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## Abstract

Recently, the light-absorbing organic carbon, i.e., brown carbon (BrC), has received an increasing attention, because they could significantly absorb the solar radiation in the range of short wavelengths rather than the purely scattering effect. BrC is ubiquitous in the troposphere. It could undergo long range transport within the atmospheric circulation. After the deposition on the surface of snow or ice in the cryospheric region, as the major light absorbing impurities with black carbon and dust, BrC could reduce the snow albedo and accelerate the glacier melting. In this context, this paper summarized the current knowledge of BrC (in aerosols and snow) in the cryospheric regions including the Arctic, Antarctic, and Alpines. Although some works have been conducted in those region, the current dataset on the optical properties of BrC like Absorption Ångström Exponent (AAE ) and Mass Absorption Efficiency (MAE ) is still limited, which hampers stimulating an accurate evaluation of its climate effects. Especially in the Himalayas and Tibetan Plateau, where very limited information concerning BrC is available. Considering biomass burning as a dominant source of BrC, a large amount of emissions from biomass burning in South Asia could reach the Himalayas and Tibetan Plateau, where the climate effect of BrC merits more investigation in the future.

## Keywords

Brown carbon ; Black carbon ; Atmospheric aerosol ; Snow ; Glacier

## 1. Introduction

Carbonaceous components in the atmospheric aerosols play an important role in the climate system, mainly due to their solar absorption and scattering properties (Seinfeld and Pandis, 2012 ). It is well established that black carbon (BC) could strongly absorb the solar radiation in visible bands (Ramanathan and Carmichael, 2008  ;  Xu et al., 2009 ), resulting in a direct radiative forcing ranging from 0.17 to 1.48 W m−2 (Bond et al., 2013 ). In contrast to BC, organic carbon in the atmospheric aerosols is traditionally considered to be purely scattering by climate models. However, recent research demonstrated that substantial fraction of organic carbon in aerosols could absorb the light in the range of short wavelengths. Such kinds of organic substances with wavelength-dependent absorption are defined as brown carbon (BrC) (Andreae and Gelencser, 2006 ). BrC has multiple primary sources such as biomass burning (Saleh et al., 2014  ;  Washenfelder et al., 2015 ), fossil fuel (e.g., coal) combustion (Bond, 2001  ;  Yang et al., 2009 ), biogenic aerosols (e.g., plant debris and fungi) and soil humic matters. Meanwhile BrC could also be secondarily formed from anthropogenic or biogenic precursors (Lack et al., 2013  ;  Zhang et al., 2011 ). Such precursors like isoprene (Limbeck et al., 2003 ) and, lignin pyrolysis products (Gelencser et al., 2003  ;  Hoffer et al., 2004 ) could yield BrC through heterogeneous or multiphase reactions in the presence of sulfuric acid or hydroxyl radicals. Although the source of BrC may varied with different locations and environment (e.g., urban, rural, forest, ocean, mountain), globally, biomass burning was identified as the most important source of BrC (Chung et al., 2012 ; Lack et al., 2012  ;  Laskin et al., 2015 ).

BC, BrC, and mineral dust are the dominant light-absorbing substances in atmospheric aerosols. After deposited on the surface of snow and glaciers through dry/wet deposition processes, those light-absorbing substances could efficiently reduce the snow albedo (surface darkening), decrease the upwelling radiation, thereby accelerating snow melting (Flanner et al., 2007 ; Kaspari et al., 2015  ;  Yang et al., 2015 ). A brief illustration of the sources, transport and deposition of BrC, as well its effects on the climate system is presented in Fig. 1 .

 Fig. 1. Schematic overview on the sources, transport, and radiative forcing of BrC, as well as its impact on the snow and glaciers after deposition.

The cryosphere, comprising snow, river and lake ice, sea ice, glaciers, ice shelves and ice sheets, and frozen ground, is crucial in the Earths climate system (IPCC, 2014 ). The cryosphere is very sensitive to climate changes and anthropogenic activities. Currently, most efforts dedicated to the light-absorbing substances were focused on BC and dust, which were well documented in recent review literatures (Bond et al., 2013 ; Gertler et al., 2016  ;  Wang et al., 2014 ). However, in this review, the state of art on the BrC in the cryosphere will be specifically summarized. Section 2 introduces the basic optical parameters of BrC, as well the current dataset available for the occurrence of BrC in aerosol and snow in the Artic, Antarctic, and mountain glaciers. Section 3 presents the current knowledge about the chemical speciation of BrC. Section 4 describes the radiative forcing of BrC in the atmosphere and snow, i.e., its impact on the cryosphere. Finally, research perspectives for the future research are proposed in Section 5 .

## 2. Optical properties of BrC

The basic theory used to describe light absorption properties of BrC is the well-known power-law relationship. It can be defined as the following equation:

 ${\displaystyle Abs_{\lambda }={\mbox{K*}}{\lambda }^{-AAE},}$
( 1)

where Absλ (unit: Mm−1 or M−1 ) refers to the absorption coefficient of BrC at wavelength λ (unit: nm). K is a constant. AAE indicates the wavelength dependence of light absorption. It could be derived by fitting the power-law relationship Eq. (1) to the scanned spectra or calculated for a given range based on the following equation (Hoffer et al., 2006 ):

 ${\displaystyle AAE=-{\frac {{\mbox{ln}}\left(A_{{\lambda }_{1}}/A_{{\lambda }_{2}}\right)}{{\mbox{ln}}\left({\lambda }_{1}/{\lambda }_{2}\right)}},}$
( 2)

where Aλ is the absorbance of BrC at wavelength λ. It can be directly measured from the spectrophotometer as the attenuation of incident light intensity.

In the cryosphere, McNaughton et al. (2011) found that AAE of BrC in the western Arctic aerosols vary between 1.5 and 3, while Corr et al. (2012) reported much larger AAE of BrC (3–7) in the same location. The AAE of BrC is larger than that of BC (typically unity), which means the light absorptivity of BrC will increase more dramatically as the wavelength become shorter (an example is provided in Fig. 2 ). Furthermore, the magnitude of AAE is closely relevant to the particle size and composition ( Bahadur et al., 2012  ;  Moosmueller et al., 2011 ), implying that the difference in AAE of BrC may be due to the various sources.

 Fig. 2. The absorption spectra of a typical BrC compound (Suwannee River Fulvic Acid (SRFA) from International Humic Substance Society) measured with a Hitachi U-3900H.

In addition to AAE , another crucial parameter for the definition of BrCs light absorption property is MAE (unit: m2  g−1 ), which demonstrate the light absorptivity for a specific mass unit of BrC. It could be derived from the following equation:

 ${\displaystyle MAE={\frac {Abs_{\lambda }}{C}},}$
( 3)

where C is the concentration of BrC. The MAE of BrC decreases sharply with an increasing wavelength, which means it has much stronger absorption in the shorter wavelength ( Alexander et al., 2008  ;  Andreae and Gelencser, 2006 ). To our best knowledge, there are very limited studies available on the MAE of BrC in the cryosphere. The MAE and AAE of BrC reported or used in the published literatures are listed in Table 1 . McNaughton et al. (2011) reported that MAE of BrC in the western Arctic aerosols is (0.83 ± 0.5) m2  g−1 at 470 nm and (0.27 ± 0.08) m2  g−1 at 530 nm, which are comparable to the MAE of BrC in some rural or even urban aerosols, for instance, (0.7 ± 0.2) m2  g−1 at 365 nm in Gosan, Korea (Kirillova et al., 2014 ) and 0.60 m2  g−1 at 365 nm in DeKalb, USA (Hecobian et al., 2010 ). While the MAE of BC and dust in snow across Northern China were 6.3 m2  g−1 and 0.9 m2  g−1 , respectively (Wang et al., 2013 ). Similarly, the MAE of BC in Arctic snow was 6.0 m2  g−1 (Hegg et al., 2010 ). Although the MAE of BrC is lower than that of BC, the global burden of BrC aerosols was simulated to be 0.65 mg m−2 , nearly 3-fold of the BC concentration (0.19 mg m−2 ) in the atmosphere (Feng et al., 2013 ). Given the large abundance of BrC in the environment, a contribution of light absorption by BrC may not be ignored.

Table 1. Summary of light-absorbing properties of BrC reported or used in the published literatures.
Site Sampling period Sample type MAE (W)a (m2  g−1 ) AAE Reference
Western Arctic 31 March–19 April, 2008 Aerosol 0.83 ± 0.15 (470 nm); 0.27 ± 0.08 (530 nm) 1.5–3 McNaughton et al., 2011
Western Arctic Spring and summer of 2008 Aerosol 3–7 Corr et al., 2012
Artic 1998 and 2005–2009 Snow 5 Doherty et al., 2010
Northern China January and February of 2010 Snow 0.3 (550 nm) 6 Wang et al., 2013
North America 10 January and 28 January–21 March, 2013 Snow 5 Doherty et al., 2014

a. W refers to the wavelength at which the MAE (mass absorption efficiency) is calculated; AAE (Absorption Ångström Exponent).

## 3. Chemical properties of BrC

The chemical composition of BrC is extremely complex. Nevertheless, in terms of chemical speciation, BrC could be generally divided into humic-like substances (HULIS) and tar ball (Chakrabarty et al., 2010 ; Cheng et al., 2016  ;  Hoffer et al., 2016 ).

### 3.1. HULIS

HULIS are termed as their similarity with terrestrial/aquatic humic substances. They are mixtures of various macromolecular organic compounds and ubiquitous in atmospheric environment (Graber and Rudich, 2006 ; Lin et al., 2010a ; Mukai and Ambe, 1986  ;  Zheng et al., 2013 ). HULIS mainly consist of aromatic and aliphatic structure bond with hydrocarbon side chains like hydroxyl, carbonyl, carboxyl and organosulfate functional groups (Chakrabarty et al., 2010 ; Graber and Rudich, 2006  ;  Mukai and Ambe, 1986 ).

#### 3.1.1. Experimental methods for extraction and determination of HULIS

Given the unclear physical-chemical characteristics of HULIS, there is no uniform extract method applicable in all conditions. It is even hard to evaluate whether an extract method valid to isolate HULIS from atmospheric media (Graber and Rudich, 2006 ). Table 2 summarizes the isolation and quantification method applied in published literatures. It is obvious that the solid phase extraction (SPE) method is the most frequently employed in isolating HULIS from aerosols and snow in cryospheric regions for its simple operation and high extract efficiency (Facchini et al., 2000  ;  Fan et al., 2012 ). Specifically, there are four different SPE methods, owing to the diverse sorbents: ENVI-18 (Fan et al., 2012 ), HLB (Varga et al., 2001 ), XAD (Duarte and Duarte, 2005 ) and DEAE (diethylaminoethyl) (Baduel et al., 2009 ). They are competent to the exaction of HULIS with low limit of detection and high recovery yield. However, they also have their particular preference and disadvantage in the isolation of HULIS (Fan et al., 2012  ;  Fan et al., 2013 ). HLB method favors polar organic matters with high recovery. It also can preserve the nature of samples while not change the properties of isolated HULIS. XAD tends to isolate hydrophobic and aromatic components due to its sorbent (non-ionic macro porous resin) used. Acidizing of solute is not necessary for DEAE and it prefers the compounds containing more aromatic signatures. However, large amount of salts could present in the eluents with DEAE extraction, which will interfere the characterization of HULIS.

Table 2. Summary of isolation and quantification method for HULIS in cryosphere.
Site Sample type Isolat. Meth. Quantif. Meth. HULIS concentration (μgC L−1 ) Reference
Min Max Average
Station Nord, Arctic Aerosol (PM10 ) HLB TOC 0.02a Nguyen et al., 2014
Barrow, Alaska, Arctic Snowpack DEAE TOC 1 16 Voisin et al., 2012
Alaska, Arctic Snowpack UV–Vis 1200b 1500b France et al., 2012
Col du DÔme, Alpine Ice core UV–Vis 50 400 Legrand et al., 2007
Col du DÔme, Alpine Ice core DEAE TOC 4 250 Guilhermet et al., 2013
Antarctic Snow, coast XAD-8 Gravimetry 16c 397c Calace et al., 2005
Antarctic Snow, inland XAD-8 Gravimetry 25c 146c Calace et al., 2005

a. in the unit of μgC m−3 .

b. the concentration of non-BC light-absorbers in snowpack.

c. in the unit of μg L−1 .

For the quantification of HULIS followed by its isolation, there are total organic carbon (TOC) measurement (Kiss et al., 2002 ; Krivacsy et al., 2008  ;  Salma et al., 2007 ), UV–Vis detection (Krivacsy et al., 2008  ;  Samburova et al., 2005 ), evaporative light-scattering detection (Emmenegger et al., 2007  ;  Lin et al., 2010b ) and gravimetric method (Kiss et al., 2002 ). TOC method is an easy and common approach to analyze the amount of HULIS-C. Hoffer et al. (2006) assumed the conversion factor is 1.90 between HULIS and HULIS-C for the aerosol samples collected in the Amazon basin, while 1.81 was used by Salma et al. (2007) for the quantity of HULIS at Budapest, Hungary. The choosing of proper conversion factor is critical during such calculation, which may arise large uncertainties.

#### 3.1.2. Abundance of HULIS in aerosols and snow

We focus on the concentration of atmospheric HULIS at the regions where there are a wide range of distributions of glacier and ice sheets (i.e., Arctic, Alps). Considering the albedo modification, glaciers are very sensitive to the increase in light-absorbing impurities like HULIS. Meanwhile, the analysis of the HULIS trapped and accumulated in polar ice sheets and mid-latitude mountain glaciers provides a unique chance to reconstruct the changing atmospheric composition in the past (Legrand et al., 2013 ).

Nguyen et al. (2014) found the annual mass concentration of HULIS in Arctic aerosols is (0.02 ± 0.01) μg m−3 and the concentration of HULIS showed an obvious seasonal variation with the highest value occurred in biomass burning influenced periods. Guilhermet et al. (2013) estimated the concentration of water-soluble atmospheric HULIS was 0.07 μg m−3 in summer, basing on the analysis of HULIS trapped in the ice core at Col du Dôme (4250 m a.s.l.), Alpine.

Legrand et al. (2007) determined HULIS in an Alpine ice core by measuring the absorbance with a UV–VIS spectrophotometer at the wavelengths of 250 and 350 nm. They found that summertime mass concentrations of HULIS were generally below 150 × 10−9 before 1945, while the contributions tended to increase after World War II (ranging from 200 × 10−9 to 900 × 10−9 ). The increased HULIS in Alpine glaciers could be attributed to the growing emission of anthropogenic precursors as well as the enhanced atmospheric oxidants (Legrand et al., 2007 ).

Using a DEAE resin, Legrand et al. (2013) extracted water-soluble HULIS in Antarctic snow samples, and then quantified it with a Shimadzu TOC analyzer. Water-soluble HULIS in central Antarctica snow was determined to be 2 × 10−9 (annual average), much lower than those in Alpine snow of 30 × 10−9 . In Arctic (Barrow, Alaska) snowpack, Beine et al. (2011) estimated that HULIS and other unknown chromophores could account for nearly half of the total absorption measured between 300 and 500 nm. Voisin et al. (2012) further calculated the MAE of HULIS in the snow from the same region as (26 ± 11) cm2  mg−1 at 250 nm. And AAE varied from 4.5 (between 300 and 350 nm) to 7.7 (between 450 and 500 nm). The high values of AAE were the typical characteristics of BrC ( Andreae and Gelencser, 2006 ), i.e., the HULIS in snow has strong absorption in the short wavelengths.

### 3.2. Tar balls

Tar ball, emitted from biomass burning (especially in smoldering condition), is an important type of BrC (Chakrabarty et al., 2010  ;  Hoffer et al., 2016 ). They are spherical (30–500 nm in diameter), amorphous, and typically not aggregated with other particles (Hand et al., 2005  ;  Posfai et al., 2004 ), which are distinct from soot in terms of morphology (i.e., BC) (Fig. 3 ). According to Alexander et al. (2008) , the light absorption of tar ball (carbon sphere) in East Asian outflow may be comparable to or even larger than soot at the individual particle level. The derived mean MAE of tar balls range from ∼3.6 m2  g−1 to 4.1 m2  g−1 at 550 nm, approximating to that of BC (Alexander et al., 2008 ). For the laboratory-generated tar balls, their MAE values were determined to be 0.8–3.0 m2  g−1 (λ  = 550 nm), with average AAE of 2.9 in the wavelength band of 467–652 nm ( Hoffer et al., 2016 ).

 Fig. 3. Secondary electron images of (a) tar ball and (b) soot in the Himalayan aerosols (6500 m a.s.l.) observed by scanning electron microscopy (Cong et al., 2010 ).

Tar ball is ubiquitous in the troposphere, especially in the regions receiving strong influence of biomass burning. According to our previous study in the Himalayas and Tibetan Plateau aerosols, tar balls accounted for 3%–27% of the particle number concentrations (Cong et al., 2009 ). Given the abundance and their optical properties, tar balls impart considerable effects in radiative forcing. However, few studies on the tar balls have been carried out in the snow surface and ice core. At the same time, the degree to which tar balls contribute to the solar absorption of BrC, especially relative to HULIS, is still unclear. Their hygroscopic properties, mixing states and aging processes urgently require more attentions to achieve accurate modeling of the radiative forcing.

## 4. Radiative forcing of BrC in the atmosphere and snow

Globally, the radiative forcing by BrC was estimated to be in the range of 0.1–0.25 W m−2 , close to 25% of that by BC (Feng et al., 2013 ). Chung et al. (2012) also highlighted the global significance of BrC. They reported that it could be responsible for 20% of carbonaceous aerosol solar absorption at 550 nm. Moreover, Lin et al. (2014) calculated the global absorption of BrC in the atmosphere ranging from +0.22 W m−2 to + 0.57 W m−2 , which is equal to 27%–70% of the BC absorption. Nevertheless, regarding the cryospheric area, such assessment in the regional scale is still absent.

BrC in the atmosphere could be transported to the surface of snow by dry or wet deposition, which will further lead to the enhancement of snow melting. For the radiative forcing of BrC in snow, most current studies focus on their contribution to the total light absorption relative to BC and dust, rather than providing an absolute value (in the unit of W m−2 ). Dang and Hegg (2014) proposed that HULIS and polar organic carbon from the western North American snow samples account for 13% of the total light absorption. While for the Arctic snow samples, Doherty et al. (2010) proportioned about 40% of the visible and ultraviolet absorption to BrC. However, it should be kept in mind that this approach must base on a given AAE values of BrC, which might be variable with different sources and environment. Apart from these regional analysis, Lin et al. (2014) applied the IMPACT (Integrated Massively Parallel Atmospheric Chemical Transport) model to assess the global radiative forcing of organic aerosols deposited over in-land snow with the range of 1.1 × 10−3  W m−2 and 3.1 × 10−3  W m−2 . These values could account for as large as 24% of radiative forcing by BC. However, they can not separate the radiative forcing of BrC from total organics yet.

Moreover, BrC, BC and dust as the major light-absorbing matters in the cryosphere, they tend to mix with each other during the transport or aging process (Kaspari et al., 2014 ; Lack and Cappa, 2010  ;  Liu et al., 2015 ), leading to another form of radiative forcing by acting as a lens, i.e., lensing effect. Lack et al. (2012) found that the absorption enhancement ascribes to internal mixing can be 1.4 times at 532 nm in biomass burning particles. The enhancement strongly dependent on the coating thickness, mixing state (Liu et al., 2015 ). Similar with BrC, dust also might mixed with BC to change its albedo reduction in snow (Kaspari et al., 2014 ). Thus BrC in the cryosphere not only absorbs solar radiation directly but also can leads to an extra radiative forcing by interaction with BC. If the lensing effect was incorporated into the modeling, the current results on the radiative forcing of BrC may be modified considerably.

## 5. Perspective

As discussed above, unlike BC, the chemical and optical properties of BrC remains poorly understood especially in the cryospheric regions. The measurements of BrC still suffer from some problems. The light absorption characteristics of BrC were commonly determined in the water extracts. However, the remaining fraction of BrC (water insoluble) is usually ignored. In particular, the water-insoluble BrC may have higher mass-specific absorption than the water-soluble fraction. Another problem is that BrC were extracted by different chemical solvents like acetone and methanol besides water in different studies. Therefore, their abundances obtained by different pretreatment and determination methods are difficult to compare with each other. For many researches, BrC is defined as the light-absorbing carbon measured by optical instruments such as Aethalometer. The relationship between the specific chemical composition of BrC and optical properties also needs more research. Another uncertainty is associated with the determination of MAE of BrC in the mixing state. In order to derive reliable result from this method, more precise optical knowledge of BrC is needed. The hydroscopic properties of BrC need more attention, since it determines the role as cloud condensation nuclei for the formation of clouds.

Discrepancy between observations and model results in a substantial uncertainties in quantifying the effect of BrC. Accurate model simulation of BrC depends on our comprehensive knowledge regarding its sources, chemical composition, optical properties, and so on. More BrC measurements are needed not only in laboratory characterization, but also in field campaigns being carried out in surface air, total atmospheric column and snow. The mixing state, aging and transformation during wet scavenging and, dry deposition on the glacier surface are also important issues.

In terms of geographic distribution of research site, although there are several works have been conducted in Alpine, Arctic and Antarctic, the knowledge about the spatial distributions of BrC is still in deficit especially in the Himalayas and Tibetan Plateau. High levels of carbonaceous aerosol from biomass/biofuel burning exist over South Asia (Bonasoni et al., 2010 ; Gustafsson et al., 2009  ;  Luthi et al., 2015 ). Increasing evidences have demonstrated that those air pollutants could reach the high altitude of Himalayas and even transport into the inland of the Tibetan Plateau (Cong et al., 2015 ). Whats more, the strong solar radiation on the Himalayas and Tibetan Plateau may also accelerates the formation of BrC through the photochemical reactions. With the vast snow and ice cover (more than 100,000 km2 (Yao et al., 2012 )), the Himalayas and Tibetan Plateau have been proved to be very sensitive to climate change. Since biomass burning and secondary formation are the dominant sources of BrC, climate effects of BrC in the Himalayas and Tibetan Plateau (not only the atmospheric warming but also the reduction of snow albedo) merit more in-depth investigation.

## Acknowledgments

This work is supported by National Science Foundation of China (41522103 , 41501082 and 41225002 ) and Strategic Priority Research Program-Climate Change: Carbon Budget and Relevant Issues (XDA05100105 ).

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# Calculating entropy, microstate/macrostate probabilities

1. Nov 3, 2013

### Flucky

Hi all, could somebody look over my answer please. I'm pulled the equation I used off the internet but can't remember where so I'm not sure what it's called.

I took a picture of my answer as I thought it would be easier to read than fiddling with symbols here.

QUESTION

Cheers,
James

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2. Nov 5, 2013

### BruceW

I don't really understand your calculation. I would agree that 0.144 is the probability of finding the system in the macrostate of "half A and half B". But then after that, you try to use this probability to calculate entropy... That doesn't really make sense to me. When you calculate entropy, you should be summing over each of the possible states, and using the probability of each of these states. I don't see how your calculation does that.

Also, I think maybe you are making things more difficult than you need to. You know Boltzmann's entropy formula, right? why not use that? The problem says to calculate the entropy associated with the macrostate "half A and half B", so Boltzmann's entropy formula is ideal.

3. Nov 7, 2013

### Flucky

Hi BruceW, yeah turned out I was using the wrong equation so sending myself in circles. You were right I did need to use the Boltzmann entropy formula which came out with the correct result.

4. Nov 7, 2013

### BruceW

cool, glad you figured it out in the end :)

format: plain


text:
## ASVAB Math Knowledge Practice Test 725758

 Questions 5 Topics Cubes, Rectangle & Square, Slope-Intercept Equation, Triangle Geometry, Two Equations

#### Study Guide

###### Cubes

A cube is a rectangular solid box with a height (h), length (l), and width (w). The volume is h x l x w and the surface area is 2lw x 2wh + 2lh.

###### Rectangle & Square

A rectangle is a parallelogram containing four right angles. Opposite sides (a = c, b = d) are equal and the perimeter is the sum of the lengths of all sides (a + b + c + d) or, comonly, 2 x length x width. The area of a rectangle is length x width. A square is a rectangle with four equal length sides. The perimeter of a square is 4 x length of one side (4s) and the area is the length of one side squared (s2).

###### Slope-Intercept Equation

A line on the coordinate grid can be defined by a slope-intercept equation: y = mx + b. For a given value of x, the value of y can be determined given the slope (m) and y-intercept (b) of the line. The slope of a line is change in y over change in x, $${\Delta y \over \Delta x}$$, and the y-intercept is the y-coordinate where the line crosses the vertical y-axis.

###### Triangle Geometry

A triangle is a three-sided polygon. It has three interior angles that add up to 180° (a + b + c = 180°). An exterior angle of a triangle is equal to the sum of the two interior angles that are opposite (d = b + c). The perimeter of a triangle is equal to the sum of the lengths of its three sides, the height of a triangle is equal to the length from the base to the opposite vertex (angle) and the area equals one-half triangle base x height: a = ½ base x height.

###### Two Equations

When presented with two equations with two variables, evaluate the first equation in terms of the variable you're not solving for then insert that value into the second equation. For example, if you have x and y as variables and you're solving for x, evaluate one equation in terms of y and insert that value into the second equation then solve it for x.


text:
# [OS X TeX] Space around equal sign

Joseph Wright joseph.wright at morningstar2.co.uk
Tue Nov 2 12:31:07 EDT 2010

On 02/11/2010 16:27, Cleo wrote:
> Hi guys,
>
> When I need to write some math, say $x=y+n$, I usually use $x~=~y~+~n$ to avoid
> it breaks between lines. However, sometimes it appears an unwanted space before
> and after the equal sign. I have tried to solve this by using $x~\!=\!~y~+~n$
> but not always it works.
>
> Is there a way to avoid an expression to break and maintain a "constant" space
> around the equal sign?
>
\hbox{$x = y + n$}


text:
시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
5 초 128 MB 54 18 12 32.432%

## 문제

The railway network of the Byteotian Railways (BR) consists of bidirectional tracks connecting certain pairs of stations. Each pair of stations is connected by at most one segment of tracks. Furthermore, there is a unique route from every station to every other station. (The route may consist of several segments of tracks, but it may not pass through any station more than once.)

Byteasar is an undercover inspector of the BR. His job is to pick one of the stations (denote it by S) for centre of his operations and to travel to all other stations. His journey should be as follows:

• Byteasar starts in station S.
• Next, he picks one of the stations he did not yet control and goes to it along the shortest path (by train, of course), inspects the station, and then goes back to S.
• The crooked employees of BR warn one another of Byteasar's comings. To deceive them, Byteasar picks the next station for control in such a way that he sets off from the station  in different direction than the last time, i.e., along a different segment of tracks leaving from S.
• Each station (except S) is inspected exactly once.
• After inspecting the last station Byteasar does not come back to S.

The travel time along every segment of tracks takes the same amount of time: one hour.

Byteasar intends to consider all the stations as the initial station S. For each of them he wants to know the order of inspecting the remaining stations that minimises the total travel time, provided that it is possible at all for that particular S.

## 입력

The first line of the standard input contains a single integer n (1 ≤ n ≤ 1,000,000) that denotes the number of stations. These are numbered from 1 to n. The following n-1 lines specify the track segments, one per line. Each of them holds two integers a,b (1 ≤ a,b ≤ n, a≠b), separated by a single space, indicating that there is a track segment connecting the stations a and b. Each track segments appears exactly once in the description.

In tests worth at least 30% of the points it holds additionally that n ≤ 2,000.

## 출력

Your program should print n lines on the standard output, each holding a single integer. The one in the i-th line should be the minimum number of hours Byteasar has to spend travelling to inspect the stations when S=i - if inspecting them all is possible for S=i; if it is not, the i-th line should hold the number -1.

## 예제 입력 1

9
3 6
2 4
2 6
2 5
1 7
2 7
8 9
7 8

## 예제 출력 1

-1
23
-1
-1
-1
-1
-1
-1
-1

## 힌트

The figure illustrates the railway network as specified in the example. It is possible to inspect all the stations only for S=2. One optimal order of inspection is: 7, 4, 8, 6, 1, 5, 3, 9. It results in 23 hours of travel.


text:
## Algebra 1

The sequence: 2, 20, 200, 2000, ... Finding the pattern: Difference between 1st and 2nd term: 2nd term-1st term=Difference 20-2= 18 Difference between 2nd and 3rd term: 3rd term-2nd term=Difference 200-20=180 We see that the difference is not constant thus the sequence is not arithmetic. Thus we need to find the ratio between consecutive terms. Ratio of first and second term $\frac{Second term}{first term}$=Ratio =$\frac{20}{2}$ =10 Ratio of Second and third term $\frac{third term}{Second term}$=Ratio =$\frac{200}{20}$ =10 Ratio of Third and Fourth term $\frac{Fourth term}{third term}$=Ratio =$\frac{2000}{200}$ =10 Therefore, the previous term is being multiplied by 10 Finding the next two terms: Fifth term= Fourth term $\times$ 10 = 2000 $\times$ 10 = 20,000 Sixth term= Fifth term $\times$ 10 = 20,000 $\times$ 10 = 200,000 Therefore the next two terms are 20,000 and 200,000


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Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than $$\frac{1}{5}$$?

(1) Reciprocal of the median is a prime number.

(2) The product of any two terms of the set is a terminating decimal.
[Reveal] Spoiler: OA

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Official Solution:

(1) Reciprocal of the median is a prime number. If all the terms equal $$\frac{1}{2}$$, then the $$median=\frac{1}{2}$$ and the answer is NO but if all the terms equal $$\frac{1}{7}$$, then the $$median=\frac{1}{7}$$ and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of $$\frac{1}{2}$$ or/and $$\frac{1}{5}$$. Thus the median could be $$\frac{1}{2}$$, $$\frac{1}{5}$$ or $$\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}$$. None of the possible values is less than $$\frac{1}{5}$$. Sufficient.

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01 Oct 2014, 15:14
Hello Bunuel. Can you please explain why do you have 7/20 and not 1/10...1/2*1/5 = 1/10..why do you have to add 1/5 and 1/2 and then divide it by 2? I did not get the rational behind it....

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02 Oct 2014, 01:58
p2bhokie wrote:
Hello Bunuel. Can you please explain why do you have 7/20 and not 1/10...1/2*1/5 = 1/10..why do you have to add 1/5 and 1/2 and then divide it by 2? I did not get the rational behind it....

Set A consist of 10 terms. The median of a set with even number of terms is the average of two middle terms, when arranged in ascending/descending order.

If two middle terms are 1/5, then the median is simply 1/5.
If two middle terms are 1/2, then the median is simply 1/2.
If two middle terms are 1/5 and 1/2, then the median is $$\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}$$.

Hope it's clear.
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02 Oct 2014, 12:35
Hi Bunuel,
i could not understand the 2nd statement

(2) The product of any two terms of the set is a terminating decimal

so what is the set looks like? Is it five 1/2 and five 1/5?

Thanks for your help.

Bunuel wrote:
Official Solution:

(1) Reciprocal of the median is a prime number. If all the terms equal $$\frac{1}{2}$$, then the $$median=\frac{1}{2}$$ and the answer is NO but if all the terms equal $$\frac{1}{7}$$, then the $$median=\frac{1}{7}$$ and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of $$\frac{1}{2}$$ or/and $$\frac{1}{5}$$. Thus the median could be $$\frac{1}{2}$$, $$\frac{1}{5}$$ or $$\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}$$. None of the possible values is less than $$\frac{1}{5}$$. Sufficient.

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02 Oct 2014, 14:26
sunita123 wrote:
Hi Bunuel,
i could not understand the 2nd statement

(2) The product of any two terms of the set is a terminating decimal

so what is the set looks like? Is it five 1/2 and five 1/5?

Thanks for your help.

Bunuel wrote:
Official Solution:

(1) Reciprocal of the median is a prime number. If all the terms equal $$\frac{1}{2}$$, then the $$median=\frac{1}{2}$$ and the answer is NO but if all the terms equal $$\frac{1}{7}$$, then the $$median=\frac{1}{7}$$ and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of $$\frac{1}{2}$$ or/and $$\frac{1}{5}$$. Thus the median could be $$\frac{1}{2}$$, $$\frac{1}{5}$$ or $$\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}$$. None of the possible values is less than $$\frac{1}{5}$$. Sufficient.

The set could be any combination of 1/2's and 1//5:
{1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2}
{1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5}
{1/5, 1/5, 1/5, 1/5, 1/5, 1/2, 1/2, 1/2, 1/2, 1/2}
....
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02 Oct 2014, 14:59
so for example if set is {1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2}
then it does not satisfy this condition (2) The product of any two terms of the set is a terminating decimal

sorry it does. Thanks Bunuel

Bunuel wrote:
sunita123 wrote:
Hi Bunuel,
i could not understand the 2nd statement

(2) The product of any two terms of the set is a terminating decimal

so what is the set looks like? Is it five 1/2 and five 1/5?

Thanks for your help.

Bunuel wrote:
Official Solution:

(1) Reciprocal of the median is a prime number. If all the terms equal $$\frac{1}{2}$$, then the $$median=\frac{1}{2}$$ and the answer is NO but if all the terms equal $$\frac{1}{7}$$, then the $$median=\frac{1}{7}$$ and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of $$\frac{1}{2}$$ or/and $$\frac{1}{5}$$. Thus the median could be $$\frac{1}{2}$$, $$\frac{1}{5}$$ or $$\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}$$. None of the possible values is less than $$\frac{1}{5}$$. Sufficient.

The set could be any combination of 1/2's and 1//5:
{1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2}
{1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5}
{1/5, 1/5, 1/5, 1/5, 1/5, 1/2, 1/2, 1/2, 1/2, 1/2}
....

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04 Nov 2014, 18:47
Bunuel wrote:
Official Solution:

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of $$\frac{1}{2}$$ or/and $$\frac{1}{5}$$. Thus the median could be $$\frac{1}{2}$$, $$\frac{1}{5}$$ or $$\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}$$. None of the possible values is less than $$\frac{1}{5}$$. Sufficient.

Hi Bunuel,

When you say that the second statement implies that the set must consist of 1/2, 1/5, or a mix of both, what exactly implies this? Are 1/2 and 1/5 the only terminating reciprocals of prime numbers and that this is just a rule we should memorize? Or is there something else that gives this away?

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05 Nov 2014, 04:41
OutOfTheHills wrote:
Bunuel wrote:
Official Solution:

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of $$\frac{1}{2}$$ or/and $$\frac{1}{5}$$. Thus the median could be $$\frac{1}{2}$$, $$\frac{1}{5}$$ or $$\frac{\frac{1}{5}+\frac{1}{2}}{2}=\frac{7}{20}$$. None of the possible values is less than $$\frac{1}{5}$$. Sufficient.

Hi Bunuel,

When you say that the second statement implies that the set must consist of 1/2, 1/5, or a mix of both, what exactly implies this? Are 1/2 and 1/5 the only terminating reciprocals of prime numbers and that this is just a rule we should memorize? Or is there something else that gives this away?

The stem says that set A consist of 10 terms, each of which is a reciprocal of a prime number. Therefore the elements of A could be 1/2, 1/3, 1/5, 1/7, 1/11, 1/13, ...

Theory:
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.

Therefore, for the product of any two terms of the set to be terminating decimal, it must consists of only of 1/2's, 1/5's, or both.

Check Terminating and Recurring Decimals Problems in our Special Questions Directory.

Hope it helps.
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06 Nov 2014, 13:53
Bunuel wrote:
OutOfTheHills wrote:
Bunuel wrote:
Official Solution:

Check Terminating and Recurring Decimals Problems in our Special Questions Directory.

Hope it helps.

I was still a bit confused at first, but the directory of problems is a huge, huge help. For anyone with similar problem is me, this is just a rule you should remember.

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24 Aug 2016, 01:54
I think this is a high-quality question and I agree with explanation.

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17 Nov 2016, 20:16
@Bunnel

For statement 2, the median is either 1/2 or 1/5, there are two different answers.
either less or equal to 1/5. How can we take B to be sufficient?

The question asks if median is less than 1/5. It's either yes with median=1/2 or no with median=1/5.

This makes the answer E, as even after combining the two options, we wouldn't get a definite answer.

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17 Nov 2016, 20:41
KaushalPandey wrote:
@Bunnel

For statement 2, the median is either 1/2 or 1/5, there are two different answers.
either less or equal to 1/5. How can we take B to be sufficient?

The question asks if median is less than 1/5. It's either yes with median=1/2 or no with median=1/5.

This makes the answer E, as even after combining the two options, we wouldn't get a definite answer.

How is 1/2 less than 1/5? 1/2 > 1/5.
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22 Nov 2017, 05:56
what if the set consists of 1/2^6-s and 1/5^6-s? In that case, does statement 2 suffice?

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NaeemHasan wrote:
what if the set consists of 1/2^6-s and 1/5^6-s? In that case, does statement 2 suffice?

We are told that "Set A consist of 10 terms, each of which is a reciprocal of a prime number". Are 1/2^6 and 1/5^6 reciprocals of prime numbers?
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Re: M28-55   [#permalink] 22 Nov 2017, 06:09
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# M28-55

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text:
# Contents

## 0.1 Introduction

ClassifyR is a framework for cross-validated classification, with the rules for functions to be used with it explained in Section 0.11 of the introductory vignette. A fully worked example is shown how to incorporate an existing classifier from

## 0.2k Nearest Neighbours

There is an implementation of the k Nearest Neighbours algorithm in the package class. Its function has the form knn(train, test, cl, k = 1, l = 0, prob = FALSE, use.all = TRUE). It accepts a matrix or a data.frame variable as input, but ClassifyR calls transformation, feature selection and classifier functions with a DataFrame, a core Bioconductor data container from S4Vectors. It also expects training data to be the first parameter, the classes of it to be the second parameter and the test data to be the third. Therefore, a wrapper for DataFrame reordering the parameters is created.

setGeneric("kNNinterface", function(measurements, ...) {standardGeneric("kNNinterface")})

setMethod("kNNinterface", "DataFrame", function(measurements, classes, test, ..., verbose = 3)
{
splitDataset <- .splitDataAndClasses(measurements, classes)
trainingMatrix <- as.matrix(splitDataset[["measurements"]])
isNumeric <- sapply(measurements, is.numeric)
measurements <- measurements[, isNumeric, drop = FALSE]
isNumeric <- sapply(test, is.numeric)
test <- test[, isNumeric, drop = FALSE]

if(!requireNamespace("class", quietly = TRUE))
stop("The package 'class' could not be found. Please install it.")
if(verbose == 3)
message("Fitting k Nearest Neighbours classifier to data and predicting classes.")

class::knn(as.matrix(measurements), as.matrix(test), classes, ...)
})

The function only emits a progress message if verbose is 3. The verbosity levels are explained in the introductory vignette. .splitDataAndClasses is an internal function in ClassifyR which ensures that classes are not in measurements. If classes is a factor vector, then the function has no effect. If classes is the character name of a column in measurements, that column is removed from the table and returned as a separate variable. The ... parameter captures any options to be passed onto knn, such as k (number of neighbours considered) and l (minimum vote for a definite decision), for example. The function is also defensive and removes any non-numeric columns from the input table.

ClassifyR also accepts a matrix and a MultiAssayExperiment as input. Provide convenience methods for these inputs which converts them into a DataFrame. In this way, only the DataFrame version of kNNinterface does the classification.

setMethod("kNNinterface", "matrix",
function(measurements, classes, test, ...)
{
kNNinterface(DataFrame(t(measurements), check.names = FALSE),
classes,
DataFrame(t(test), check.names = FALSE), ...)
})

setMethod("kNNinterface", "MultiAssayExperiment",
function(measurements, test, targets = names(measurements), ...)
{
tablesAndClasses <- .MAEtoWideTable(measurements, targets)
trainingTable <- tablesAndClasses[["dataTable"]]
classes <- tablesAndClasses[["classes"]]
testingTable <- .MAEtoWideTable(test, targets)

.checkVariablesAndSame(trainingTable, testingTable)
kNNinterface(trainingTable, classes, testingTable, ...)
})

The matrix method simply involves transposing the input matrices, which ClassifyR expects to store features in the rows and samples in the columns (customary in bioinformatics), and casting them to a DataFrame, which dispatches to the kNNinterface method for a DataFrame, which carries out the classification.

The conversion of a MultiAssayExperiment is more complicated. ClassifyR has an internal function .MAEtoWideTable which converts a MultiAssayExperiment to a wide DataFrame. targets specifies which assays to include in the conversion. By default, it can also filters the resultant table to contain only numeric variables. The internal validity function .checkVariablesAndSame checks that there is at least 1 column after filtering and that the training and testing table have the same number of variables.

## 0.3 Verifying the Implementation

Create a data set with 10 samples and 10 features with a clear difference between the two classes. Run leave-out-out cross-validation.

classes <- factor(rep(c("Healthy", "Disease"), each = 5), levels = c("Healthy", "Disease"))
measurements <- matrix(c(rnorm(50, 10), rnorm(50, 5)), ncol = 10)
colnames(measurements) <- paste("Sample", 1:10)
rownames(measurements) <- paste("mRNA", 1:10)

library(ClassifyR)
trainParams <- TrainParams(kNNinterface)
predictParams <- PredictParams(NULL)
classified <- runTests(measurements, classes, datasetName = "Example",
classificationName = "kNN", validation = "leaveOut", leave = 1,
params = list(trainParams, predictParams))
classified
## An object of class 'ClassifyResult'.
## Data Set Name: Example.
## Classification Name: kNN.
## Feature Selection Name: Unspecified.
## Features: List of length 10 of feature identifiers.
## Validation: Leave 1 Out.
## Predictions: List of data frames of length 1.
## Performance Measures: None calculated yet.
cbind(predictions(classified)[[1]], known = actualClasses(classified))
##       sample   class   known
## 1   Sample 1 Healthy Healthy
## 2   Sample 2 Healthy Healthy
## 3   Sample 3 Healthy Healthy
## 4   Sample 4 Healthy Healthy
## 5   Sample 5 Healthy Healthy
## 6   Sample 6 Disease Disease
## 7   Sample 7 Disease Disease
## 8   Sample 8 Disease Disease
## 9   Sample 9 Disease Disease
## 10 Sample 10 Disease Disease

NULL is specified instead of a function to PredictParams because one function does training and prediction. As expected for this easy task, the classifier predicts all samples correctly.


text:
# Deducing the Gaussian Formula

How would one prove the following formula: For all $n \in \mathbb{N}$, $$\sum_{m=0}^n (-1)^m \binom{n}{m}_q = \begin{cases} \displaystyle \mathop{\prod_{k \text{ odd}}}_{1 \le k\leq n} (1-q^k), & \text{if } n \text{ is even}\\[30pt]0, &\text{if } n \text{ is odd}. \end{cases}$$

I know the odd case follows swiftly from the symmetry property of $q$-binomial coefficients, but I am having trouble working out the even case. Any help would be appreciated!

-

The recurrence of the $q-$ binomial coefficients gives $\binom{2n}{m}_q$= $q^m \binom{2n-1}{m}_q+ \binom{2n-1}{m-1}_q$ and therefore $\sum_{m=0}^{2n} (-1)^m \binom{2n}{m}_q =\sum_{m=0}^{2n-1} (-1)^m q^m\binom{2n-1}{m}_q .$ Formula $(1-q^m) \binom{2n-1}{m}_q=(1-q^{2n-1}) \binom{2n-2}{m-1}_q$ implies therefore $\sum_{m=0}^{2n} (-1)^m \binom{2n}{m}_q =(1-q^{2n-1})\sum_{m=0}^{2n-2} (-1)^{m-1}\binom{2n-2}{m-1}_q .$

Edit

A more elegant proof is the following: Let $r(n)=\sum_{m=0}^n (-1)^m \binom{n}{m}_q.$

Consider the $q-$ exponential function $e(x) = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{(1 - q) \cdots (1 - {q^n})}}} .$

Comparing coefficients gives $\sum\limits_{n = 0}^\infty {\frac{{r(n)}}{{(1 - q) \cdots (1 - {q^n})}}} = e( - x)e(x).$

Now $e(x) = \frac{1}{{(1 - x)(1 - qx)(1 - {q^2}x) \cdots }}$ and therefore $e( - x)e(x) = \frac{1}{{(1 - {x^2})(1 - {q^2}{x^2})(1 - {q^4}{x^2}) \cdots }} = \sum\limits_{}^{} {\frac{{{x^{2n}}}}{{(1 - {q^2})(1 - {q^4}) \cdots (1 - {q^{2n}})}}.}$

Comparing coefficients of the last two series gives

$r(2n) = \frac{{(1 - q)(1 - {q^2}) \cdots (1 - {q^{2n}})}}{{(1 - {q^2})(1 - {q^4}) \cdots (1 - {q^{2n}})}} = (1 - q)(1 - {q^3}) \cdots (1 - {q^{2n - 1}}).$

-


text:
# Question

Control charts for x and R are maintained for an important quality characteristic. The sample size is n= 7; x and R are computed for each sample. After 35 samples we have found that
(a) Set up x and R charts using these data.
(b) Assuming that both charts exhibit control, estimate the process mean and standard deviation.
(c) If the quality characteristic is normally distributed and if the specifications are 220  35, can the process meet the specifications? Estimate the fraction nonconforming.
(d) Assuming the variance to remain constant, state where the process mean should be located to minimize the fraction nonconforming. What would be the value of the fraction nonconforming under these conditions?

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