a declining discount rate in project analysis? Rev. Environ. Econ. Policy 8, 145–163 (2014).

35. Newell, R. G. & Pizer, W. A. Discounting the distant future: how much do uncertain rates increase valuations? J. Environ. Econ. Manage. 46, 52–71 (2003).

36. Groom, B., Koundouri, P., Panopoulou, E. & Pantelidis, T. Discounting the distant future: how much does model selection affect the certainty equivalent rate? J. Appl. Econ. 22, 641–656 (2007).

37. Greenstone, M., Kopits, E. & Wolverton, A. Developing a social cost of carbon for us regulatory analysis: a methodology and interpretation. Rev. Environ. Econ. Policy 7, 23–46 (2013).

38. Quiggin, J. Stern and his critics on discounting and climate change: an editorial essay. Clim. Change 89, 195–205 (2008).

39. Henderson, J. V., Storeygard, A. & Weil, D. N. Measuring economic growth from outer space. Am. Econ. Rev. 102, 994–1028 (2012).

40. Jean, N. et al. Combining satellite imagery and machine learning to predict poverty. Science 353, 790–794 (2016).

## Acknowledgements

We thank L. Goulder, S. Hsiang and D. Lobell for comments. We thank the Erol Foundation for funding.

### Reviewer information

Nature thanks R. Kopp and the other anonymous reviewer(s) for their contribution to the peer review of this work.

## Author information

Authors

### Contributions

Author contributions: M.B. and N.S.D. designed the research. M.B., W.M.D. and N.S.D. analysed data, interpreted results and wrote the paper.

### Corresponding author

Correspondence to Marshall Burke.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Extended data figures and tables

### Extended Data Fig. 1 Discount rate scenarios used in calculation of cumulative discounted impacts of future warming.

a, Projected global average annual growth rates under SSP1 with and without climate change; estimates are averaged across bootstraps and climate models. Projected growth rates with climate change are used to define future consumption growth in Ramsey-based discount rates. b, Evolution of discount rates under different schemes through 2099. Ramsey-based schemes are Stern30, Weitzman31 and Nordhaus32, with corresponding assumptions about the pure rate of time discount ρ and the elasticity of marginal utility of consumption η shown in parentheses. Dashed lines are versions of these Ramsey-based discounting schemes that account for growth-rate uncertainty. Non-Ramsey schemes are Newell and Pizer35 and Groom36. c, Projected average annual growth rates separately for rich and poor countries under SSP1, with and without climate change. d, Corresponding Ramsey-based discount rates calculated separately for rich and poor countries, using income-specific growth rates from c.

### Extended Data Fig. 2 Global GDP impacts can be negative at +1 °C but positive at +2 °C for some high-temperature-optimum bootstrap runs.

a, b, Country share of global GDP at baseline (a) and by the end of the century (b) under SSP1, assuming no climate change. c, Distribution of global GDP by temperature, under baseline (black) and the end of the century SSP1 without climate change (red dashed); absent climate change, a substantial portion of global GDP is projected to be produced in countries with hotter average temperatures. d, Climate-model-predicted average global warming under RCP2.6 by the end of the century (x axis) versus the correlation between country-level baseline average temperature and country-level predicted warming in each model. In models that warm less at the global scale, countries that are currently warm tend to exhibit relatively larger warming, while in models that warm more at the global scale, countries that are currently cool tend to exhibit relatively larger warming. Future impacts on global GDP are a sum of country-specific impacts, which are a function of where each country is on the temperature response function (Fig. 1a) and the projected amount of future warming in that country; a given percentage impact in a country with a large GDP has a larger effect on global GDP than the same percentage impact in a country with small GDP. For high-temperature-optimum response functions (for example, Fig. 1g), impacts can be negative at +1 °C but positive at +2 °C because (i) absent climate change, a much larger proportion of total global GDP is projected by SSP1 to be produced in countries that are currently warmer than the optimum, and (ii) climate models with lower overall global warming projections under RCP2.6 tend to have higher relative warming in countries that are currently warm. This generates negative impacts at about 1 °C, where impacts are dominated by negative effects in warm countries (largely in the developing world), but positive impacts at about 2 °C, where high-latitude countries instead warm disproportionately and experience benefits that outweigh the damages in tropical countries.

### Extended Data Fig. 3 Change in cumulative global GDP under 1.5 °C versus 2 °C global warming by the end of the century under different discounting schemes.

Positive values indicate benefits (reduced losses) at 1.5 °C versus 2 °C. Each vertical line corresponds to a bootstrap estimate of benefits under each discounting scheme30,31,32,35,36. Red lines indicate median across bootstraps for each discounting scheme. Uniform schemes correspond to those in Extended Data Table 1; other schemes are described in Methods.

### Extended Data Fig. 4 Robustness of results to alternative specifications.

Change in global GDP per capita in 2049 and 2099 based on regression models that include 0, 1 or 5 lags (a and b); bootstrap schemes that sample by country, five-year block or single year (c and d); or alternative SSPs (e and f). Top panels show percentage changes in global GDP per capita under 1.5 °C versus 2 °C; the bottom panels show change in cumulative global GDP in US\$ trillions under a 3% discount rate.

### Extended Data Fig. 5 Robustness under alternative warming paths.

Benefit—in terms of per capita GDP (a) and cumulative GDP (b)—of 1.5 °C versus 2 °C by end of century under the baseline assumption that overall projected warming occurs linearly between the baseline year and 2099 (pink), versus projected benefit assuming that all projected warming occurs by 2049 and temperatures remain constant thereafter (blue). Both scenarios have the same projected global warming by the end of the century. For the same level of overall warming by the end of the century, scenarios with rapid initial warming worsen the overall impacts of climate change and increase the cumulative benefits of limiting warming to 1.5 °C versus 2 °C.

### Extended Data Fig. 6 Projected change in global GDP (%) under global warming by the end of the century, for each SSP.

Panels ae show the change in GDP for different climate models under different RCP forcing scenarios, relative to a no-warming baseline (median bootstrap) for SSPs 1–5, respectively. Results are as in Fig. 4a, but for each SSP. Each dot represents an RCP-climate model projected change in global GDP under a given SSP; colours represent the four RCPs. Lines are least-squares fits to the points corresponding to the different RCPs with matching colour scheme. The three vertical black lines denote the 1.5 °C target, the 2 °C target and the median-estimated warming expected under current Paris commitments (2.9 °C)7. Warming is relative to pre-industrial levels.

## Rights and permissions

Reprints and Permissions

Burke, M., Davis, W.M. & Diffenbaugh, N.S. Large potential reduction in economic damages under UN mitigation targets. Nature 557, 549–553 (2018). https://doi.org/10.1038/s41586-018-0071-9

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• Published:

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• DOI: https://doi.org/10.1038/s41586-018-0071-9

• ### Contrasting influences of biogeophysical and biogeochemical impacts of historical land use on global economic inequality

• Shu Liu
• Yong Wang
• Le Yu

Nature Communications (2022)

• ### Assessing climate change impacts on tourism demand in Turkey

• Aysun Aygün Oğur
• Tüzin Baycan

Environment, Development and Sustainability (2022)

• ### National attribution of historical climate damages

• Christopher W. Callahan
• Justin S. Mankin

Climatic Change (2022)

• ### Informing environmental health and risk priorities through local outreach and extension

• Khara Grieger
• Christopher L. Cummings

Environment Systems and Decisions (2022)

• ### Climate action with revenue recycling has benefits for poverty, inequality and well-being

• Mark Budolfson
• Francis Dennig
• Stéphane Zuber

Nature Climate Change (2021)


text:
Question

# Luka and Anja each measured the height of their friend three times. Their friend is 59 inches tall. They recorded their measurements as shown. Luka: 5

Measurement
Luka and Anja each measured the height of their friend three times. Their friend is 59 inches tall. They recorded their measurements as shown. Luka: 59 in., 58, in., 58 in. Anja: 59.3 in., 59.6 in., 58.2 in. Which statement is true?
Luka’s measurements are more precise and more accurate. Anja’s measurements are more precise and more accurate. Luka’s measurements are more precise, but Anja’s measurements are more accurate. Luka’s measurements are more accurate, but Anja’s measurements are more precise

2020-12-16

Step 1
Define accuracy and precision of values.
An accurate value is judged to be closer to true to value, whereas precision is the consistency between values obtained by repeated measurements.
Here, Luka's measurements are 59 in,58 in, 58 in and Anja's measurements are 59.3 in, 59.6 in,58.2 in
Step 2
The average of Luka's measurements is,
$$A=\frac{59+58+58}{3}=58.33$$in.
The average of Anja's measurements is,
$$A=\frac{59.3+59.6+58.2}{3}=59.03$$in.
The Anja's measurements are more close towards true value. Thus, it is more accurate.
The recordings of Luka is more precise, because the degree of closeness is high compared to Anja.
Therefore, Option (C) is correct.

format: plain


text:
# Sever-sort an array

## Challenge

Given a non-empty array of integers, e.g.:

[5, 2, 7, 6, 4, 1, 3]

First sever it into arrays where no item is larger than the previous (i.e. non-ascending arrays):

[5, 2] [7, 6, 4, 1] [3]

Next, reverse each array:

[2, 5] [1, 4, 6, 7] [3]

Finally, concatenate them all together:

[2, 5, 1, 4, 6, 7, 3]

This should be what your program outputs/function returns. Repeat this procedure enough times and the array will be fully sorted.

## Rules

• Input and output may be given through any standard methods, and may be in any reasonable array format.
• The input array will never be empty, but may contain negatives and/or duplicates.
• The absolute value of each integer will always be less than 231.

## Test cases

Hopefully these cover all edge cases:

[1] -> [1]
[1, 1] -> [1, 1]
[1, 2] -> [1, 2]
[2, 1] -> [1, 2]
[2, 3, 1] -> [2, 1, 3]
[2, 1, 3] -> [1, 2, 3]
[2, 1, 2] -> [1, 2, 2]
[2, 1, 1] -> [1, 1, 2]
[3, 1, 1, 2] -> [1, 1, 3, 2]
[3, 2, 1, 2] -> [1, 2, 3, 2]
[3, 1, 2, 2] -> [1, 3, 2, 2]
[1, 3, 2, 2] -> [1, 2, 2, 3]
[1, 0, 5, -234] -> [0, 1, -234, 5]
[1, 0, 1, 0, 1] -> [0, 1, 0, 1, 1]
[1, 2, 3, 4, 5] -> [1, 2, 3, 4, 5]
[5, 4, 3, 2, 1] -> [1, 2, 3, 4, 5]
[2, 1, 5, 4, 3] -> [1, 2, 3, 4, 5]
[2, 3, 1, 5, 4] -> [2, 1, 3, 4, 5]
[5, 1, 4, 2, 3] -> [1, 5, 2, 4, 3]
[5, 2, 7, 6, 4, 1, 3] -> [2, 5, 1, 4, 6, 7, 3]
[-5, -2, -7, -6, -4, -1, -3] -> [-5, -7, -2, -6, -4, -3, -1]
[14, 5, 3, 8, 15, 7, 4, 19, 12, 0, 2, 18, 6, 11, 13, 1, 17, 16, 10, 9] -> [3, 5, 14, 8, 4, 7, 15, 0, 12, 19, 2, 6, 18, 11, 1, 13, 9, 10, 16, 17]

## Scoring

This is , so the shortest code in bytes wins.

• What's the big-o of this sorting method? Dec 21, 2016 at 22:08
• @mbomb007 I don't understand big-o notation very well, but I think a single iteration is O(n). Multiply that by worst-case n iterations and you get O(n^2) (worst-case; best-case would be O(n), I think, for a single iteration). Dec 21, 2016 at 22:23
• That sounds right to me, however it's worth pointing out that reversing an array isn't a very efficient operation, so it's a slow O(n^2) Dec 21, 2016 at 22:45
• @WheatWizard reversing an array doesn't require room for a copy of the array, only room for a single element. and is O(n). swap first and last elements then swap second and second last elements etc, when you get to the middle stop. Dec 22, 2016 at 11:27
• Reversing is O(n), but reversing can be built right into the algorithm (that's what my JS answer does); since each iteration loops over each item in the array once, a single iteration is O(n). (I think...) Dec 22, 2016 at 14:27

# JavaScript (ES6), 64 bytes

f=([n,...a],z=[],q=[n,...z])=>a+a?n<a[0]?[...q,...f(a)]:f(a,q):q

Recursion FTW! The basic algorithm in use here is to keep track of the current non-ascending run in an array, "returning" it whenever an ascending element is found. We do this recursively, continually concatenating the results, until we run out of items. By creating each run in reverse ([n,...z] instead of [...z,n]), we can avoid the lengthy .reverse() at no cost.

### Test snippet

f=([n,...a],z=[],q=[n,...z])=>a+a?n<a[0]?[...q,...f(a)]:f(a,q):q

g=a=>console.log("Input:",[${a}],"Output:",[${f(a)}])

g([1])
g([1,1])
g([1,2])
g([2,1])
g([3,2,1])
g([3,1,2])
g([3,1,1,2])
g([5,2,7,6,4,1,3])
<input id=I value="5, 2, 7, 6, 4, 1, 3">
<button onclick="g((I.value.match(/\d+/g)||[]).map(Number))">Run</button>

• Can you explain how your array gets parsed into your first parameter [n,...a]. What is n? Is that just the first item in your array? Dec 22, 2016 at 0:05
• @obarakon Correct. n is the first item in the array, and a is the rest of the array. You can find more info here. Dec 22, 2016 at 0:12
• Thank you. That was very helpful. Since your first parameter is an array, why do you need to include the ...a? Is that just so you can take advantage of n? One more thing, when you call f(a,q), does q get set to the parameter z? Dec 22, 2016 at 0:21
• @obarakon Well, f=([n])=>... would only capture the first element, and f=([n,a])=>... would capture only the first in n and the second in a. Another way to do what f=([n,...a])=>,,, does would be f=a=>(n=a.unshift(),.... Dec 22, 2016 at 4:49
• And since z is the second parameter in the function, when f(a,q) is called, f sees it as z. Hope this helps! Dec 22, 2016 at 4:50

# Python 3, 63 bytes

f=lambda x,*p:x[:1]>p>()and p+f(x)or x and f(x[1:],x[0],*p)or p

Try it online!

# Jelly, 8 bytes

Ṁ;<œṗ³UF

Try it online!

## Explanation:

Ṁ;         Prepend the list [a1, a2… an] with its maximum.
<        Elementwise compare this with the original list:
[max(a) < a1, a1 < a2, …, a(n-1) < an, an]
The first element is always 0.
œṗ³     Partition the original list (³) at the indices
of the non-zero values in the working list.
(The spurious an at the end of the left argument,
resulting from comparing lists of different sizes,
is ignored by this operation, thankfully.)
U    Reverse each part.
F   Flatten.

• I was on the verge of hitting Save Edits when I saw your answer... Well done. Dec 22, 2016 at 0:09
• @Dennis Heh, so you added Dyalog partitioned enclose, but what about APL2 partition?
Dec 22, 2016 at 9:09

## JavaScript (ES6), 70 bytes

Sure, this is already beaten by ETHproductions answer, but that's the best I could come up with so far without using recursion.

a=>a.map((n,i)=>a[x=[...o,...r=[n,...r]],i+1]>n&&(o=x,r=[]),r=o=[])&&x

Note: Initializing both r and o to the exact same object with r = o = [] may look like a hazardous idea. But it is safe to do so here because r is immediately assigned its own instance (containing the first element of a) on the first iteration with r = [n, ...r].

### Test cases

let f =

a=>a.map((n,i)=>a[x=[...o,...r=[n,...r]],i+1]>n&&(o=x,r=[]),r=o=[])&&x

console.log(JSON.stringify(f([1])));
console.log(JSON.stringify(f([1, 1])));
console.log(JSON.stringify(f([1, 2])));
console.log(JSON.stringify(f([2, 1])));
console.log(JSON.stringify(f([2, 3, 1])));
console.log(JSON.stringify(f([2, 1, 3])));
console.log(JSON.stringify(f([2, 1, 2])));
console.log(JSON.stringify(f([2, 1, 1])));
console.log(JSON.stringify(f([3, 1, 1, 2])));
console.log(JSON.stringify(f([3, 2, 1, 2])));
console.log(JSON.stringify(f([3, 1, 2, 2])));
console.log(JSON.stringify(f([1, 3, 2, 2])));
console.log(JSON.stringify(f([1, 0, 5, -234])));
console.log(JSON.stringify(f([1, 0, 1, 0, 1])));
console.log(JSON.stringify(f([1, 2, 3, 4, 5])));
console.log(JSON.stringify(f([5, 4, 3, 2, 1])));
console.log(JSON.stringify(f([2, 1, 5, 4, 3])));
console.log(JSON.stringify(f([2, 3, 1, 5, 4])));
console.log(JSON.stringify(f([5, 1, 4, 2, 3])));
console.log(JSON.stringify(f([5, 2, 7, 6, 4, 1, 3])));
console.log(JSON.stringify(f([-5, -2, -7, -6, -4, -1, -3])));
console.log(JSON.stringify(f([14, 5, 3, 8, 15, 7, 4, 19, 12, 0, 2, 18, 6, 11, 13, 1, 17, 16, 10, 9])));

• No worries, I love seeing different approaches. And one often ends up becoming shorter than another after golfing :-) Dec 21, 2016 at 23:17

# MATL, 15 bytes

lidO>vYsGhXSOZ)

Input is a column vector, with the format [5; 2; 7; 6; 4; 1; 3] (semicolon is the row separator).

Try it online!

Take input [5; 2; 7; 6; 4; 1; 3] as an example.

### Explanation

l     % Push 1
% STACK: 1
i     % Push input
% STACK: 1, [5; 2; 7; 6; 4; 1; 3]
d     % Consecutive differences
% STACK: 1, [-3; 5; -1; -2; -3; 2]
O>    % Test if greater than 0, element-wise
% STACK: 1, [0; 1; 0; 0; 0; 1]
v     % Concatenate vertically
% STACK: [1; 0; 1; 0; 0; 0; 1]
Ys    % Cumulative sum
% STACK: [1; 1; 2; 2; 2; 2; 3]
G     % Push input again
% STACK: [1; 1; 2; 2; 2; 2; 3], [5; 2; 7; 6; 4; 1; 3]
h     % Concatenate horizontally
% STACK: [1 5; 1 2; 2 7; 2 6; 2 4; 2 1; 3 3]
XS    % Sort rows in lexicographical order
% STACK: [1 2; 1 5; 2 1; 2 4; 2 6; 2 7; 3 3]
OZ)   % Get last column. Implicitly display
% STACK: [2; 5; 1; 4; 6; 7; 3]

• I translated your answer to Octave saved me 31 bytes! Dec 23, 2016 at 4:47

# Mathematica, 30 27 bytes

3 bytes saved due to @Martin Ender.

Join@@Sort/@Split[#,#>#2&]&

Anonymous function. Takes a list of numbers as input and returns a list of numbers as output.

• Beat me to it! :) Dec 22, 2016 at 18:55

# Python 2, 100 bytes

A really terrible golf, but I wanted to post my solution (one does not simply outgolf Dennis)...

d=input();L=[];x=0;d+=-~d[-1],
for i in range(1,len(d)):
if d[i]>d[i-1]:L+=d[x:i][::-1];x=i
print L

Test on repl.it!

Input should be given as a Python list literal, such as [5, 3, 4, 2, 6, 1].

The basic idea is to make heavy use of Python's slicing syntax, slicing each necessary section from the array, reversing it, and adding it to the new array.

• I think the first line can be d,L,x=input(),[],0;d+=.... Dec 22, 2016 at 6:16
• @Dopapp that's exactly the same byte count Dec 22, 2016 at 6:51

## Pyke, 11 8 bytes (old version)

$0m<fm_s  Try it here! (works on latest version) $        -     delta(input)
0m<     -    map(i<0 for i in ^)
f    -   split_at(input, ^)
m_  -  map(reverse, ^)
s - sum(^)

# Retina, 163 bytes

Yes, I know how horrid this is. Supporting zeros and negatives was super fun. Byte count assumes ISO 8859-1 encoding.

\d+
$* (?<=-1*)1 x - x,1 x¶1 \b(1+),(1+\1)\b$1¶$2 ,,1 ,¶1 x,(¶|$)
x¶¶
(?<=\b\1x+(?=,(x+))),\b
¶
O%$#.(?=(.*))$.1
+¶
,
\bx
-x
(\w+)
$.1 ^, 0, ,$
,0
,,
,0,
^$0  Try it online Explanation: \d+ # Convert to unary$*
(?<=-1*)1                   # Replace negatives with x's instead of 1's
x
-                           # Remove minus sign

x,1                         # Separate if negative before positive
x¶1
\b(1+),(1+\1)\b             # or greater positive follows a positive
$1¶$2
,,1                         # or positive follows a zero
,¶1
x,(¶|$) # or zero follows a negative x¶¶ (?<=\b\1x+(?=,(x+))),\b # or negative follows a negative of greater magnitude. ¶ O%$#.(?=(.*))              # Swear at the input, then reverse each line
$.1 +¶ # Remove breaks, putting commas back , \bx # Put the minus signs back -x (\w+) # Replace unary with length of match (decimal)$.1
^,                          # Do a bunch of replacements to resurrect lost zeros
0,
,$,0 ,, ,0, ^$
0

# 05AB1E, 191816 14 bytes

Saved 2 bytes using Luis Mendo's sorting trick

ü‹X¸ì.pO¹)ø{ø¤

Try it online!

Explanation

Example input [5, 2, 7, 6, 4, 1, 3]

ü‹               # pair-wise less-than
# STACK: [0, 1, 0, 0, 0, 1]
X¸ì            # prepend a 1
# STACK: [1, 0, 1, 0, 0, 0, 1]
.p          # prefixes
O         # sum
# STACK: [1, 1, 2, 2, 2, 2, 3]
¹        # push input
# STACK: [1, 1, 2, 2, 2, 2, 3], [5, 2, 7, 6, 4, 1, 3]
)       # wrap stack in list
# STACK: [[1, 1, 2, 2, 2, 2, 3], [5, 2, 7, 6, 4, 1, 3]]
ø      # zip
# STACK: [[1, 5], [1, 2], [2, 7], [2, 6], [2, 4], [2, 1], [3, 3]]
{     # sort
# STACK: [[1, 2], [1, 5], [2, 1], [2, 4], [2, 6], [2, 7], [3, 3]]
ø    # zip
# STACK: [[1, 1, 2, 2, 2, 2, 3], [2, 5, 1, 4, 6, 7, 3]]
¤   # tail
# OUTPUT: [2, 5, 1, 4, 6, 7, 3]

Previous 16 byte solution

Dü‹X¸ì.pO.¡€g£í˜

• Those linebreaks explained it wonderfully... :-P Dec 21, 2016 at 23:09
• @StewieGriffin: Yeah, I changed up the code and posted before I had rewritten the explanation :P Dec 21, 2016 at 23:10

# JavaScript (ECMA 6), 121128125119 108 bytes

f=a=>{p=a[0],c=[],b=[];for(e of a){e>p&&b.push(c.reverse(c=[]));c.push(p=e)}return[].concat.call([],...b,c)}

Lambda expression takes a single Array parameter, a.

Thanks to @ETHproductions for helping me see my first mistake.

• Nice! I think you can do return(b+","+c).split,  to save a few bytes at the end. Dec 21, 2016 at 23:27
• Better yet, you can use c.unshift instead of c.push to remove the need to reverse c. After doing this, I got 94 bytes. Dec 21, 2016 at 23:31

# Brachylog, 10 bytes

~c:{>=r}ac

Try it online!

### Explanation

~c            Deconcatenate the Input
:{>=r}a     Each resulting sublist must be non-increasing, and then reverse it
c    Concatenate

• Does Brachylog's c, when run in reverse, necessarily try splits into fewer lists first?
– user62131
Dec 25, 2016 at 3:26
• @ais523 yes, it does. Dec 25, 2016 at 13:07

# Ruby, 60 55 bytes

s=->x{x.slice_when{|p,q|p<q}.map{|z|z.reverse}.flatten}

Pretty much what the challenge asked for. I defined a lambda s, that takes an array x, and sever (slices) it into smaller pieces where the following element would be greater than. This gives back an enumerator, which we can call map on and reverse the order of the pieces, before finally bringing it all together with flatten, which concatenates the elements in the defined order into one array.

Tests

p s[[1]]===[1]
p s[[1, 1]]===[1, 1]
p s[[1, 2]]===[1, 2]
p s[[2, 1]]===[1, 2]
p s[[2, 3, 1]]===[2, 1, 3]
p s[[2, 1, 3]]===[1, 2, 3]
p s[[2, 1, 2]]===[1, 2, 2]
p s[[2, 1, 1]]===[1, 1, 2]
p s[[3, 1, 1, 2]]===[1, 1, 3, 2]
p s[[3, 2, 1, 2]]===[1, 2, 3, 2]
p s[[3, 1, 2, 2]]===[1, 3, 2, 2]
p s[[1, 3, 2, 2]]===[1, 2, 2, 3]
p s[[1, 0, 5, -234]]===[0, 1, -234, 5]
p s[[1, 0, 1, 0, 1]]===[0, 1, 0, 1, 1]
p s[[1, 2, 3, 4, 5]]===[1, 2, 3, 4, 5]
p s[[5, 4, 3, 2, 1]]===[1, 2, 3, 4, 5]
p s[[2, 1, 5, 4, 3]]===[1, 2, 3, 4, 5]
p s[[2, 3, 1, 5, 4]]===[2, 1, 3, 4, 5]
p s[[5, 1, 4, 2, 3]]===[1, 5, 2, 4, 3]
p s[[5, 2, 7, 6, 4, 1, 3]]===[2, 5, 1, 4, 6, 7, 3]
p s[[-5, -2, -7, -6, -4, -1, -3]]===[-5, -7, -2, -6, -4, -3, -1]
p s[[14, 5, 3, 8, 15, 7, 4, 19, 12, 0, 2, 18, 6, 11, 13, 1, 17, 16, 10, 9]]===[3, 5, 14, 8, 4, 7, 15, 0, 12, 19, 2, 6, 18, 11, 1, 13, 9, 10, 16, 17]

• Welcome, nice <s>first</s> second answer, check this: codegolf.stackexchange.com/questions/363/…
– G B
Dec 22, 2016 at 8:40
• Thanks a lot. Turned this into a lambda, as suggested in the link you provided, and saved 5 bytes that way. Dec 22, 2016 at 16:05

# Husk, 4 bytes

ṁ↔ġ≤

Try it online!

I've ġot ṁyself the right tools for this job.

### Explanation

ġ is "groupby": it divides a list into sublists where in each sublist the given function returns a truthy value for any pair of elements. Used with ≤, this divides the list into non-increasing sequences.

ṁ is "concatmap": it maps a function to each element of a list, and concatenates the results together. With ↔ (reverse), this performs the reverse-and-concatenate final part of the algorithm.

# Dyalog APL, 7 15 bytes

Requires ⎕ML←3, which is default on many systems.*

{∊⌽¨⍵⊂⍨1+⍵-⌊/⍵}

∊ enlist (flatten)

⌽¨ each-reversed

⍵⊂⍨ the argument partitioned* by cutting where each corresponding element is larger than its predecessor in

1+ one plus

⍵- the argument minus

⌊/⍵ the smallest element of the argument

Old 7 byte solution fails with non-positive integers:

Requires ⎕ML←3, which is default on many systems.*

∊⌽¨⊆⍨⎕

∊ enlist (flatten) the

⌽¨ each-reversed

⊂⍨ self-partitioned*

* Partition (⊂) is a function which cuts its right argument where the corresponding left argument is larger than the preceding one. (Unfortunately it only accepts non-negative integers, and zero has special meaning.) From version 16, this functionality of ⊂ is available on all systems (even those where ⎕ML≠3), using the glyph ⊆.

# Jelly, 6 bytes

$- To A: Ɲ - For each overlapping pair of elements [a, b] in A: < - Is a < b? k - Partition A at the truthy elements U - Reverse each subarray F - Flatten  ## Haskell, 49 bytes (a:b)%l|any(<a)l=l++b%[a]|1<2=b%(a:l) _%l=l (%[])  Usage example: (%[]) [5,2,7,6,4,1,3] -> [2,5,1,4,6,7,3]. Recursive approach. The function % takes the input list as its first parameter and an accumulator l which keeps track of the non-ascending chunk so far (in reverse order). The base case is reached when the input list is empty and then the result is the accumulator. If the input list is not empty and the first element a doesn't fit in the current chunk (any(<a)l), return the accumulator and append a recursive call on the rest of the list and a as the new accumulator (l++b%[a]). Else, make a recursive call on the rest of the list and a prepended to tha accumulator (b%(a:l)). The main function (%[]) calls % with an empty accumulator. ## Retina, 98 bytes Byte count assumes ISO 8859-1 encoding. $

-(\d+)
$1$*-/
\d+
$* S-(?<=(-+)/ )(?!\1)|(?=\b(1+))(?<!\2 ) O%\S* ¶ ((-)|1)*/?$2$#1$

Try it online!

## R, 64 bytes

cat(unlist(lapply(split(x<-scan(),cumsum(c(F,diff(x)>0))),rev)))

Reads input from stdin. We split the input into a list of vectors using split() which requires a factor variable that groups the input. The factor is created by taking the cumulative sum of the the logical vector for which the difference is positive.

Consider the vector:

x=c(5, 2, 7, 6, 4, 1, 3)

Now taking the difference and prepending F by running y=c(F,diff(x)>0) would produce the following logical vector:

[1] FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE

Taking the cumulative sum cumsum(y) produces a vector where each group is represented by a unique factor upon which we can combine with the split function:

[1] 0 0 1 1 1 1 2

• 60 bytes using diffinv rather than cumsum. Nov 20, 2018 at 21:57

# Octave, 75 44 bytes

Based on MATL answer of @LuisMendo

@(a)sortrows([cumsum([1;diff(a)>0]),a])(:,2)

Try it Online!

@(a)[fliplr(mat2cell(f=fliplr(a),1,diff(find([1,diff(f)<0,numel(a)])))){:}]

Try it Online!

reverse the array

f=fliplr(a)

take first difference of f

d = diff(f);

find position of where the next element is less than the previous element

p=find([1,diff(f)<0,numel(a)])

first difference of the positions returns length of each sub array

len=diff(p)

use length of each sub array in mat2cell to split the array to nested list of arrays

nest = mat2cell(f,1,len);

reverse the nested list

rev_nest = fliplr(nest)

flatten the nested list

[rev_nest{:}]

# R, 41 bytes

ave(x,diffinv(diff(x<-scan())>0),FUN=rev)

Try it online!

# K (ngn/k), 17 bytes

{,/|'(&0N>':x)_x}

Try it online!

Chunks the input on indices where the value is greater than the previous value, reversing each chunk and concatenating back into a single array.

• (&0N>':x) get indices where the value is greater than the value of the prior element. Seed with 0N, which is the smallest possible value (i.e. INT_MIN).
• (...)_x split the input on these indices
• |' reverse each chunk
• ,/ flatten the chunks back into a single array

# Husk, 12 bytes

Completely and utterly outgolfed by Leo while I was typing in my answer...!

ṁ↔CẊ-Θ:W<¹L¹

Try it online!

ṁ              # map this function:
↔             # reverse
# across all the elements of:
C            # cut the input into sublists given by
Ẋ-          # differences between elements of
W<¹     # list of indices of the input that are less than their successor
Θ:   L¹   # prepended with zero and appended with the length of the input

• Let me introduce you to ġroupby :) codegolf.stackexchange.com/a/218911/62393
– Leo
Feb 11, 2021 at 23:01
• @Leo - I pressed 'submit', only to find your answer staring back at me! Feb 11, 2021 at 23:04

# Pyth - 15 bytes

s_McQf<@QtT@QTU

# Perl 6, 59 bytes

{map |+«*.[0].reverse,m/:s([(\-?\d+)<?{[>=] $0}>] +)+/[0]}  Regex-based solution. Because this is Sparta Perl!! • m/ /: Stringify the input array, and match a regex against it. • (\-? \d+): Match a number, and capture it as $0.
• <?{ [>=] $0 }>: Zero-width assertion that only matches if all $0 captured so far in the current sub-match are in non-ascending order.
• ([ ] +)+: Repeat the last two steps as often as possible, otherwise start a new sub-match.
• map , [0]: Iterate over the sub-matches.
• |+«*.[0].reverse: For each one, take the list of values matched by $0, reverse it, coerce the values to numbers (+«), and slip them into the outer list (|). # Perl 6, 63 bytes sub f(\a){flat$_,f a[+$_..*]with first {[<=]$_},:end,[\R,] a}

Recursive list processing solution.
More laborious than I had hoped.
Even though the language has lots of convenient built-ins, there seem to be none for list partitioning (e.g. like Ruby's slice_when or Haskell's takeWhile).

# Stacked, noncompeting, 34 bytes

Still constantly developing this language.

{e.b:e b last<}chunkby$revmap flat  Argument lies on TOS. Try it here! chunkby takes a function and collects arrays of contiguous data that satisfy the function. The function then is: {e.b:e b last<} {e.b: } function with arguments [e, <unused>, b]--the element, <the index>, and the chunk being built e < check if e is less than b last the last element of b  This gives a strictly decreasing array. $revmap is basically [rev]map and reverses each item.

flat finally flattens the array.

Some fun for actually sorting the array:

[{e.b:e b last<}chunkby$revmap flat] @:sortstep [$sortstep periodloop] @:sort

10:> @arr
arr out
arr shuf @arr
arr out
arr sort out

This outputs (for example):

(0 1 2 3 4 5 6 7 8 9)
(4 5 1 0 6 7 2 8 9 3)
(0 1 2 3 4 5 6 7 8 9)

# Python, 151 139 Bytes

Saved 12 Bytes thanks to @Flp.Tkc!

Nowhere near @Flp.Tkc, let alone...

def s(l):
r=[];i=j=0
while j<len(l)-1:
if l[j+1]>l[j]:r+=l[i:j+1][::-1],;i=j+1
j+=1
r+=l[i:j+1][::-1],;return[i for s in r for i in s]

• Instead of using append, use += data, the trailing comma implicitly constructs a tuple, which is then concatenated with the list, adding the data as the last element in the list. In this context, do r+=l[i:j+1][::-1], Dec 22, 2016 at 16:22

# Python 2, 74 bytes

b=[];c=[];a+=9e9,
for i in a[:-1]:
b=[a.pop(0)]+b
if b[0]<a[0]:c+=b;b=[]

Input in a, output in c

# Python 3, 191 Bytes

a=[int(i)for i in input().split()]
while a!=sorted(a):
b=[[]]
for i,j in enumerate(a):
if a[i-1]<j:b+=[[j]]
else:b[-1]+=[j]
a=[]
for l in[k[::-1]for k in b]:a+=[k for k in l]
print(a)

I'm not sure if using the sorted function to check is allowed here, but I couldn't think of a good reason against it, and it brought down my byte count by ~30 Bytes.

## Clojure, 105 bytes

#(filter number?(mapcat reverse(partition-by not(mapcat(fn[[a b]][a(< b a)])(partition 2 1(conj % 1))))))

Partitions into pairs on consecutive numbers, puts true or false between them, partitions on not to true and numbers become false and false true, reverses partitions and keeps numerical values.


text:
We will finish this module by introducing and briefly discussing some additional statistical analysis topics (ANOVA, ANCOVA, MANOVA, multivariate statistics, goodness-of-fit tests). Another common application, linear regression and its variants, will receive a thorough treatment in subsequent modules.

### 4.5.1 Analysis of Variance

Analysis of variance (ANOVA) is a statistical method that partitions a dataset’s variability into explainable variability (model-based) and unexplained variability (error) using various statistical models, to determine whether (multiple) treatment groups have significantly different group means.

The total sample variability of a feature $$y$$ in a dataset is defined as $\text{SS}_{\textrm{tot}}=\sum_{k=1}^{N}(y_{k}-\overline{y})^{2},$ where $$\overline{y}$$ is the overall mean of the data.

Let us return to the teaching method example of Hypothesis Testing with R.

The mean of the grades, for all students, is:

(mu = mean(teaching$Grade)) [1] 77.0625 The plot below shows all the students’ scores, ordered by participant ID; the overall mean is displayed for comparison. plot(teaching$ID,teachingGrade, xlab="ID", ylab="Grade") abline(h = mu) Since the assignment of ID is arbitrary (at least, in theory), we do not observe any patterns – if we were to guess someone’s score with no knowledge except for their participant ID, then picking the sample mean is as good a guess as any other reasonable guesses. Statistically speaking, this means that the null model $y_{i,j}=\mu+\varepsilon_{i,j},$ where $$\mu$$ is the overall mean, $$i= {A,B}$$, and $$j=1,\ldots,40$$, does not explain any of the variability in the student scores (as usual, $$\varepsilon_{i,j}$$ represents the departure or noise from the model prediction). But the students DID NOT all receive the same treatment: $$40$$ randomly selected students were assigned to group $$A$$, and the other $$40$$ to group $$B$$, and both group were taught using a different method. When we add this information onto the plot, we clearly see that the two study groups show different characteristics in term of their average scores. ggplot2::ggplot(teaching, ggplot2::aes(x=ID,y=Grade,colour=Group,shape=Group)) + ggplot2::geom_point() + ggplot2::geom_hline(ggplot2::aes(yintercept = y.bar.B),col="#00BFC4") + ggplot2::geom_hline(ggplot2::aes(yintercept = y.bar.A),col="#F8766D") + ggplot2::theme_bw()  With the group assignment information, we can refine our null model into the treatment-based model $y_{i,j}=\mu_{i}+\varepsilon_{i,j},$ where $$\mu_i$$, $$i={A,B}$$ represent the group means. Using this model, we can decompose $$\text{SS}_{\textrm{tot}}$$ into between-treatment sum of squares and error (within-treatment) sum of squares as \begin{aligned} \text{SS}_{\textrm{tot}}&=\sum_{i,j}(y_{i,j}-\overline{y})^{2}=\sum_{i,j}(y_{i,j}-\overline{y}_{i}+\overline{y}_{i}-\overline{y})^{2}\\ &=\sum_{i}N_{i}(\overline{y}_{i}-\overline{y})^{2}+\sum_{i,j}(y_{i,j}-\overline{y}_{i})^2=\text{SS}_{\textrm{treat}}+\text{SS}_{\textrm{e}}\end{aligned} The $$\text{SS}_{\textrm{treat}}$$ component looks at the difference between each of the treatment means and the overall mean, which we consider to be explainable49; the $$\text{SS}_{\textrm{e}}$$ component, on the other hand, looks at the difference between each observation and its own group mean, and is considered to be random.50 Thus, $$\text{SS}_{\textrm{treat}}/\text{SS}_{\textrm{tot}}\times 100$$% of the total variability can be explained using a treatment-based model. This ratio is called the coefficient of determination, denoted by $$R^{2}$$. Formally, the ANOVA table incorporates a few more items – the table below summarizes all the information that it contains. Source Sum of Squares df Mean Square $$\mathbf{F_{0}}$$ $$\mathbf{p-}$$value Treatment (Model) $$\text{SS}_{\textrm{treat}}$$ $$p-1$$ $$\text{MS}_{\textrm{treat}}=\text{SS}_{\textrm{treat}}/(p-1)$$ $$\text{MS}_{\textrm{treat}}/\text{MS}_{\textrm{e}}$$ $$P(F_0>F^*)$$ Error $$\text{SS}_{\textrm{e}}$$ $$N-p$$ $$\text{MS}_{\textrm{e}}=\text{SS}_{\textrm{e}}/(N-p)$$ Total $$\text{SS}_{\textrm{tot}}$$ $$N-1$$ The specific table for the teaching methodology dataset can be obtained directly from the lm() function. model.lm <- lm(Grade ~ Group, data = teaching) SS.Table <- anova(model.lm) SS.Table Analysis of Variance Table Response: Grade Df Sum Sq Mean Sq F value Pr(>F) Group 1 300.31 300.312 49.276 7.227e-10 *** Residuals 78 475.38 6.095 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Or, in a prettier format: Source Sum of Squares df Mean Square $$\mathbf{F_{0}}$$ $$\mathbf{p-}$$value Treatment (Model) $$300.31$$ $$1$$ $$300.31$$ $$49.28$$ $$7.2\times 10^{-10}$$ *** Error $$475.38$$ $$78$$ $$6.095$$ Total $$775.69$$ $$79$$ The test statistic $$F_{0}$$ follows an $$F$$-distribution with $$(\text{df}_{\textrm{treat}}, \text{df}_{\textrm{e}})=(1,78)$$ degrees of freedom. At a significance level of $$\alpha=0.05$$, the critical value $$F^*=F_{0.95, 1, 78}=3.96$$ is substantially smaller than the test statistic $$F_{0}=49.28$$, implying that the two-treatment model is statistically significant. This, in turn, means that the model recognises a statistically significant difference between the students’ scores, based on the teaching methods. (R2 = summary(model.lm)r.squared)
[1] 0.3871566

The coefficient of determination $$R^2$$ provides a way to measure the model’s significance. From the ANOVA table for the teaching example, we compute $R^{2}=\frac{\text{SS}_{\textrm{treat}}}{\text{SS}_{\textrm{tot}}}=\frac{300.31}{775.69}\approx 0.39,$ which means that $$39$$% of the total variation in the data can be explained by the two-treatment model.

Is this good enough? That depends on the specifics of the situation (in particular, on the researcher’s or the client’s needs).

#### Diagnostic Checks

As with most statistical procedures, ANOVA relies on certain assumptions for its result to be valid. Recall that the model is given by $y_{i,j}=\mu_{i}+\varepsilon_{i,j}.$ What assumptions are made?

The main assumption is that the error terms follow independently and identically distributed (iid) normal distributions (i.e., $$\varepsilon_{i,j}\stackrel{\text{iid}}{\sim}\mathcal{N}(0,\sigma^{2})$$).

Assuming independence, we are required to verify three additional assumptions:

• normality of the error terms;

• constant variance (within treatment groups), and

• equal variances (across treatment groups).

Normality of the errors can be tested visually with the help of a normal-QQ plot, which compares the standardized residuals quantiles against the theoretical quantiles of the standard normal distribution $$\mathcal{N}(0,1)$$ (a straight line indicates normality).

In other words, if the errors are normally distributed with mean $$0$$ and variance $$\sigma^2$$, we would expect that the $$80$$ standardized residuals $$r_{i,j}=\frac{\varepsilon_{i,j}-0}{\sigma}$$ should behave as though they had been drawn from $$\mathcal{N}(0,1)$$.

plot(model.lm, which = c(1,2,3,4))

The plots above show some departure in the lower tail, however, moderate departure from normality is usually acceptable as long as it is mostly a tail phenomenon.

To test the assumption of constant variance, we can run visual inspection using

• residuals vs.fitted values, and/or

• residuals vs.order/time.

The standardized residuals in both groups should be approximately distributed according to $$\mathcal{N}(0,1)$$. The plots also show that variability from the mean in each treatment group is reasonably similar.51

More formally, equality of variance is often tested for using Bartlett’s test (when normality of the residuals is met) or the modified Levene’s test (when it is not).

Assuming that we felt the evidence of normal residuals was warranted in the two-treatment model of the teaching dataset, we get a $$p-$$value of $$0.57$$ for Bartlett’s test:

(B.T <- bartlett.test(Grade~Group, teaching))

Bartlett test of homogeneity of variances

Bartlett's K-squared = 0.32192, df = 1, p-value = 0.5705

Otherwise, we get a $$p-$$value of $$0.76$$ for Levene’s test.

(L.T <- lawstat::levene.test(teaching$Grade, teaching$Group, location="median", correction.method="zero.correction"))

Modified robust Brown-Forsythe Levene-type test based on the absolute
deviations from the median with modified structural zero removal
method and correction factor

Test Statistic = 0.095106, p-value = 0.7586

In either case, the $$p-$$value falls above reasonable significance levels (0.05, say), which means that we cannot reject the null hypothesis of equal variance.

When there are $$p>2$$ treatment groups, ANOVA provides a test for $H_{0}: \mu_{1} = \cdots = \mu_{p}\quad\text{vs.}\quad H_{1}: \mu_{i} \neq \mu_{j} \text{ for at least one } i \neq j.$

A significant $$F_0$$ value indicates that there is at least one group which differs from the others, but it does not specify which one(s) that may be.

Specialized methods such as Scheffe’s method and Tukey’s test can be used to identify the statistically different treatments.

Finally, while ANOVA can accommodate unequal treatment group sizes, it is recommended to keep those sizes equal across all groups – this makes the test statistic less sensitive to violations of the assumption of equal variances across treatment groups, providing yet another reason to involve the analysts/consultants in the data collection process.

### 4.5.2 Analysis of Covariance

In a previous section, we looked at the effectiveness of new teaching method by assigning each group to a specific treatment and comparing the mean test scores. A crucial assumption for that model is that subjects in each group have similar background knowledge about statistics prior to the three week lectures.

If this assumption is wrong, however, we may be making incorrect decisions based on the model. Even if each group had similar background knowledge on average, there may be large variability from person-to-person, masking the true treatment effect.

#### Paired Comparison

One way to avoid such subject-to-subject variability is to administer both treatments to each individual, and then compare treatment effects by looking at the difference in the outcomes. For instance, if a grocery chain is interested in measuring the effectiveness of two advertising campaigns, it could be reasonable to assume that there is a large variability in total sales, as well as popular items sold, at each store.

It may then be preferable to run both campaigns in each store and analyze the resulting data rather than to split the stores into two groups (in each of which a different advertising campaign is run) and then to compare the mean outcomes in the two groups.

Formally, let $$X_{i,1}$$ denote the total sales with campaign $$A$$ and $$X_{i,2}$$ the total sales with campaign $$B$$. The quantity of interest is the difference $$D_{i}=X_{i,1}-X_{i,2}$$ for each store $$i=1,\ldots,N$$.

Assuming that the differences $$D_{i}$$ follow an iid normal distribution with mean $$\delta$$ and variance $$\sigma^{2}_{d}$$, then we test for $H_{0}: \delta=0\quad\mbox{against}\quad H_{1}: \delta \neq 0$ using the test statistic $t_0=\sqrt{N}\frac{\overline{D}}{s_{d}},$ which follows a Student’s $$t$$ distribution with $$N-1$$ degrees of freedom; thus we reject $$H_{0}$$ if the observed test statistic $$t_{0}$$ has $$p$$-value less than the significance level $$\alpha/2$$.

#### ANOVA vs. ANCOVA

ANOVA compares multiple group means and tests whether any of the group means differ from the rest, by breaking down the total variability into a treatment (explainable) variability component and an error (unexplained) variability component, and building a ratio $$F_0$$ to determine whether or not to reject $$H_{0}$$.

Analysis of covariance (ANCOVA) introduces concomitant variables (or covariates) to the ANOVA model, splitting the total variability into 3 components: $$\text{SS}_{\textrm{treat}}$$, $$\text{SS}_{\textrm{con}}$$, and $$\text{SS}_{\textrm{e}}$$, aiming to reduce error variability.

The choice of covariates is thus crucial in running a successful ANCOVA. In order to be useful, a concomitant variable must be related to response variable in some way, otherwise it not only fails to reduce error variability, but it also increases the model complexity:

• in the teaching method example, we could consider administering a pre-study test to measure the prior knowledge level of each participant and use this score as a concomitant variable;

• in the advertising campaign example, we could have used the previous month’s sales as a covariate;

• in medical studies, we could use the age and weight of subjects, say.

Importantly, concomitant variables should not be affected by treatments. As an example, suppose that the patients in a medical study were asked:

How strongly do you believe that you were given actual medication rather than a placebo?

If the treatment is indeed effective, then a participant’s response to this question could be markedly different in the treatment group than in the placebo group.52

This means that true treatment effect may be masked by concomitant variable due to unequal effects on treatment groups. Note that qualitative covariates (such as gender, say) are not part of the ANCOVA framework – indeed, such covariates create new ANOVA treatment groups instead.

When moving from an ANOVA to an ANCOVA model, the error variability is further split into a pure error and a covariate component, while the treatment variability remains unchanged.

#### ANCOVA Model and Assumptions

Suppose that we are testing the effect of $$p$$ treatments, with $$N_{j}$$ subjects in each group. Then the ANCOVA model takes the form $y_{i,j}=\mu+\tau_{j}+\gamma (x_{i,j}-\overline{x})+\varepsilon_{i,j}$ where

• $$y_{i,j}$$ is the response of the $$i^{\text{th}}$$ subject in the $$j^{\text{th}}$$ treatment group;

• $$\mu$$ is the overall mean;

• $$\tau_{j}$$ is the $$j^{\text{th}}$$ treatment effect, subject to a constraint $\sum_{j=1}^{p}\tau_{j}=0;$

• $$\gamma$$ is the coefficient for the covariate effect;

• $$(x_{i,j}-\overline{x})$$ is the covariate value of the $$i^{\text{th}}$$ subject in the $$j^{\text{th}}$$ treatment group, adjusted by the mean, and

• $$\varepsilon_{i,j}$$ is the error of $$i^{\text{th}}$$ subject in the $$j^{\text{th}}$$ treatment group.

Additionally, four assumptions must be satisfied:

• independence and normality of residuals – the residuals follow an $${iid}$$ normal distribution with mean of $$0$$ and variance $$\sigma^{2}_{\varepsilon}$$;

• homogeneity of residual variances – the variance of the residuals is uniform across treatment groups;

• homogeneity of regression slopes – the regression effect (slope) is uniform across treatment groups, and

• linearity of regression – the regression relationship between the response and the covariate is linear.

The first of these assumptions can be tested with the help of a QQ-plot and a scatter-plot of residuals vs.fitted values, while the second may use the Bartlett or the Levene test. The final assumption is not as crucial as the other three assumptions, however. Various remedial methods can be applied should any of these assumptions fail.

The third assumption, however, is crucial to the ANCOVA model; it can be tested with the equal slope test, which requires an ANCOVA regression with an additional interaction term $$x \times \tau$$. If the interaction is not significant, the third assumption is satisfied.

In the event that the interaction term is statistically significant, a different approach (e.g. moderated regression analysis, mediation analysis) is required since using the original ANCOVA model is not prescribed.

An in-depth application of an ANCOVA model is highlighted in Case Study: IBS.

### 4.5.3 Basics of Multivariate Statistics

To this point, we have only considering situations where the response has been univariate. In applications, the situation often calls for multivariate responses, where the response variables are thought to have some relationship to one another (e.g., a correlation structure).

It remains possible to analyze each response variable independently, but the dependence structure can be exploited to make joint (or simultaneous) inferences.

#### Properties of the Multivariate Normal Distribution

The probability density function of a multi-dimensional random vector $$\mathbf{X}\in\mathbb{R}^p$$ that follows a multivariate normal distribution with mean vector $$\mathbf{\mu}$$ and covariance matrix $$\mathbf{\Sigma}$$, denoted by $$\mathbf{X}\sim \mathcal{N}_p(\mathbf{\mu},\mathbf{\Sigma})$$, is given by $f(\mathbf{X})=\frac{1}{(2\pi)^{p/2}\det(\mathbf{\Sigma})^{1/2}}\exp\left(-\frac{1}{2}(\mathbf{X}-\mathbf{\mu})^{\!\top}\mathbf{\Sigma}^{-1}(\mathbf{X}-\mathbf{\mu})\right),$ where $\mathbf{\Sigma}=\begin{bmatrix} \sigma_{1,1} & \sigma_{1,2} & \cdots & \sigma_{1,p}\\ \sigma_{2,1} & \sigma_{2,2} & \cdots & \sigma_{2,p}\\ \vdots & \vdots & \ddots & \vdots\\ \sigma_{p,1} & \sigma_{p,2} & \cdots & \sigma_{p,p}\\ \end{bmatrix}.$

For such an $$\mathbf{X}$$, the following properties hold:

1. any linear combination of its components are normally distributed;

2. all subsets of components follow a (modified) multivariate normal distribution;

3. a diagonal covariance matrix implies the independence of its components;

4. conditional distributions of components follow a normal distribution, and

5. the quantity $$(\mathbf{X}-\mathbf{\mu})^{\!\top}\mathbf{\Sigma}^{-1}(\mathbf{X}-\mathbf{\mu})$$ follows a $$\chi^{2}_{p}$$.

These properties make the multivariate normal distribution attractive, from a theoretical point of view (if not always entirely realistic).

For instance:

• using property 1, we can use contrasts to test which components are distinct from the others;

• property 5 is the multivariate analogue of the square of a standard normal random variable $$Z\sim \mathcal{N}(0,1)$$ following a $$Z^2\sim \chi^2_1$$ distribution;

• but two univariate normal random variables with zero covariance are not necessarly independent (the joint p.d.f. of two such variables is not necessarily the p.d.f. of a multivariate normal distribution).

#### Hypothesis Testing for Mean Vectors

When the sample comes from a univariate normal distribution, we can test $H_{0}: \mu=\mu_{0}\quad\mbox{against}\quad H_{1}: \mu \neq \mu_{0}$ by using a $$t-$$statistic. Analogously, if the sample comes from a $$p-$$variate normal distribution, we can test $H_{0}: \mathbf{\mu}=\mathbf{\mu_{0}}\quad\mbox{against}\quad H_{1}: \mathbf{\mu} \neq \mathbf{\mu_{0}}$ by using Hotelling’s $$\mathbf{T}^2$$ test statistic $T^{2}=N\cdot (\mathbf{\overline{X}}-\mathbf{\mu})^{\!\top}\mathbf{S}^{-1}(\mathbf{\overline{X}}-\mathbf{\mu}),$ where $$\mathbf{\overline{X}}$$ denotes the sample mean, $$\mathbf{S}$$ the sample covariance matrix, and $$N$$ the sample size.

Under $$H_{0}$$, $T^{2}\sim \frac{(N-1)p}{(N-p)}F_{p, N-p}.$ Thus, we do not reject $$H_{0}$$ at a significance level of $$\alpha$$ if $N\cdot (\mathbf{\overline{X}}-\mathbf{\mu_{0}})^{\!\top}\mathbf{S}^{-1}(\mathbf{\overline{X}}-\mathbf{\mu_{0}}) \leq \frac{(N-1)p}{(N-p)}F_{p, N-p}(\alpha)$ and reject it otherwise.

#### Confidence Region and Simultaneous Confidence Intervals for Mean Vectors

In the $$p-$$variate normal distribution, any $$\mathbf{\mu}$$ that satisfies the condition $N\cdot (\mathbf{\overline{X}}-\mathbf{\mu})^{\!\top}\mathbf{S}^{-1}(\mathbf{\overline{X}}-\mathbf{\mu}) \leq \frac{(N-1)p}{(N-p)}F_{p, N-p}(\alpha)$ resides inside a $$(1-\alpha)100$$% confidence region (an ellipsoid in this case).

Simultaneous Bonferroni confidence intervals with overall error rate $$\alpha$$ can also be derived, using $(\overline{x}_{j}-\mu_{j})\pm t_{N-1}(\alpha/p)\sqrt{\frac{s_{j,j}}{N}} \text{ for }j=1,\ldots, p.$

Another approach is to use Hotelling’s $$\mathbf{T}^2$$ simultaneous confidence intervals, given by $(\overline{x}_{j}-\mu_{j})\pm \sqrt{\frac{p(N-1)}{N-p}F_{p,N-p}(\alpha)} \sqrt{\frac{s_{j,j}}{N}} \text{ for }j=1,\ldots, p.$

Figure @(fig:testA7) shows these regions for a bivariate normal random sample. Note that the Hotelling’s $${T}^{2}$$ simultaneous confidence intervals form a rectangle (in grey) that confines the confidence region, while the Bonferroni confidence intervals (in blue) are slightly narrower.

Given that all the components of the mean vector are correlated (since the covariance matrix is generally non-diagonal), the confidence region should be used if the goal is to study the plausibility of the mean vector as a whole, while Bonferroni confidence intervals may be more suitable when component-wise confidence intervals are of needed.

#### Multivariate Analysis of Variance

ANOVA is often used as a first attempt to determine whether the means from every sub-population are identical.

ANOVA can test means from more than two populations; the multivariate ANOVA (MANOVA) is quite simply a multivariate extension of ANOVA which tests whether the mean vectors from all sub-populations are identical.

Assume there are $$I$$ sub-populations in the population, from each of which $$N_i$$ $$p-$$dimensional responses are drawn, for $$i=1,\ldots,I$$.

Each observation can be expressed as: $\mathbf{X}_{i,j}=\mathbf{\mu}+\mathbf{\tau}_{i}+\mathbf{\varepsilon}_{ij},$ where $$\mathbf{\mu}$$ is the overall mean vector, $$\mathbf{\tau}_{i}$$ is the $$i^{\text{th}}$$ population-specific treatment effect, and $$\mathbf{\varepsilon}_{ij}$$ is the random error, which follows a $$N_{p}(\mathbf{0},\mathbf{\Sigma})$$ distribution.

It is important to note that the covariance matrix $$\mathbf{\Sigma}$$ is assumed to be the same for each sub-population, and that $\sum_{i=1}^{I}N_{i}\mathbf{\tau}_{i}=\mathbf{0}$ to ensure that the estimates are uniquely identifiable.

To test the hypothesis $H_{0}: \mathbf{\tau}_{1}=\cdots=\mathbf{\tau}_{I}=\mathbf{0}\quad\mbox{against}\quad H_{1}: \text{some } \mathbf{\tau}_{i}\neq \mathbf{0},$ we decompose the total sum of squares and cross-products $$\textrm{SSP}_{\textrm{tot}}$$ into $\textrm{SSP}_{\textrm{tot}}=\textrm{SSP}_{\textrm{treat}}+\textrm{SS }_{\textrm{e}}.$

Based on this decomposition, we compute the test statistic known as Wilks’ lambda $\Lambda^{*}=\frac{|\mathbf{W}|}{|\mathbf{B}+\mathbf{W}|},$ where $$\mathbf{B},\mathbf{W}$$ are as in the MANOVA table below:

Source SSP df MSP $$\mathbf{F_0}$$
Treatment $$\mathbf{B}$$ $$I-1$$ $$\mathbf{B}/(I-1)$$ $$\mathbf{W}^{-1}\mathbf{B}$$
Error $$\mathbf{W}$$ $$\sum_{i=1}^{I}N_{i}-I$$ $$\mathbf{W}/\sum_{i=1}^I(N_i-1)$$
Total $$\mathbf{B+W}$$ $$\sum_{i=1}^{I}N_{i}-1$$ $$(\mathbf{B+W})/(\sum_{i=1}^{I}N_{i}-1)$$

We have $\mathbf{B}=\sum_{i=1}^{I}N_{i}(\mathbf{\bar{X}}_{i}-\mathbf{\bar{X}})(\mathbf{\bar{X}}_{i}-\mathbf{\bar{X}})^{\!\top}$ and $\mathbf{W}=\sum_{i=1}^{I}\sum_{j=1}^{n_{i}}(\mathbf{X}_{ij}-\mathbf{\bar{X}}_{i})(\mathbf{X}_{ij}-\mathbf{\bar{X}}_{i})^{\!\top};$ we reject $$H_{0}$$ if $$\Lambda^{*}$$ is below some preagreed upon threshold, which depends on $$p$$, $$I$$, and $$N_i$$, $$i=1,\ldots, I$$.

### 4.5.4 Goodness-of-Test Fits

A (fictitious) 2017 survey asked a sample of $$N=200$$ adults between the age of 25 to 35 about their highest educational achievement. The result is summarized below.

Year $$\mathbf{<}$$HS HS CU CU$$\mathbf{+}$$
$$2017$$ $$16$$ $$55$$ $$83$$ $$46$$

In a 1997 survey, it was also found that

• $$p_1=13$$% of adults had not completed high school;

• $$p_2=32$$% had obtained a high school degree but not a post-secondary degree;

• $$p_3=37$$% had either an undergraduate college or university diploma but no post-graduate degree, and

• $$p_4=18$$% had at least one post-graduate degree in 19977.

Based on the result of this survey, is there sufficient evidence to believe that educational backgrounds of the population have changed between 1997 and 2007?

Since each respondent’s educational achievement can only be classified into one of these categories, they are mutually exclusive. Furthermore, these categories cover all possibilities on the educational front, so they are also exhaustive.

We can thus view the distribution of educational achievements as being multinomial. For such a distribution, with parameters $$p_{1},\cdots,p_{k}$$, the expected frequency in each category is $$m_{j}=Np_{j}$$.

Let $$O_{j}$$ denote the observed frequency for the $$j^{\text{th}}$$ category. If there has been no real change since 1997, we would expect the sum of squared differences between the observed 2017 frequencies and the expected frequencies based on 1997 data to be small.

We can use this information to test the goodness-of-fit between the observations and the expected frequencies via Pearson’s $$\chi^{2}$$ test statistic $X^{2}=\sum_{j=1}^{k}\frac{(O_{j}-m_{j,0})^{2}}{m_{j,0}}$ which follows a $$\chi^{2}$$ distribution with $$k-1$$ df.

In the above example, the hypotheses of interest are $H_{0}: \mathbf{p}=\mathbf{p}^*=(0.13,0.32,0.37,0.18)\quad\mbox{vs}\quad H_{1}: \mathbf{p}\neq \mathbf{p}^*.$

The table below summarizes the information under $$H_{0}$$.

Category $${O}_{j}$$ $${p}_{j,0}$$ $${m}_{j,0}$$ $$({O}_{j}-{m}_{j,0})^2/{m}_{j,0}$$
$$1$$ $$16$$ $$0.13$$ $$26$$ $$3.846$$
$$2$$ $$55$$ $$0.32$$ $$64$$ $$1.266$$
$$3$$ $$83$$ $$0.37$$ $$74$$ $$1.095$$
$$4$$ $$46$$ $$0.18$$ $$36$$ $$2.778$$
Total $$200$$ $$1$$ $$200$$ $$7.815$$

Pearson’s test statistic is $$X^{2}=7.815$$, with an associated $$p-$$value of $$0.0295$$, which implies that there is enough statistical evidence (at the $$\alpha=0.05$$ level) to accept that the population’s educational achievements have changed over the last 20 years.


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# Did the GMAT ever have 3 sections?

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Did the GMAT ever have 3 sections? [#permalink]

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21 Jul 2008, 17:12
I have a friend that says he took the GMAT in the late 90's and it had 3 sections. A Quant, Analytical and Verbal. He says he took it on the computer and then had blue books to do the writing. Is this BS?
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Director Joined: 20 Feb 2008 Posts: 797 Location: Texas Schools: Kellogg Class of 2011 Followers: 6 Kudos [?]: 146 [0], given: 9 Re: Did the GMAT ever have 3 sections? [#permalink] ### Show Tags 21 Jul 2008, 17:34 Check this out: June 21, 1997 was the last of the traditional "paper and pencil" GMAT. A record number of people took the GMAT on that day. These test takers were not motivated by feelings of nostalgia. For many, June 21 was the final chance to avoid the new Computer Adaptive GMAT - commonly known as the CAT. Better the devil you know than the one you don't! Those in the test prep industry marketed the idea that the CAT should be avoided and that June 21 was the last chance to avoid it. Predictably, enrollments in GMAT prep courses were very high during the spring of 1997. Between June 21, 1997 and October 11, 1997 the test prep industry worked hard to determine how the upcoming CAT should be taught. GMAT provided very little information. In July 1997 the new GMAT Bulletin was published. The new Official Guide For GMAT Review and Computer Adaptive PowerPrep Software were released in August of 1997. Appointments to take the CAT were made starting August 15, 1997. October 11, 1997 was the first day that the new Computer Adaptive GMAT was administered. On January 1, 1980 New York's "Truth In Testing" law took effect. This law forced those in the business of standardized testing to disclose the questions that contributed to a test taker's score. GMAT CAT questions will not be subject to disclosure laws. Individual GMAT's are now constructed by computer, from a pool of questions, to meet the ability of specific test takers. Questions in the pool will be reused and will not be disclosed The "paper and pencil" GMAT continues to be administered in a small number of countries. Obviously test disclosure laws would not apply to tests administered outside the U.S. Indications are that GMAT may not be generating new "paper and pencil" GMAT for these locations. I know there is a history of the GMAT in the front of the Princeton Review book and they talk about how the sections change, so its certainly possible. SVP Joined: 30 Apr 2008 Posts: 1888 Location: Oklahoma City Schools: Hard Knocks Followers: 39 Kudos [?]: 537 [0], given: 32 Re: Did the GMAT ever have 3 sections? [#permalink] ### Show Tags 21 Jul 2008, 17:55 This is the same person that says he got a 743 and was waitlisted at Texas. A mutual friend of ours also said "If I had a nickel for every stupid thing he claimed, I'd be rich." I take most of what he says with a block of salt. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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22 Jul 2008, 06:15
jallenmorris wrote:
This is the same person that says he got a 743 and was waitlisted at Texas. A mutual friend of ours also said "If I had a nickel for every stupid thing he claimed, I'd be rich."

I take most of what he says with a block of salt.

743 I doubt, but I do remember reading about there being a separate section for logical reasoning ala LSAT.
Re: Did the GMAT ever have 3 sections?   [#permalink] 22 Jul 2008, 06:15
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# American Institute of Mathematical Sciences

March  2010, 13(2): 375-391. doi: 10.3934/dcdsb.2010.13.375

## On the time decay of solutions in porous-thermo-elasticity of type II

 1 Departament de Matemàtica Aplicada 2, ETSEIAT–UPC, C. Colom 11, 08222 Terrassa, Barcelona, Spain, Spain, Spain

Received  November 2008 Revised  July 2009 Published  December 2009

In this paper we investigate the asymptotic behaviour of the solutions of the linear theory of thermo-porous-elasticity. That is, we consider the theory of elastic materials with voids when the heat conduction is of type II. We assume that the only dissipation mechanism is the porous dissipation. First we prove that, generically, the solutions are exponentially stable on time or, in other words, the decay of solutions can be controlled by a negative exponential for a generic class of materials. The reason lies in the fact that the temperature is strongly coupled with both the microscopic and macroscopic structures of the materials and plays the role of a "driving belt" between the dissipation at the microscopic structure and the macroscopic one. Later we note that the decay of solutions cannot be fast enough to make the solutions be zero in a finite period of time. Finally, we show that when the coupling term between the microscopic (or macroscopic) structure and the thermal variable vanishes, the solutions do not decay exponentially (generically).
Citation: M. Carme Leseduarte, Antonio Magaña, Ramón Quintanilla. On the time decay of solutions in porous-thermo-elasticity of type II. Discrete & Continuous Dynamical Systems - B, 2010, 13 (2) : 375-391. doi: 10.3934/dcdsb.2010.13.375
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# Solve the following differential equation by undetermined coefficients $(1-x^2)y''-2xy'+6y=0$

I'm trying to obtain a recursion formula relating $$a_{n+2}$$ to $$a_n$$ using the method of undetermined coefficients. Then determining $$a_n$$ explicitly for each $$n$$ and finding the sum of the series.

$$(1-x^2)y''-2xy'+6y=0$$

Taking the first and second derivative of the power series gives me:

$$(1)\space y = \sum_{n=0}^{\infty}a_nx^n$$ $$(2)y'\space = \sum_{n=1}^{\infty} na_nx^{n-1}$$ $$(3) \space y'' = \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$$

Then plugging this into the differential equation:

$$(a) \space \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}+\sum_{n=2}^{\infty}n(n-1)a_nx^{n}-2\sum_{n=1}^{\infty} na_nx^{n} + 6 \sum_{n=0}^{\infty}a_nx^n$$

Then simplifying the exponents: $$(b) \space \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n} + \sum_{n=0}^{\infty}n(n-1)a_{n}x^{n} - 2\sum_{n=0}^{\infty} na_nx^{n} + 6 \sum_{n=0}^{\infty}a_nx^n$$

Then simplifying with $$x^n$$:

$$(c) \space \left([(n+2)(n+1)a_{n+2}] + [n(n-1-2)+6]a_n\right)x^n$$

Finally, equating coefficients of $$x^n$$ to obtain the relation:

$$(d) \space [(n+2)(n+1)a_{n+2}] + [n(n-1-2)+6]a_n=0$$

Or what amounts to: $$(4) \space a_{n+2} = \frac{n(n-1-2)a_n}{(n+2)(n+1)}$$ $$(5) \space a_n = \frac{(n+2)(n+1)}{n(n-1-2)}$$

I would really appreciate a check on my calculations and whether I missed something out or went wrong along the way. Additionally, how do I proceed from here?

• Seems to me that in (a) the second term should be with a minus sign. Apr 24 at 15:40

This is a special case of Legendre's differential equation at $$\ell=2$$.
The series expansion should be $$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\color{blue}-\sum_{n=2}^{\infty}n(n-1)a_nx^{n}-2\sum_{n=1}^{\infty} na_nx^{n} + 6 \sum_{n=0}^{\infty}a_nx^n=0$$ from which $$(n+2)(n+1)a_{n+2}-(n+3)(n-2)a_n=0.$$ Note that $$a_2=-3a_0$$ and $$12a_4=0$$ so $$a_{2N}=0$$ for all $$N>1$$. Iterating gives \begin{align}a_{n+2}&=\frac{(n+3)(n-2)}{(n+2)(n+1)}a_n=\frac{(n+3)(n-2)}{(n+2)(n+1)}\frac{(n+1)(n-4)}{n(n-1)}a_{n-2}=\cdots\\&=(n+3)(n+1)\frac{(n-1)!(n-2k)(n-2k-2)}{(n-2k+1)!}\frac{(n-2k)!}{(n+2)!}a_{n-2k}\\&=\frac{n+3}{(n+2)n}\frac{(n-2k)(n-2k-2)}{n-2k+1}a_{n-2k}.\end{align} When $$n=2N+1$$ choosing $$k=N$$ gives $$a_{2N+3}=-\frac{N+2}{(2N+3)(2N+1)}a_1$$ so the solution is $$y(x)=a_0\left(1-3x^2\right)+a_1x\left(1-\frac23x^2-\frac15x^4-\frac4{35}x^6-\cdots\right).$$ Indeed, taking $$a_0=1/2$$ and $$a_1=0$$ yields the Legendre polynomial of order two.


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# Can the CPT theorem be valid if Lorentz invariance is only spontaneously broken?

Earlier, I asked here whether one can have spontaneous breaking of the Lorentz symmetry and was shown a Lorentz invariant term that can drive the vacuum to not be Lorentz invariant. How relaxed are the assumptions in the CPT theorem? Can one have the theorem to be valid for a Lorentz invariant theory with a vacuum that breaks it? Can someone point me to the original and alternate proofs of this theorem?

Also, is there any connection between CPT invariance and renormalizability?

-
Original Article: arxiv.org/abs/hep-ph/0201258 –  Benjamin Horowitz Oct 4 '11 at 22:20
@dbrane if Lorentz invariance is only spontaneously broken if, maybe, perhaps... this is not physics sounds like wish –  user8784 May 24 '12 at 21:10

The CPT theorem is a consequence of the statement that the Euclidean theory is invariant under 180 degree rotations in a plane involving (the analytic continuation of) the time axis. The result of this rotation flips the sense of Euclidean time, and flips one space axis. The analytic continuation that defines the Minkowski theory is a PT transformation (it reverses T and also P), and it maps each field to it's reflection positivity conjugate.

So any theory will have a CPT whenever the Lorentz breaking leaves a Euclidean subgroup with at least one 180 degree rotation involving time. For an explicit but somewhat trivial example, consider an effective field theory of a massive vector field with a field potential $-m^2|V|^2 + \lambda|V|^4$. The field acquires a spacelike expectation value, breaking the SO(3,1) symmetry down to SO(2,1). so the Euclidean symmetry is SO(3) and you have a 180 degree rotation. In this case, the CPT invariance is obvious, because the remaining SO(2,1) symmetry is enough to show it directly.

Other examples of this sort are string reductions, where you compactify a large dimensional space to a smaller dimension space, leaving the Lorentz group on the smaller space. The result is automatically CPT invariant, because of the lower dimensional Lorentz invariance.

For a less trivial but purely mathematical example, consider an effective symmetric tensor condensate T which has a vacuum value with no special symmetry. You can choose orthogonal coordinates to diagonalize T, and this determines a preferred frame. The Lorentz group is broken to a discrete subgroup, but this subgroup includes 180 degree Euclidean rotations, so that the result is a CPT invariant theory.

For another example, completely physical, and breaking translation invariance, consider a zero temperature crystal made out of antimatter together with the same crystal at another position, reflected along the line of separation. You can ask if the theory of this double-crystal system has CPT. The 180 rotation in this case is around the center of mass position, with one axis time, and the other axis the line of separation between the crystals. The system is obviously symmetric under this transformation, so there is a CPT. In this case, the CPT reflection is physically obvious: a quasi-particle in the crystal is a p-reversed quasi-particle in the reflected crystal. This doesn't work if the two crystals are not exactly reflected in both the matter/antimatter sense, and in the spatial sense--- if you rotate one of the crystal, but not the other, the CPT invariance will be broken by electromagnetic interactions between the two crystals.

For another physical example, this one more speculative in nature but perhaps realized, consider a gravitationally bound cluster of massive neutrinos, at zero temperature. The neutrinos are widely believed to be their own antiparticle, and they are the only known massive particle with this property. If the resulting configuration is symmetric under rotations (or even in the ordinarily absurd case, perhaps realizable at extremely high densities, that the neutrinos form a crystal which is symmetric under a crystal group that includes a reflection) then a time rotation by 180 degrees preserves the Euclidean theory in this state, so all excitations of this medium, despite breaking Lorentz symmetry, have a CPT invariance. So an electron moving or scattering off this crystal behaves in exactly the time reversed way as a positron with a reflected wavefunction moving the opposite way.

The general rule is that, any condensate which is CPT invariant with respect to at least one reflection plane for P preserves CPT. If a condensate is not invariant with respect to any P plane, it's interactions break CPT.

-


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3.4k views

In a college, there are three student clubs, $60$ students are only in the Drama club, $80$ students are only in the Dance club, $30$ students are only in Maths club, $40$ students are in both Drama and Dance clubs, $12$ students are in both Dance and Maths clubs, $7$ students are in both Drama and Maths clubs, and $2$ students are in all clubs. If $75 \%$ of the students in the college are not in any of these clubs, then the total number of students in the college is _____.

1. $1000$
2. $975$
3. $900$
4. $225$

edited | 3.4k views

Read question carefully. We cannot directly use the formula

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A\cap B\cap C)$

because of the only keyword which means we do not directly know what is $P(A), P(B)$ or $P(C)$

Here, according to question statement, we can calculate total number of students belonging to at least one club :
$P(A \cup B \cup C) = 60 + 80 + 30 + (40-2) + (12-2) + (7-2) + 2 = 225$

This is just $25\%$ of the total no of students since $75\%$ do not belong to any club.

So, total number of students$= 225 \div 25\% = 225 \times 4 = 900.$

by Boss (10.7k points)
edited by
0

will you please explain more about why not to use this formula and various scenarios to avoid this.

0

@Asim Siddiqui 4 refer my answer :)

$900$, Hence option C is the answer.
by Boss (23.9k points)
edited by
0
the answer is 900. but in Pragy Sir's MyMarks app, the answer is given as 975. please correct it
0

@Sukhbir Singh It is given as 900 . Check again

Let us solve it using the conventional formula :

(A U B U C) = A + B + C - (A ∩ B) - (B ∩ C) - (C ∩ A) + (A ∩ B ∩ C)

Notice that in this formula,

A means students in Drama club (but not Drama club only).

B means students in Dance club (but not Dance club only).

C means students in Maths club (but not Maths club only).

So, A = students in Drama club ONLY + students in Drama and Dance clubs + students in Drama and Maths clubs - students in all clubs.

A= 60 + 40 + 7 - 2 = 105.

Similarly,

B = 80 + 12 + 40 - 2 = 130.

C = 30 + 7 + 12 - 2 = 47.

Also, Given that (A ∩ B) = 40, (B ∩ C) = 12, (C ∩ A) = 7 &  (A ∩ B ∩ C) = 2

Now just substitute these values in the above formula.

(A U B U C) = 105 + 130 + 47 - 40 - 12 - 7 + 2 = 282 - 57 = 225.

But given that 75% of the students are not in any of these clubs ,i.e, 25% of the students are in at least one of these clubs.

0.25 * x = 225 ==> x = 900.

So, Total number of students in the college = 900.

by Active (1.9k points)
edited
60 + (40-2) + 80 + (7-2) + 30 + (12-2) + 2 = 225 present in the clubs.
75 % are not in any of the clubs i.e. 25% are in clubs.
Therefore ,  x/4 = 225 => x =900
Ans is C
by (93 points)


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# Mathematica generated DAE file can not be opened by other softwares [closed]

I am building a 3D model in mathematica, but the exported dae file can not be opened other softwares except mathematica, ie., xcode and preview on mac. As a comparison, the same dae file can be reimported to mathematica just as fine. Don't know what to know.

Here is my code:

test = MeshRegion[{p1, p2, p3, p4, p5, p6, p7, p8}, {Polygon[{1, 2, 3, 4}],
Polygon[{3, 2, 6, 7}], Polygon[{3, 4, 8, 7}],
Polygon[{1, 2, 6, 5}], Polygon[{1, 4, 8, 5}]}]

Export["test.dae", test]
Import["test.dae"]

## closed as off-topic by Henrik Schumacher, MarcoB, m_goldberg, halirutan♦Jun 18 '18 at 3:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – Henrik Schumacher, m_goldberg
If this question can be reworded to fit the rules in the help center, please edit the question.

• I'm voting to close this question as off-topic because it seems most likely that Wolfram Support would have to look into this, rather than end users. – MarcoB Jun 17 '18 at 16:35


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# Definition:Wedge Product

## Definition

Let $\alpha$ and $\beta$ be two differential forms.

Let $\alpha$ be an $x$-form.

Let $\beta$ be a $y$-form.

The wedge product $\alpha \wedge \beta$ is defined as the linear antisymmetric map from $F^x \times F^y \to F^{x + y}$, where $F^a$ is the set of $a$-forms in some manifold.

Let $x_0$ be a specific point in a manifold $X$.

Let $\alpha$ be an $x$-form.

Let $\phi$ be a $1$-form.

Let there be a set of vectors $\mathbf u_1, \mathbf u_2, \dotsc, \mathbf u_x, \mathbf v \in T_{x_0} \left({X}\right)$.

The wedge product is defined as:

$\alpha \wedge \phi \left({\mathbf u_1, \mathbf u_2, \dotsc, \mathbf u_x, \mathbf v}\right) := \sum_P \varepsilon \left({P}\right) \alpha \left({P_1}\right) \phi \left({P_2}\right)$

where:

$P$ is some permutation of $\mathbf u_1, \mathbf u_2, \dots, \mathbf u_x, \mathbf v$
$P_1$ is the first $x$ terms of the permutation $P$
$P_2$ the final term of permutation $P$
$\varepsilon$ is the permutation symbol of $P$.

The sum is taken over all possible permutations.

This definition extends to wedge products of arbitrary forms through the linearity and antisymmetric conditions.


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Help on log derivation

• November 19th 2010, 07:39 AM
softballchick
Help on log derivation

Moderator edit: Restored deleted question. Thread closed.
• November 19th 2010, 07:47 AM
harish21
what have you tried?

Hint: write $x^3(x-2)^3 = (x(x-2))^3=(x^2-2x)^3$

so have $\dfrac{x^3(x-2)^3}{(x^2+8)^7} = (x^2-2x)^3\; (x^2+8)^{-7}$

use the product rule!
• November 19th 2010, 07:48 AM
TheCoffeeMachine
Quote:

Originally Posted by softballchick

Are we to assume that you're required to find the derivative by logarithmic differentiation? It isn't clear!
• November 19th 2010, 12:37 PM
HallsofIvy
If $f(x)= \frac{x^3(x- 2)^3}{(x^2+ 8)^7}$

then $ln(f(x))= 3 ln(x)+ 3ln(x- 2)- 7 ln(x^2+ 8)$

So that $\frac{1}{f(x)}\frac{df}{dx}=$ the derivative on the right.

To find $\frac{df}{dx}$ differentiate on the right and multiply by $f(x)= \frac{x^3(x- 2)^3}{(x^2+ 8)^7}$.
• November 19th 2010, 02:55 PM
softballchick
thanks I got it!
• November 19th 2010, 03:04 PM
TheCoffeeMachine
Quote:

Originally Posted by softballchick
thanks I got it!

1. Be clear as to what you mean next time so as not to waste people's time.
2. Editing out posts after getting help isn't allowed, and it's a sign of cheating.


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# Simple Geometric Proof of Intersecting Circles

The following problem has been bothering quite a while. I guess there is a gap in my school knowledge of geometry, but I do not know how to show that:

Prove or disprove that given two circles with centers $$O_1, O_2$$ of the same radius $$r$$, which intersect at two points $$A, B$$, the intersection of the circles is entirely contained in the circle with the center $$M$$ at the midpoint of the line $$O_1O_2$$ and radius $$AM$$.

In the picture below: $$M$$ would be the origin, and $$A$$ and $$B$$ are two points, where the circles intersect. The desired circle that is supposed to contain the intersection is drawn in green.

• Have you drawn a picture? – kccu Apr 23 at 13:21
• Yes, it looks obvious from the picture somehow. But I cannot put any coherent reasoning on paper. – Renat Sergazinov Apr 23 at 13:24
• Are they always on a horizontal line? That is, are the $y$ coordinates of their centers always the same? This problem is a LOT easier if so – Michael Stachowsky Apr 23 at 14:14
• The problem did not come with the coordinate system. But you can impose it anyway you like. – Renat Sergazinov Apr 23 at 14:32

With the help of the coordinate system, and to simplify calculation, we can draw two circles as you did, where the centres are on the $$x$$-axis, both circles are symmetrical about $$y$$-axis, with individual radius being $$r$$, and distance between the two radii being $$d$$. We can obtain three functions for each circle: $$(x+a)^2+y^2=r^2$$ $$(x-a)^2+y^2=r^2$$ $$x^2+y^2=r^2-a^2$$
From the three equations, we compare the $$x$$ value of these three functions for $$-\sqrt{r^2-a^2}≤y≤\sqrt{r^2-a^2}$$, and it is then easy to see that the two curves are in between the two half circles, i.e. the two curves are contained in the circle. Thus the intersection is contained in the circle.


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# Is there a tool for finding Nash equilibria in parametric games?

There are a few tools, either online or offline, that could solve (find equilibrium) in a game explicitly given as a real-valued matrix.

Such tools are Game Theory Explorer and Gambit.

However, as far as I know, these can not be used for finding equilibrium in a game with parameters.

For example, consider a two player game symmetric game, parametrized by some $x\in[0,1]$, which is given by the following payoff matrix for the row player (and $A^t(x)$ for the column player): $$A(x)= \left( \begin{array}{ccc} \frac{x}{2} & x \\ 1-x & \frac{1-x}{2} \\ \end{array} \right)$$

Each equilibrium in this game can be represented as $<p_a,p_b>\in[0,1]^2$ where $p_a$ is the probability of the first player playing the first strategy and $p_b$ is the probability the second player plays the first strategy.

Analytically we can find all the equilibriums for this game

$$EQ(x)= \begin{cases} \{<1,1>\}&\mbox{if } x>\frac{2}{3}\\ \{<3x-1,3x-1>,<1,0>,<0,1>\}&\mbox{if } \frac{1}{3}\leq x\leq\frac{2}{3}\\ \{<0,0>\}& \mbox{else} \end{cases}$$

I'm looking for a solver which is able to understand inputs like $A(x)$ and produce $EQ(x)$.

Is there a tool which solves parametric games?

• How can the entries depend on $x$? Jan 1 '15 at 21:20
• @KristofferArnsfeltHansen - what do you mean? For every $x$, $A(x)$ is a normal game given in matrix form. A parametric solver would be able to produce the output with respect to $x$.
– R B
Jan 2 '15 at 7:41
• I understand that. But how is the input specified? For instance, in your example, the entries are simple linear functions of $x$. Jan 2 '15 at 10:43
• @KristofferArnsfeltHansen - assume you can simply write the matrix as $<f_{1,1}(x),f_{1,2}(x);f_{2,1}(x),f_{2,2}(x)>$, where $f_{i,j}$ is a some function of $x$, e.g. $<x/2, x; 1-x, (1-x)/2>$. Such tool would be very useful (for me anyways) even for a single parameter, small matrices (up to $3\times 3$) and linear functions of the parameter.
– R B
Jan 2 '15 at 11:40
• For small matrices you can simply enumerate over square submatrices and use the characterization of Shapley and Snow. Jan 2 '15 at 12:39

The best suggestion I have if you don't find such a tool (and I doubt one exists) is to do a parameter sweep and solve a bunch of individual instantiations and see what the resulting sets of equilibria say about $EQ()$. Gambit is very tightly integrated with python, so you can easily define your game as a python function with the parameters as arguments and solve a set of instances via a parameter sweep.


text:
Sales Toll Free No: 1-855-666-7446

# Mole Unit

Top
 Sub Topics The amount of gas in a gas sample is most conveniently indicated by specifying the number of molecules in the sample. This number is expressed by the mole number unit. When the mole unit is used, the particles being counted must be indicated. We can be deal with a mole of atoms of a particular element, a mole of electrons, a mole of photons and so forth.   From Avogradro's law it is known that equal volumes of all gases contain equal number of molecules under similar conditions. In other words we can say that different samples of gases containing equal number of molecules occupy same volume under similar conditions. Now, since one mole molecules of all gases contain same number of molecules therefore, they occupy same volume under similar conditions of temperature and pressure.

## Standard Temperature and Pressure

It is very important to understand the general aspects of gas pressure, its measurements and calibrations. Pressure is force excerted over unit area. The unit of pressure is pascal (Pa) which is equivalent to kg/ms2. In chemistry, we usually use a mercury based barometer to measure the pressure. The unit is millimeter of mercury (mmHg). This is the same as the unit torr. Another unit commonly used to denote pressure is atmosphere (atm). The conversion of these units back and forth as

1 atm = 760 mmHg = 760 torr
1 atm = 1.01 $\times$ 105 Pa

Gases can be cmpressed or expanded by adjusting the temperature and other conditions that prevail. In order to standardize the quantities of gases measured and conveyed, arbitrary reference conditions called standard temperature and pressure (STP) have been chosen and internationally accepted.

Standard temperature and pressure (STP)

Temperature = 0oC or 273K
Pressure = 1 atm or 760 mmHg
There is no such thing as standard volume, although molar volume is often used in studying gas behavior. The relationship of a certain mass of a sample gas to its volume, temperature and pressure was investigated by Amadeo Avogadro. The ideal gas law of Avogadro uses the molar mass of the sample gas and compares it to the actual sample mass of the gas sample being investigated. This yields the numbers of moles of gas in the system.