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## Abstract

Background. In sub-Saharan Africa, 112 million people are infected with Schistosoma haematobium, with the most intense infections in children 5–15 years old.

Methods. We describe a longitudinal epidemiological study that evaluates the relationship between S. haematobium infection and associated morbidity in children before and after the large-scale administration of praziquantel for schistosomiasis and albendazole for soil-transmitted helminths.

Results. At baseline, higher intensities of S. haematobium infection were observed in children with anemia and/or severe microhematuria, but there was no apparent association between the risk of undernutrition and intensity of S. haematobium infection. Significant reductions in the prevalence and intensity of S. haematobium infection 1 year after treatment were, however, observed. Children who benefited the most from anthelmintic treatment in terms of increased hemoglobin concentrations were those who had anemia at baseline and those with highly positive microhematuria scores at baseline.

Conclusions. This study suggests that even a single round of mass chemotherapy can have a substantial impact on S. haematobium infection and its associated morbidity in children.

Improving the health of school-aged children, particularly in developing countries, has emerged as a policy priority in international health [1, 2]. Over the past 2 decades, significant progress has been made in improving child survival, resulting in more children reaching primary school age. However, human infections with 1 of the 5 parasitic helminths of the family Schistosomatidae still represent a significant segment of the global burden of illness, with ∼200 million people infected and with the highest intensities in children 5–15 years old [3]. Schistosomiasis causes granuloma formation and both reversible and irreversible damage to the urinary and intestinal tracts [4]. Furthermore, new estimates of schistosomiasis-related disability have indicated the need to reassess priorities for treating this chronic infection in areas where it is endemic [5].

Praziquantel has been established in several controlled trials as a safe and effective drug for the treatment of infection with all human schistosome species [6–9]. The dramatic reduction in its price since 1990—by >90%, from US$4 to treat a person to approximately US$0.30—has led to the resolution of many of the challenges surrounding large-scale chemotherapy campaigns [10]; recently, through the Schistosomiasis Control Initiative (SCI), >20 million treatments were administered in 2005–2006 in 6 sub-Saharan African countries [11, 12]. One of the primary objectives of these SCI-supported control programs is to achieve, and hence also to demonstrate, a quantifiable reduction in schistosomeassociated morbidity as a consequence of chemotherapeutic intervention.

The aim of the present study was to evaluate the relationship between Schistosoma haematobium infection and associated morbidity in children before and after the large-scale administration of praziquantel and albendazole (against soil-transmitted helminths) by the national Burkinabé helminth control program. A secondary aim was to identify those individuals whom one may predict to show the greatest improvements in nutritional status and hemoglobin (Hb) concentrations after chemotherapy.

## Methods

Control program, study sites, sampling, and cohort design. Both Schistosoma mansoni and S. haematobium are endemic throughout Burkina Faso [13]. The SCI-supported schistosomiasis control program was implemented during 2004 and had treated 3,322,564 school-aged children in the 13 regions of the country through October 2006. Further details about the national Burkinabé helminth control program have been described elsewhere [14].

For the present study, parasitological and morbidity data were collected from a cohort of 1727 Burkinab é children 6–14 years old, randomly sampled from 16 schools before and 1 year after chemotherapy (2004 and 2005, respectively). The schools included in these surveys were randomly selected from all schools in 4 Regional Health Directorates known a priori to be places where schistosomiasis is highly endemic. Details concerning sample-size calculations and cohort design have been described elsewhere [15, 16].

Parasitological and morbidity measures. All children enrolled in the study were interviewed by appropriately trained personnel at the Ministry of Health, Burkina Faso. Ethical clearance was obtained from the Ministry of Health and Imperial College London.

Stool examination. A single stool sample was collected from each child, and 41.7 mg was processed to make duplicate Kato-Katz slides for microscopic determination of intestinal helminth infection. Individual egg output was expressed as eggs per gram of feces, calculated as the arithmetic mean of the 2 individual slide counts whenever these were available.

Urine examination. One urine specimen was collected from each child to determine the prevalence and intensity of S. haematobium infection by the filtration method. The intensity of S. haematobium infection was expressed as the number of eggs per 10 mL of urine. To determine the presence and severity of microhematuria, all urine specimens were tested for presence of detectable blood using urine reagent strips (Bayer Hemastix). The results were recorded semiquantitatively: negative, trace hemolyzed, weakly positive (+), moderately positive (++), and highly positive (+++).

Nutritional assessment. Heights and weights were measured to determine height-for-age z scores (HAZ) and body-mass-index z scores (BMIZ). All measures were obtained using height poles and electronic balances in the morning, and children were barefoot, wearing only light indoor clothing. A low BMIZ is the index of choice for the assessment of recent undernutrition resulting in thinness or wasting, whereas a low HAZ represents long-term growth and nutritional status resulting in shortness or stunting [17, 18]. z scores for each nutritional index were calculated from Centre for Disease Control (National Center for Health Statistics; year 2000) reference values using EpiInfo (version 2000; US Centers for Disease Control and Prevention) [19].

Anemia assessment. Blood samples for Hb concentrations were obtained from each individual by the fingerprick method using a photometer (Hemocue) [20]. Anemia was defined according to World Health Organization (WHO) guidelines [21].

Statistical analyses. Differences between dropouts and children successfully followed up were tested by univariate analysis using a Wilcoxon 2-sample test for means and a χ2 test or Fisher's exact test if there was a small value for proportions. SAS software was used (version 8; SAS Institute).

Hierarchical models are often applicable to modeling data from complex surveys of a population, with a hierarchical structure used to explain relations between individual and supraindividual determinants. A 2-level linear hierarchical model assuming normally distributed errors was fitted to the logarithmically transformed baseline S. haematobium egg counts (ln[χ + 1]), using Gibbs sampling for all parameters [22] to quantify any associations with anemia, measures of nutritional status, and microhematuria while adjusting at the same time for demographic factors such as age and sex. In this way, we tested whether children with pathology potentially induced by S. haematobium infection had higher S. haematobium egg counts before treatment. We used a similar model to quantify changes in S. haematobium egg counts from baseline to 1 year of follow-up. Mlwin software (version 2.01; Multilevel Models Project, Institute of Education, University of London) was used. Box plots of the S. haematobium egg counts at baseline were used for validation and comparison of the significant findings from the model described above.

Changes in Hb concentration and in HAZ and BMIZ scores over time were evaluated using 2-level linear hierarchical models of raw change scores between baseline and the 1-year follow-up time point. With this approach, we aimed to compare the average change in each of the studied outcomes over the studied period between different groups of children. All of the models presented were also adjusted for age and sex. P < .05 was considered to be significant.

An additional 3-level hierarchical linear modeling analysis was performed to determine any change in the children's Hb concentrations. With this model, we aimed to quantify the adjusted overall change in Hb concentration from baseline to follow-up and to quantify average Hb concentrations of different groups of children at baseline as well as to examine whether the intensity of S. haematobium infection was associated with lower Hb concentrations. Logistic random-intercepts regression models were fitted to examine whether the intensity of S. haematobium infection was associated with an increased risk of thinness and shortness at baseline while adjusting for potential confounders.

## Results

A total of 1727 children from 16 schools were recruited at baseline. Of these, 1131 (65%) were successfully retraced at the 1-year follow-up time point, and 321 new children were recruited into the cohort during the second year of the study (data not shown). There were significant differences between the children who were successfully followed up and the dropouts, according to their demographic characteristics and nutritional status as defined by baseline thinness and shortness as well as by baseline Hb concentrations (table 1). The children who dropped out were of an older mean age than those successfully followed up, and boys proved more difficult to recruit into the cohort during the second year of the study. All children who dropped out were wasted and stunted at baseline and had slightly higher mean Hb concentrations. No other baseline characteristic measured varied significantly between children followed up and those who dropped out.

Table 1

Baseline characteristics of Burkinabé schoolchildren who were followed up for 1 year or dropped out.

Table 1

Baseline characteristics of Burkinabé schoolchildren who were followed up for 1 year or dropped out.

Table 2 presents the health indicators of children surveyed at baseline and successfully followed up 1 year after treatment. During the 12 months between examinations, the overall prevalences of S. haematobium, S. mansoni, and hookworm infections decreased significantly (P < .001). For both years examined, Ascaris lumbricoides infection was absent, and the prevalence of Trichuris trichura infection was estimated to be 1.1% at baseline and totally absent 1 year later. Because prevalences and coinfections with S. haematobium were so low for the intestinal helminth species at both time points, such data were not analyzed further here.

Table 2

Health characteristics of children successfully followed up for 1 year (2004–2005), at baseline and after treatment.

Table 2

Health characteristics of children successfully followed up for 1 year (2004–2005), at baseline and after treatment.

A significant increase in mean Hb concentration (P < .001) and a significant decrease in the prevalence of anemia (P = .021) were also observed between 2004 and 2005. Finally, the unadjusted observed changes in both recent and chronic undernutrition from baseline to follow-up were not significant (P = .135 and P = .093, respectively).

Table 3 presents the results of the model of the change in S. haematobium egg counts for 1 year after treatment as well as differences in S. haematobium egg counts between different groups of children at baseline. This model indicated that, compared with baseline counts, there was on average an overall significant decrease in S. haematobium egg counts by 52% 1 year after treatment (P < .001). At baseline, only children 10 and 12 years old had significantly higher S. haematobium egg counts, compared with those who were 14 years old (P = .035 and P = .007, respectively) after controlling for sex, nutritional and anemia status, and microhematuria test scores. Children with +++, ++, +, and trace microhematuria scores had on average significantly higher S. haematobium egg counts than those of children with negative scores at baseline, by 4107%, 1470%, 462%, and 242%, respectively. Additionally, children with anemia at baseline had significantly higher S. haematobium egg counts (by 11%) than children without anemia (P = .026). Boys also had significantly higher S. haematobium egg counts (by 11%) than girls at baseline (P = .020).

Table 3

Estimates from 2-level hierarchical model for Schistosoma haematobium egg counts among 1130 Burkinabé schoolchildren at baseline and after treatment (2004–2005).

Table 3

Estimates from 2-level hierarchical model for Schistosoma haematobium egg counts among 1130 Burkinabé schoolchildren at baseline and after treatment (2004–2005).

Figure 1A shows that children with the most severe microhematuria scores harbored higher intensities of S. haematobium infection than children who were negative for microhematuria at baseline. Children with anemia at baseline also harbored slightly higher intensities of S. haematobium infection than those without anemia (figure 1B).

Figure 1

Box plots for the logarithmically transformed (base e) Schistosoma haematobium egg counts of positive subjects only (n = 613 Burkinabé schoolchildren, 2004). A, Box plot for log S. haematobium egg counts with respect to different microhematuria test scores at baseline. B, Box plots for log S. haematobium egg counts with respect to anemia status at baseline. Data are smallest observations, lower quartiles (Q1), medians, upper quartiles (Q3), and largest observations. +, weakly positive; ++, moderately positive; +++, highly positive.

Figure 1

Box plots for the logarithmically transformed (base e) Schistosoma haematobium egg counts of positive subjects only (n = 613 Burkinabé schoolchildren, 2004). A, Box plot for log S. haematobium egg counts with respect to different microhematuria test scores at baseline. B, Box plots for log S. haematobium egg counts with respect to anemia status at baseline. Data are smallest observations, lower quartiles (Q1), medians, upper quartiles (Q3), and largest observations. +, weakly positive; ++, moderately positive; +++, highly positive.

Results from the 2-level logistic regression model for the probability of being wasted did not suggest associations with any baseline characteristics examined other than age. In particular, there was a trend toward younger children being less likely to be wasted, although none of the other odds ratios (ORs), except for those for 10-year-old children, were significantly different from the ORs for 14-year-old children (P = .022). Furthermore, the 2-level logistic regression model of the probability of being stunted at baseline suggested a trend toward younger children being less likely to be stunted; the OR for 6–12-year-old children was significantly different from that for 14-year-olds. In addition, children with anemia were almost 1.5 times more likely than those without anemia to be stunted at baseline (P = .034; data not shown).

Table 4 contains estimates from the two 2-level linear multilevel models for changes in HAZ and BMIZ during the period studied. The effect for the 14-year-old comparison group was a decrease in BMIZ of 0.519 units, which was of borderline significance (P = .053). This suggests that there was no dramatic change over the period studied in this group. However, children who were 7, 8, and 10 years old at baseline had a greater decrease in BMIZ, compared with 14-year-olds. The coefficients of the continuous variables that refer to the baseline HAZ and BMIZ scores indicated that BMIZ increased for wasted children, whereas, BMIZ decreased for stunted children more than that of children with no nutritional problems. These results can be obtained if one multiplies the coefficients of HAZ and BMIZ by −2 (i.e., the cutoff score that defines wasting and stunting, respectively).

Table 4

Estimates from 2-level hierarchical model for the change of body-mass-index z score (BMIZ) and height-for-age z score (HAZ) (n = 1130).

Table 4

Estimates from 2-level hierarchical model for the change of body-mass-index z score (BMIZ) and height-for-age z score (HAZ) (n = 1130).

The effect for the 14-year-old comparison group was a non-significant (P = .214) increase in HAZ of 0.143 units. Compared with 14-year-olds, 6-year-olds had a significantly greater increase, whereas the 12-year-olds had a decrease in HAZ (significantly different from that of 14-year-olds). Boys had a significantly smaller increase in HAZ than girls. Likelihood ratio tests indicated a better fit in both models mentioned above when we included HAZ and BMIZ as explanatory continuous variables and not the relevant categorical ones that would denote wasting and stunting if HAZ or BMIZ was, respectively, less than −2 SD.

Table 5 contains the estimates of two 3-level hierarchical models for Hb concentrations before and after chemotherapeutic treatment, taking into account adjustment and nonadjustment for microhematuria scores. From the former model it was estimated that an overall increase of 0.092 g/dL in Hb concentration after treatment was not significant (P = .146). Children with +++ microhematuria scores had significantly lower Hb concentrations (0.318 g/dL; P = .016) than micro-hematuria- negative children at baseline. Children who were 6, 7, 8, and 10 years old at baseline had significantly lower Hb concentrations (P = .024, P < .001, P = .019, and P = .008, respectively) than those who were 14 years old, after controlling for intensity of S. haematobium infection, sex, hematuria, wasting, and thinness. From the alternative model, which did not adjust for microhematuria scores, estimates of the parameters mentioned previously remained approximately the same, but the effect of S. haematobium infection intensity became significant, such that children who were heavily infected with S. haematobium at baseline had significantly lower Hb concentrations (0.220 g/dL; P = .024) than uninfected children. This suggests that because the variable of microhematuria scores is related to both Hb concentration and the intensity of S. haematobium infection, 2 different causal effects of S. haematobium infection on Hb concentrations are indicated. Two-way interaction terms were also tested, but because they were not significant, they were omitted from the models presented. Also, variances in all 3 levels of the second model were higher than the corresponding ones in the first model, which implies that the microhematuria scores explain some of the variability in the studied outcome across all 3 levels of the models presented.

Table 5

Estimates from 3-level hierarchical models for hemoglobin concentration before and after treatment, taking into account adjustment and nonadjustment for microhematuria scores (n = 1124).

Table 5

Estimates from 3-level hierarchical models for hemoglobin concentration before and after treatment, taking into account adjustment and nonadjustment for microhematuria scores (n = 1124).

Finally, table 6 contains estimates from the 2-level linear multilevel model for the change in Hb concentration over the course of the period studied. This model suggested that the change varied significantly as a function of the following baseline characteristics: anemic status, +++ microhematuria score, and age. The effect for the comparison group (baseline uninfected, male, 14 years old, not anemic, negative microhematuria score, not wasted, and not stunted) was a decrease in Hb concentration of 0.128 g/dL, which was not significant (P = .747). Significant increases in Hb concentration during the period studied were observed among children with anemia at baseline (increase by 1.360 g/dL [that is, 1.488-0.128 g/dL]; P < .001) and among children with +++ microhematuria scores at baseline (increase by 0.233 g/dL [that is, 0.361-0.128 g/dL]; P = .039).Marginally significant decreases inHb concentration during the same period were observed in children who were 12 years old at baseline; these children had a greater decrease, by 0.733 g/dL (that is, −0.605-0.128 g/dL; P = .055), in Hb concentration than 14-year-old children. In addition, 9-year-old children had a significantly greater decrease in Hb concentration, by 0.847 g/dL (that is, −0.719-0.128 g/dL; P = .019).

Table 6

Estimates from a 2-level hierarchical model for changes in hemoglobin concentration (n = 1130).

Table 6

Estimates from a 2-level hierarchical model for changes in hemoglobin concentration (n = 1130).

## Discussion

Among the different schistosomes infecting humans, S. haematobium is responsible for the largest number of infections. It has been estimated that, in sub-Saharan Africa, 112 million people are infected with S. haematobium, compared with 54 million infected with S. mansoni [23]. However, S. haematobium remains largely unstudied, particularly in comparison to S. mansoni, primarily because of the more demanding conditions for its laboratory maintenance. This is also reflected in the paucity of research examining the potential effectiveness of praziquantel against S. haematobium under various experimental and clinical circumstances [24]. To our knowledge, the present study represents the first longitudinal study in Africa that reports on the relationship between S. haematobium infection and its associated morbidity as a whole, by use of a uniquely detailed large data set from 16 randomly selected schools across the entire national territory of Burkina Faso. Moreover, these data have the potential to provide important evidence characterizing urinary schistosomiasis-associated morbidity, particularly in a population such as that of Burkinabé, where the prevalence of hookworms and other soil-transmitted helminthiasis is estimated to be very low. Although previous studies have tended to focus on the impact of large-scale control programs on intense transmission of S. haematobium infection with regard to Hb concentrations and anemia only (Tohon Z, Boubacar Mainassara H, Elhaj Mahamane A, et al., submitted) [25], to parasitological measures only [26, 27], or to S. haematobium morbidity indicators before and after treatment [28], the present study examined all these issues together and also adjusted for potential differences in demographic characteristics as well as potential confounders.

We have demonstrated that children with anemia or children with more severe microhematuria scores at baseline had higher S. haematobium infection intensities (table 3 and figure 1), which suggests that heavy intensities of S. haematobium infection can be associated with anemia and hematuria. We also provide evidence that heavy infections of urinary schistosomiasis are associated with lower Hb concentrations and, as a consequence, with potential anemia, given that the models in table 5 indicate that S. haematobium infection might be associated with anemia in 2 different ways. More precisely, hematuria—which, as demonstrated previously (Tohon Z, Boubacar Mainassara H, Elhaj Mahamane A, et al., submitted) [29], is associated with S. haematobium infection—can be an important cause of blood and iron loss, which also may lead to anemia. Our data suggest significant reductions in the prevalence and, more importantly, intensity of infection of S. haematobium as well as in the percentages of children with positive microhematuria scores 1 year after treatment (tables 2 and 3).

The children who most benefited from anthelmintic treatment in terms of increased Hb concentration were those with anemia at baseline and those who had +++ microhematuria scores at baseline (table 6), which confirms similar findings presented elsewhere [30, 31]. The mechanisms by which treatment for S. haematobium infection allows Hb concentrations to increase in children with anemia may be the decrease in blood in urine that results from a reduction in the intensity of S. haematobium infection [32].

Growth and nutritional status have been proposed to represent the most sensitive indicators of health in children [33]. Furthermore, one of the factors emphasized in the World Development Report 1993 is the relationship between parasitic infections and malnutrition [2]. We examined whether greater S. haematobium egg counts were associated with increased risks of wasting or stunting at baseline, adjusting for demographic characteristics and other potential predictors such as microhematuria and anemia status. The results of this study did not suggest any significant association between the risk of undernutrition and intensity of S. haematobium infection, with only age being revealed as a significant factor. Younger children tended to be less likely to be wasted or stunted than 14-year-old children, which could imply prior malnutrition in these older children, as has been reported elsewhere [1]. In the aforementioned study, the authors also reported that, in Zanzibari boys, the association between microhematuria and poor growth was only marginally significant [1], which is in line with our findings that, in the Burkinabé children, changes in the BMIZ scores depended only on age, whereas changes in the HAZ scores depended on age and sex (table 4). A more plausible explanation for the lack of statistical association between the intensity of S. haematobium infection and stunting may relate to dropout bias toward stunted children (table 1); this means that it is difficult to make a definitive conclusion regarding the relationship between urinary schistosomiasis and stunting in our study population [34].

Nevertheless, it must be considered that essential methodological constraints inherent in the present study design—in particular the lack of a control group, which was necessary for ethical reasons, could result in some potential bias toward the estimation of the absolute impact of the treatment, thereby allowing only the relative impact of the treatment in different groups of children to be calculated. However, even though these data were obtained from a large-scale control program and studies such as ours are generally difficult to execute in terms of design, methodology, and sampling, we believe that the results will be of substantial value in estimating the total benefit that control of schistosomiasis can provide to communities [35].

This study shows that praziquantel can have a substantial impact on S. haematobium infection and associated disease when delivered as part of a large-scale control program in a country such as Burkina Faso, which was the first country in the WHO African Region to achieve nationwide coverage with anthelminthic drugs against 3 major neglected tropical diseases: lymphatic filariasis, schistosomiasis, and soil-transmitted helminthiasis [36]. Our findings suggest a dramatic reduction in the prevalence and intensity of S. haematobium infection, an improvement in Hb concentration in certain groups of children, and reductions in schistosome-related morbidity in a cohort of Burkinabé schoolchildren, which demonstrate that mass chemotherapy can offer a practical strategy for the control of S. haematobium infection and its associated morbidity.

## Acknowledgments

We extend our sincere thanks to the schoolchildren, teachers, and community members who participated in the study and the staff of Institute de Recherches Scientifiques de Santé and Centre National de la Recherche Scientifique et Technologique in Burkina Faso. We particularly thank Hamado Ouedraogo, Césaire Ky, and Malachie Yaogho for their contribution.

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 [2014] The secondary fixpoint issue. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/15/2014, 08:17 PM (06/15/2014, 08:07 PM)sheldonison Wrote: (06/15/2014, 06:31 PM)tommy1729 Wrote: .... Seems unlikely that sexp contains none of these n-ary fixpoints ?! And as for the functional equation f(x+1) = exp(f(x)) + 2pi i that is on another branch. So that does not seem to help. Conjecture: if $\text{sexp}(z)=L$, than for some positive integer n, there is a non-zero integer m such that $\text{sexp}(z-n)=L+2m\pi i$ This would apply for L equals any fixed point of exp(z). This conjecture would apply to both the Kneser solution, and the secondary fixed point solution. I think it can be proven by showing for these two solutions, that sexp(z-n)<>L as n goes to infinity. Wait a minute , we have sexp(R + oo i) = L or conj(L) for all real R and all real oo !? I thought we all agreed on that for years ? But R + oo i - n is also of the form R + oo i. ?? Now Im even more confused. Thanks for the reply though. regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread [2014] The secondary fixpoint issue. - by tommy1729 - 06/15/2014, 06:31 PM RE: [2014] The secondary fixpoint issue. - by sheldonison - 06/15/2014, 08:07 PM RE: [2014] The secondary fixpoint issue. - by tommy1729 - 06/15/2014, 08:17 PM

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# Math Help - Help me...

1. ## Help me...

1) $|2x+1|<4x-2$

2) $|2x-1|>x+2$

3) $|\frac {x-2}{x+1}|<3$

4) $|2x-1|>\frac {1}{x}$

(Show me as many methods as possible. Thanks)

2. When you have inequalities involving moduluses you need to separate it into the greater than and less than 0 cases as follows

1) |2x+1|<4x-2
first do the case $2x+1 \geq 0$

We want the intersection of $2x+1 \geq 0$ and 2x+1<4x-2
$2x \geq -1$
$x \geq -1/2$

2x+1<4x-2
3<2x
x>3/2

So the intersection is x > 3/2

Next we look at 2x+1 < 0

So we want the intersection of 2x+1<0 and -(2x+1)<4x-2
solving this gives x<-1/2 and x>1/6. This has no intersection so the final answer is simply x > 3/2.
If the intersection did exist, we would take the union of it and the first solution to get the final result.

3. 1) $|2x+1|<4x-2$

$2x+1<4x-2$
$3<2x$
$3/2 ...... i

$2x+1>2-4x$
$6x>1$
$x>1/6$ ...... ii

To fulfill both i and ii, therefore $3/2.

Can I do like that?

4. For
$|2x+1|<4x-2$

It would be
$2x+1<4x-2$

and
$2x+1>2-4x$

Note the change from < to > in the second expression.

-Dan

5. 2) $|2x-1|>x+2$

$2x-1>x+2$
$x>3$ ... i

$2x-1<-x-2$
$3x<-1$
$x<-1/3$ ... ii

6. Originally Posted by SengNee
2) $|2x-1|>x+2$

$2x-1>x+2$
$x>3$ ... i

$2x-1<-x-2$
$3x<-1$
$x<-1/3$ ... ii

Looks good to me. Can you plug in some numbers from your solution set to see if it looks good to you?

-Dan

7. ## linear inequalility solution

Originally Posted by SengNee
1) $|2x+1|<4x-2$

2) $|2x-1|>x+2$

3) $|\frac {x-2}{x+1}|<3$

4) $|2x-1|>\frac {1}{x}$

(Show me as many methods as possible. Thanks)
1) 2x +1< 4x -2
Collect like terms
2x-4x < -2-1
-2x < -3
Divide both sides by -2
X > 1½
2) 2x – 1 > x +2
2x- x > 2+1
x> 3

from clement
cokhale@yahoo.com

8. 1.

set the postive mod to equal the straightline graph to get point of intersection

cant go wrong

9. ## inequality solution

[QUOTE=cokhale;112884]1) 2x +1< 4x -2
Collect like terms
2x-4x < -2-1
-2x < -3
Divide both sides by -2
X > 1½
2) 2x – 1 > x +2
2x- x > 2+1
x> 3

3) X -2\ x +1 < 3
Solution
-----------
Multiple both sides by (x +1)²
X -2\ x +1 (x +1)² < 3(x +1)² = (x -2)(x +1)< 3(x +1)(x +1) = (x -2)(x +1) < 3( x² +2x +1) = (x -2)(x +1)<3x² +6x +3 = x² -x-2< 3x² +6x +3
= 2x² +7x +5<0
2x² +7x +5 =0
2x² + 5x +2x +5
X(2x +5) +1(2x +5)
(2x +5)(x +1)<0
2x +5 =0
2x = -5
Divide both sides by 2
X = -5\2 = -2½
Or
X +1 =0
X= -1
_+____-_______+__
-2.5 0 1 2

Since,< is a negative solution.
-2½ < x < 2.


text:
# Regarding Irreducibility of two variable polynomial

Following is an example taken from Dummit Foote - Abstract Algebra after Proposition $9.4.12$

The idea of reducing modulo an ideal to determine irreducibility can be used also in several variables, but some extra care must be exercised. For example, the polynomial $x^2+xy+1$ in $\mathbb{Z}[x,y]$ is irreducible since modulo the ideal $(y)$ it is $x^2+1$ in $\mathbb{Z}[x]$ which is irreducible and of same degree. In this sort of argument it is necessary to be careful about collapsing. For example the polynomial $xy+x+y+1$ (which is $(x+1)(y+1)$) is reducible but appears irreducible modulo both $(x)$ and $(y)$.The reason for this is that non unit polynomials in $\mathbb{Z}[x,y]$ can reduce to units in quotient ring. To take account of this it is necessary to determine which elements in the original ring becomes units in the quotient.

I do not really understand this...

I realize that $xy+x+y+1$ is reducible in $\mathbb{Z}[x,y]$

going modulo $(y)$ gives $x+1$ which is irreducible.

I do not understand what does he mean when he says The reason for this is that non unit polynomials in $\mathbb{Z}[x,y]$ can reduce to units in quotient ring.

$f(x,y)=xy+x+y+1$ after going modulo $(y)$ is $x+1$ which is clearly not a unit...

So, I do not understand what does he actually wants me to see in this.

Thank you.

EDIT : The proposition that i was referring to and which was used in this example is :

Let $I$ be proper ideal in an integral domain $R$ and let $p(x)$ be a non constant monic polynomial in $R[x]$. If the image of $p(x)$ in $(R/I)[x]$ is irreducible then $p(x)$ is irreducible in $R[x]$

Perhaps we should begin by discussing the proof of the proposition in detail. If we emphasize a particularly important step, we will then see why when we consider $$\mathbb{Z}[x,y]$$, we are lead to the remark that you're asking about.

The proof of the proposition (paraphrasing):

Assume $$p(x)$$ is reducible in $$R[x]$$ but irreducible in $$(R/I)[x]$$. Then $$p(x)=f(x)g(x)$$ with $$f(x), g(x)$$ monic, non-constant polynomials. Then reducing $$f(x)$$ and $$g(x)$$ mod $$I$$ gives a factorization in $$(R/I)[x]$$. Contradiction.

Notice that it was critical that both $$f(x)$$ and $$g(x)$$ were monic (or at least that their leading coefficients are units, which can be shown to be equivalent to choosing them to be monic). Why? Because otherwise $$f(x)$$ could reduce to a unit in $$(R/I)[x]$$ (this could happen easily, think of $$3x+1 \in \mathbb{Z}[x]$$ and then consider $$\mathbb{Z}_3[x]$$). Then we wouldn't have found found that $$p(x)$$ was reducible in $$(R/I)[x]$$ and there would be no contradiction!

For completeness: the key point in this proposition is that we are assuming that $$p(x)$$ is monic and $$I$$ is a proper ideal. Then the leading coefficients of $$f(x)$$ and $$g(x)$$ are units (we argue that we can always multiply by a constant to arrange the leading coefficients to be $$1$$) and since $$I$$ is a proper ideal, these coefficients are not in $$I$$. Then the leading coefficients of $$f(x)$$ mod $$I$$ and $$g(x)$$ mod $$I$$ are not zero and so we have a nontrivial factorization in $$(R/I)[x]$$.

So really we need to emphasize why $$f(x)$$ and $$g(x)$$ couldn't reduce to a unit. Then it becomes more clear why we need to also worry about factors reducing to units when we consider polynomial rings of several variables.

Some further remarks:

Let's say we try to extend this proposition to $$R[x,y]$$. But we immediately have a problem: we know we need some kind of "leading coefficient" condition if we hope to repeat the proof as before (or else we need some other condition and strategy to show that reducible $$q(x,y)$$ cannot become irreducible in a quotient by a proper ideal!). But what we mean by "leading coefficient" is unclear for something like

\begin{align} q(x,y) = x^3y-6xy^3+x^2y^2+x^2y+3xy^2+x-2y+1 \end{align}

Maybe we mean coefficients of terms of highest degree ($$x^3y-6xy^3+x^2y^2$$ in this case). Then is it enough to ask that at least one of them have coefficient $$1$$? All of them? But then notice that

\begin{align} q(x,y) = (x^2y+3xy^2+1)(x-2y+1) \end{align}

and so one of the maximum degree terms $$x^2y^2$$ doesn't arise as simply the product of two lower degree terms - it is the sum of two products. So $$x^2y^2$$ having coefficient $$1$$ didn't mean that it couldn't come from terms that are $$0$$ mod some ideal. The situation can get worse for higher degrees and higher numbers of variables: terms with coefficient 1 could come from a large number of sums of terms. This means we have to think carefully before trying to apply our previous proof.

One such strategy is to recognize that $$R[x,y] \cong (R[x])[y]$$. In other words the multivariable case can be thought of as the single variable case with coefficients coming from the ring $$R[x]$$ and so then the condition is simply that the term with the highest power of $$x$$ (or $$y$$) should be monic (more generally: check that it's enough to require that the coefficient of the highest power of say $$y$$ is not an element of the ideal $$I[x] \subset R[x]$$ that you want to reduce by). If this is the case, then we can simply apply the original proposition of the single variable case. For example

\begin{align} p(x,y) = y^2 + (7x^2+14x)y+17 \end{align}

has this property and so we have $$p(x,y) = y^2 + 3$$ mod $$7$$, which is irreducible and so $$p(x,y)$$ was irreducible.

Notice that Dummit and Foote's example of $$xy+x+y+1$$ does not meet this criteria (neither the coefficient of the highest power of $$x$$ nor $$y$$ is $$1$$), so while this wouldn't have proved it was reducible, it would have prevented us from making the mistake they warn against. This criteria also would have been applicable to their other example $$x^2 + xy+1$$ (and we would have found that it is irreducible).

Since they have not yet introduced the idea of reducing mod some $$I[x,y]$$, hopefully combining the two parts of my answer fully resolves what we're trying to do, why we can do it, and where we can't, at least with respect to the context of what they're trying to point out at this point in the book.

I think the other factor is what you are being asked to consider.

Yes, mod $y$ the factor $x+1$ remains irreducible but the factor $y+1$ becomes a unit.

Hence the product $(x+1)(y+1)$ winds up mod $y$ being an irreducible polynomial.

• Is it ? I am not sure... This does makes sense to me though :O
– user87543
Mar 13, 2014 at 10:37
• Note also the consideration of $x^2 + xy + 1$ mod $y$ being $x^2 + 1$ of the same degree. This eliminates the possibility of one factor "collapsing" to a unit, as your example would illustrate. Mar 13, 2014 at 12:45


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## Intermediate Algebra (6th Edition)

100% of a number $=$ the number
100% of a number $a$ is equivalent to $1a=a$. Therefore, 100% of a number will be equal to that number. Example, $a=10$, so $1a=a=10$. $10=10$


text:
Why isn’t SHA-3 in wider use?

SHA-3 was released by NIST just over 4 years ago this week. In my experience it does not seem to be as widely used as I might have expected. I see SHA-2 and even SHA-1 more often. What are your opinions on why this is the case?:

• This question/answer and those directly linked cover #1: crypto.stackexchange.com/questions/15727/… – Richie Frame Aug 10 at 17:27
• @RichieFrame Perhaps I'm completely missing it, but how does that say anything about security concerns about SHA-3? It seems to be about changing some parameters by NIST, particularly in relation to padding, which supposedly has no adverse affect on security... but it does not mention anything about Bruce Schneier or that there were any security concerns with SHA-3 to begin with. What do you mean exactly? (Note that the content your comment refers to has moved to this answer: crypto.stackexchange.com/a/72527/2512 ) – Luc Aug 13 at 10:59
• @Luc The other question linked in that question listed the "proposed" changes regarding capacity and preimage resistance, as well as the rationale by both NIST and the Keccak team. My answer states that those changes never made it into SHA-3 – Richie Frame Aug 14 at 23:37

First, you're taking the question backwards. Inertia is the default position. You shouldn't be looking for reasons not to switch, but for reasons to switch. If there are no strong reasons to switch, nobody will switch.

Security is not a reason. Between SHA-2 and SHA3, there is no reason to believe that one is more secure than the other. It isn't like when switching from MD5 to SHA-1 or from SHA-1 to SHA-2, where in each case the older function has structural weaknesses that did lead to attacks, and had a smaller output size that was starting to raise concerns over brute-force attacks. SHA-2 and SHA3 have the same sizes and no known structual weaknesses.

Performance is not a reason for most applications. Indeed, performance is a reason not to switch. SHA3 is slower than SHA-2 on a general-purpose processor. It was one of the slowest finalists of the SHA3 competition across various processors. This is not a big reason not to switch, because hashes are pretty much never a bottleneck, but it's certainly not an incentive to switch.

SHA3 does have a performance benefit, which is that it's cheap to implement in specialized hardware. It's very fast on a dedicated circuit. This is especially important for low-power devices: SHA3 costs fewer Joules per byte than SHA-2 (or any other SHA3 finalist) when implemented in hardware. So IoT may drive the adoption of SHA3. However, hardware design takes time, and hardware designers are conservative (you can't fix bugs in hardware), so this won't happen any time soon. Furthermore, a hardware implementation of SHA3 uses more gates than SHA-2, so it costs more to build a SHA3 device, even if it costs less to use afterwards, therefore there isn't a clear cost incentive to switch even when a low power requirement is critical.

SHA3 also has a versatility benefit. The same core primitive (the Keccak sponge) can not only be used as a hash, but also as a MAC (KMAC, at lower cost than HMAC), as a key derivation function (SHAKE with a partially-secret input, at a lower cost than constructions such as HKDF), etc. It's not that SHA3 can do things that nobody else can do, but it can do things that otherwise require more complex combinations of primitives. That's an incentive to switch, but only in circumstances where 1. you need those extra things and 2. it is a problem to combine more code. Again, IoT (which wants low-cost, low-power devices) may look towards SHA3, but only once protocols that use more than SHA3's hash functionality are defined and start being deployed.

Coming back to security, a benefit of SHA3 is that it's very different from SHA-2. Indeed this is one of the main reasons it was chosen over other SHA3 proposals: since its construction is different, if a new class of attacks breaks SHA-2, it's unlikely to apply to SHA3, and vice versa. The security benefit would not be in switching to SHA3, but in deploying software and protocols that support both, so that if a weakness is found in SHA-2, the world can quickly and cheaply transition to SHA3.

• a bit of trivia from the hardware side, I've yet to see anyone ask for SHA-3. There's a lot of BLAKE2 implementations, which I will guess is driven by silicon space alone. – b degnan Aug 10 at 21:43
• You consistently write SHA3 without a dash where for SHA-1 and SHA-2 you always use the dash. The NIST spec also writes it as SHA-3 but knowing you as rather precise I assume it is not a typo when you write SHA3. Care to enlighten me if there is a difference between SHA3 and SHA-3? – rene Aug 11 at 12:43
• @FutureSecurity The question was about SHA-2 vs SHA-3 (which is Keccak). I would not recommend using other SHA-3 finalists because they're unlikely to get as much cryptanalysis going forward. – Gilles Aug 11 at 20:58
• @rene The official name of the family is indeed SHA-3, but the name of individual algorithms is SHA3-224 etc. I should have used SHA-3, but given that to use one of the SHA-3 algorithms, you need to write “SHA3-…”, I expect the SHA3 spelling to be even more prevalent than SHA1 and SHA2. – Gilles Aug 11 at 21:01
• @Luc Die area. The smallest implementations of BLAKE2 are smaller than Keccak. See e.g. pdfs.semanticscholar.org/649a/… table 4. – Gilles Aug 13 at 11:20

Since this question is asking about opinions, it's hard to give the correct answer (alternatively, all possible answers are correct, because they're an opinion). However, my opinion:

I believe that there are several aspects contributing to it:

Most application designers (that is, the people who use crypto to actually solve a problem) generally don't actually specify something as low level as the hash function; instead, they specify things at a higher level (e.g. perhaps the public key signature method, or the encryption protocol), and use whatever hashes those higher level primitives use.

As for designers of the lower level primitives, they tend to update things at a much slower rate - and hence, they tend to take a while to pick up things like SHA-3 (especially if there isn't a strong perceived need to make the change).

As for perceived need, SHA-2 is mostly seen as perfectly adequate technology, which solves the problems we give it, and at generally reasonable cost, and so there is no immediate need to spend effort at moving to SHA-3. The big advantage to SHA-3 is that it can be implemented faster, and for most applications, the speed of the hash function just isn't a major factor.

Now, we will likely see more adoption over the next several years; if you go through the NIST postquantum candidates, they tend to use SHAKE (at least, the ones that need a hash function). That means that, as they are taken up by the applications (and used in real products), they'll take SHA-3 with them (and one would expect that, given that there is now a SHA-3 implementation in the crypto library already) it'll find more general use.

However, as for becoming 'the standard' (that is, people are actually discouraged from using SHA-2) I don't see that for a very long time (barring a cryptanalytic discovery); as far as we can tell, it doesn't have any weakness, and it does have some practical advantages over SHA-3 (at least, I believe it's easier to build with a minimal number of gates)

• “SHA-3 is that it can be implemented faster” actually mostly no, SHA3 is consistently faster in hardware and slower in software; it's only faster in a dedicated circuit. – Gilles Aug 10 at 14:43
• As someone who helps maintain a few security libraries, I would prefer SHA-256 over SHA-3 because there's more hardware support. SHA-256 acceleration is available on x86 (Goldmont), Aarch64 (Armv8.1-a), PowerPC (Power8) and MIPS. SHA-3 acceleration is only available on Aarch64 (Armv8.4-a), as far as I know. SHA-3 is also slower in software when hardware acceleration is not available. – jww Aug 13 at 5:15
• ... there's SHA-3 acceleration for aarch64? I had never heard from commodity hardware with an actual Keccak implementation! – Ruben De Smet Aug 14 at 12:12
• @jww Does it implement SHA-3 itself, or the Keccak-f permutation function? – forest Aug 21 at 21:34
• @forest - I believe it is SHA3 specifically since there is also SM3, SM4 and SHA512. But I could be wrong. I don't know anyone who has implemented it (yet). I just checked OpenSSL and Andy has not cut it in yet. I noticed Jack Lloyd merged a PR for ARMv8.4's SM4 at Add support for ARMv8 SM4 instructions. As soon as I can buy hardware with the ISA, then I will cut it in. I require testing on real hardware for an algorithm. – jww Aug 22 at 3:20

Mostly option 2. People are lazy. Furthermore, those using SHA-2 have no good reason to switch. Those using SHA-1 are lazy, and have been lazy for a while. Another reason is compatibility, If I use SHA3 in my certificate some software may not be able to use it, and if SHA2 is fast and as secure for all practical purposes why should I opt for the less compatible option. I don't think there are serious security concerns against SHA-3, some think it wasn't the best candidate (I'm one of them) but that doesn't mean they think objectively it is an insecure hash function. Most developers do not make an informed choice about cryptographic primitives and inertia is a good choice. I think very few are holding out for more cryptanalysis.

• My favorite SHA-1 fact is that the algorithm was broken in 2005. The collision published a couple years ago required about as much computing time as estimated for the 2005 break. Like with Heartbleed and Dual EC DRBG, it was a "don't say we didn't warn you" event. Except developers point at those three things and say "Why should we listen to you now when you didn't warn us about these things?" – Future Security Aug 11 at 15:41
• SHA-2 isn't so bad though. If it weren't for length extension problems the best argument for upgrading to something else would just be for performance reasons. SHA-2's use today isn't as facepalm worthy as SHA-1 or RC4 use was years ago. – Future Security Aug 11 at 15:49

Below are some suggestions I originally included in the question but fit better as an answer.

1) There are still some concerns about SHA-3. I recall that there was some controversy when released, e.g. Bruce Schneier and others raised some concerns. Are these concerns still lingering? [Based on other answers I now see that this is not really valid. I also researched Bruce Schneider’s criticism further and it was in relation to NIST-enforced changes to the capacity optionality to balance security and performance, not security as such]

2) People are just slow/lazy when it comes to adopting new standards and four years is not a long time. Adoption will continue to grow slowly until SHA-2 is shown as weak or withdrawn by NIST.

3) Many developers wait to see a crypto function in use for 5 years and widely research and attacked before they adopt (ie more conscious/systematic approach than 2). SHA-3 usage will likely increase significantly in the next couple of years and become the standard.

• The controversy was entirely political because the finalization was concurrent with an onslaught of Snowden disclosures including Project BULLRUN. There is no technical foundation to the controversy—the result is that the fixed-size hash functions SHA3-256 etc. are slower than they need to be for their security goals, because of bad timing with Snowden. Fortunately SHAKE128 avoids this mistake. – Squeamish Ossifrage Aug 21 at 16:47
• @SqueamishOssifrage The controversy was about the capacity (and thus collision resistance), right? Not about the number of rounds or anything else that would improve resistance against cryptanalysis? – forest Aug 21 at 21:31
• @forest It was about the capacity, yes. Specifically, the capacity for (e.g.) SHA3-256, namely $c = 512$, is twice what it needs to be, namely $c = 256$, for a 128-bit security level, at substantial cost to performance and no meaningful improvement in security. – Squeamish Ossifrage Aug 21 at 22:06


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# Ideal solution

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 Title: Ideal solution Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

### Ideal solution

In chemistry, an ideal solution or ideal mixture is a solution with thermodynamic properties analogous to those of a mixture of ideal gases. The enthalpy of solution (or "enthalpy of mixing") is zero[1] as is the volume change on mixing; the closer to zero the enthalpy of solution is, the more "ideal" the behavior of the solution becomes. The vapor pressure of the solution obeys Raoult's law, and the activity coefficient of each component (which measures deviation from ideality) is equal to one.[2]

The concept of an ideal solution is fundamental to chemical thermodynamics and its applications, such as the use of colligative properties.

## Contents

• Physical origin 1
• Formal definition 2
• Thermodynamic properties 3
• Volume 3.1
• Enthalpy and heat capacity 3.2
• Entropy of mixing 3.3
• Consequences 4
• Non-ideality 5
• References 7

## Physical origin

Ideality of solutions is analogous to ideality for gases, with the important difference that intermolecular interactions in liquids are strong and cannot simply be neglected as they can for ideal gases. Instead we assume that the mean strength of the interactions are the same between all the molecules of the solution.

More formally, for a mix of molecules of A and B, the interactions between unlike neighbors (UAB) and like neighbors UAA and UBB must be of the same average strength, i.e., 2 UAB = UAA + UBB and the longer-range interactions must be nil (or at least indistinguishable). If the molecular forces are the same between AA, AB and BB, i.e., UAB = UAA = UBB, then the solution is automatically ideal.

If the molecules are almost identical chemically, e.g., 1-butanol and 2-butanol, then the solution will be almost ideal. Since the interaction energies between A and B are almost equal, it follows that there is a very small overall energy (enthalpy) change when the substances are mixed. The more dissimilar the nature of A and B, the more strongly the solution is expected to deviate from ideality.

## Formal definition

Different related definitions of an ideal solution have been proposed. The simplest definition is that an ideal solution is a solution for which each component (i) obeys Raoult's law p_i=x_ip_i^* for all compositions. Here p_i is the vapor pressure of component i above the solution, x_i is its mole fraction and p_i^* is the vapor pressure of the pure substance i at the same temperature.[3][4][5]

This definition depends on vapor pressures which are a directly measurable property, at least for volatile components. The thermodynamic properties may then be obtained from the chemical potential μ (or partial molar Gibbs energy g) of each component, which is assumed to be given by the ideal gas formula

\mu(T,p_i) = g(T,p_i)=g^\mathrm{u}(T,p^u)+RT\ln {\frac{p_i}{p^u}}.

The reference pressure p^u may be taken as P^0 = 1 bar, or as the pressure of the mix to ease operations.

On substituting the value of p_i from Raoult's law,

\mu(T,p_i) =g^\mathrm{u}(T,p^u)+RT\ln {\frac{p_i^*}{p^u}} + RT\ln x_i =\mu _i^*+ RT\ln x_i.

This equation for the chemical potential can be used as an alternate definition for an ideal solution.

However, the vapor above the solution may not actually behave as a mixture of ideal gases. Some authors therefore define an ideal solution as one for which each component obeys the fugacity analogue of Raoult's law f_i=x_if_i^*,

Here f_i is the fugacity of component i in solution and f_i^* is the fugacity of i as a pure substance.[6][7] Since the fugacity is defined by the equation

\mu(T,P) = g(T,P)=g^\mathrm{u}(T,p^u)+RT\ln {\frac{f_i}{p^u}}

this definition leads to ideal values of the chemical potential and other thermodynamic properties even when the component vapors above the solution are not ideal gases. An equivalent statement uses thermodynamic activity instead of fugacity.[8]

## Thermodynamic properties

### Volume

If we differentiate this last equation with respect to P at T constant we get:

\left(\frac{\partial g(T,P)}{\partial P}\right)_{T}=RT\left(\frac{\partial \ln f}{\partial P}\right)_{T}

but we know from the Gibbs potential equation that:

\left(\frac{\partial g(T,P)}{\partial P}\right)_{T}=v

These last two equations put together give:

\left(\frac{\partial \ln f}{\partial P}\right)_{T}=\frac{v}{RT}

Since all this, done as a pure substance is valid in a mix just adding the subscript i to all the intensive variables and changing v to \bar{v_i}, standing for Partial molar volume.

\left(\frac{\partial \ln f_i}{\partial P}\right)_{T,x_i}=\frac{\bar{v_i}}{RT}

Applying the first equation of this section to this last equation we get

v_i^*=\bar{v_i}

which means that in an ideal mix the volume is the addition of the volumes of its components.

### Enthalpy and heat capacity

Proceeding in a similar way but derivative with respect of T we get to a similar result with enthalpies

\frac{g(T,P)-g^\mathrm{gas}(T,p^u)}{RT}=\ln\frac{f}{p^u}

derivative with respect to T and remembering that \left( \frac{\partial \frac{g}{T}}{\partial T}\right)_P=-\frac{h}{T^2} we get:

-\frac{\bar{h_i}-h_i^\mathrm{gas}}{R}=-\frac{h_i^*-h_i^\mathrm{gas}}{R}

which in turn is \bar{h_i}=h_i^*.

Meaning that the enthalpy of the mix is equal to the sum of its components.

Since \bar{u_i}=\bar{h_i}-p\bar{v_i} and u_i^*=h_i^*-pv_i^*:

u_i^*=\bar{u_i}

It is also easily verifiable that

C_{pi}^*=\bar{C_{pi}}

### Entropy of mixing

Finally since

\bar{g_i}=\mu _i=g_i^\mathrm{gas}+RT\ln \frac{f_i}{p^u}=g_i^\mathrm{gas}+RT\ln \frac{f_i^*}{p^u}+RT\ln x_i=\mu _i^*+ RT\ln x_i

Which means that

\Delta g_{i,\mathrm{mix}}=RT\ln x_i

and since

G=\sum_i x_i{g_i}

then

\Delta G_\mathrm{mix}=RT\sum_i{x_i\ln x_i}

At last we can calculate the entropy of mixing since g_i^*=h_i^*-Ts_i^* and \bar{g_i}=\bar{h_i}-T\bar{s_i}

\Delta s_{i,\mathrm{mix}}=-R\sum _i \ln x_i
\Delta S_\mathrm{mix}=-R\sum _i x_i\ln x_i

## Consequences

Solvent-Solute interactions are similar to solute-solute and solvent-solvent interactions

Since the enthalpy of mixing (solution) is zero, the change in Gibbs free energy on mixing is determined solely by the entropy of mixing. Hence the molar Gibbs free energy of mixing is

\Delta G_{\mathrm{m,mix}} = RT \sum_i x_i \ln x_i

or for a two component solution

\Delta G_{\mathrm{m,mix}} = RT (x_A \ln x_A + x_B \ln x_B)

where m denotes molar, i.e., change in Gibbs free energy per mole of solution, and x_i is the mole fraction of component i.

Note that this free energy of mixing is always negative (since each x_i is positive and each \ln x_i must be negative), i.e., ideal solutions are always completely miscible.

The equation above can be expressed in terms of chemical potentials of the individual components

\Delta G_{\mathrm{m,mix}} = \sum_i x_i \Delta\mu_{i,\mathrm{mix}}

where \Delta\mu_{i,\mathrm{mix}}=RT\ln x_i is the change in chemical potential of i on mixing.

If the chemical potential of pure liquid i is denoted \mu_i^*, then the chemical potential of i in an ideal solution is

\mu_i = \mu_i^* + \Delta \mu_{i,\mathrm{mix}} = \mu_i^* + RT \ln x_i

Any component i of an ideal solution obeys Raoult's Law over the entire composition range:

\ P_{i}=(P_{i})_{pure} x_i

where

(P_i)_{pure}\, is the equilibrium vapor pressure of the pure component
x_i\, is the mole fraction of the component in solution

It can also be shown that volumes are strictly additive for ideal solutions.

## Non-ideality

Deviations from ideality can be described by the use of Margules functions or activity coefficients. A single Margules parameter may be sufficient to describe the properties of the solution if the deviations from ideality are modest; such solutions are termed regular.

In contrast to ideal solutions, where volumes are strictly additive and mixing is always complete, the volume of a non-ideal solution is not, in general, the simple sum of the volumes of the component pure liquids and solubility is not guaranteed over the whole composition range. By measurement of densities thermodynamic activity of components can be determined.

## References

1. ^ A to Z of Thermodynamics Pierre Perrot ISBN 0-19-856556-9
2. ^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version:  (2006–) "ideal mixture".
3. ^ P. Atkins and J. de Paula, Atkins’ Physical Chemistry (8th edn, W.H.Freeman 2006), p.144
4. ^ T. Engel and P. Reid Physical Chemistry (Pearson 2006), p.194
5. ^ K.J. Laidler and J.H. Meiser Physical Chemistry (Benjamin-Cummings 1982), p.180
6. ^ R.S. Berry, S.A. Rice and J. Ross, Physical Chemistry (Wiley 1980) p.750
7. ^ I.M. Klotz, Chemical Thermodynamics (Benjamin 1964) p.322
8. ^ P.A. Rock, Chemical Thermodynamics: Principles and Applications (Macmillan 1969), p.261
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1. # Plot memory usage as a function of time

Fri 07 November 2014

One of the lesser known features of the memory_profiler package is its ability to plot memory consumption as a function of time. This was implemented by my friend Philippe Gervais, previously a colleague at INRIA and now working for Google.

With this feature it is possible to generate very easily ...

2. # Surrogate Loss Functions in Machine Learning

Fri 20 June 2014

TL; DR These are some notes on calibration of surrogate loss functions in the context of machine learning. But mostly it is an excuse to post some images I made.

9. # Memory plots with memory_profiler

Fri 04 January 2013

Besides performing a line-by-line analysis of memory consumption, memory_profiler exposes some functions that allow to retrieve the memory consumption of a function in real-time, allowing e.g. to visualize the memory consumption of a given function over time.

The function to be used is memory_usage. The first argument specifies what ...


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## Precalculus (6th Edition) Blitzer

The solution to the system of equations is $x=2,y=1,z=-1$
Let us consider the following system of equations: $x+2y-z=5$ $2x-y+3z=0$ $2y+z=1$ Multiply equation $x+2y-z=5$ by $2$ and equation $2x-y+3z=0$ by $1$ and now, subtract: \begin{align} & 2\left( x+2y-z \right)-1\left( 2x-y+3z \right)=2\left( 5 \right)-1\left( 0 \right) \\ & 2x+4y-2z-2x+y-3z=10-0 \\ & 5y-5z=10 \\ & y-z=2 \end{align} Add equations $2y+z=1$ and $y-z=2$ and simplify: \begin{align} & 2y+z+y-z=1+2 \\ & 3y=3 \\ & y=1 \end{align} Put the value of y in equation $y-z=2$ and get $z=-1$ Then, substitute the values of y and z in equation $x+2y-z=5$ and simplify: \begin{align} & x+2\left( 1 \right)-\left( -1 \right)=5 \\ & x+2+1=5 \\ & x=2 \end{align} Therefore, consider $x+2y-z=5$ and substitute $x=2,y=1,z=-1$ as given below: \begin{align} & \left( 2 \right)+2\left( 1 \right)-\left( -1 \right)=5 \\ & 2+2+1=5 \\ & 5=5,\text{ true}\text{.} \end{align} Therefore, consider $2x-y+3z=0$ and substitute $x=2,y=1,z=-1$ as given below: \begin{align} & 2\left( 2 \right)-\left( 1 \right)+3\left( -1 \right)=0 \\ & 4-1-3=0 \\ & 0=0,\text{ true}\text{.} \end{align} So, consider $2y+z=1$ and put $x=2,y=1,z=-1$ as given below: \begin{align} & 2\left( 1 \right)+\left( -1 \right)=1 \\ & 2-1=1 \\ & 1=1,\text{ true}\text{.} \end{align} Thus, the solution of the system of equations is $x=2,\,\,y=1,\,\,z=-1$


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# Homework Help: I need help very confused and frustrated

1. Apr 25, 2005

### Jayhawk1

I asked this question earlier and I didn't get any significant help, atleast beyond what I already knew. I do not know how to calculate this, nor do my classmates who I have seen also posed this question to the forum. Please help!! I know the concept, but I can not calculate it. I have worked on it for a few days now, and it is due in about an hour.

2. Apr 25, 2005

### HallsofIvy

You are given the coefficient of linear expansion of iron so you can calculate how much the linear dimensions of the ring have increased and so the volume at each temperature. From that you can calculate the inside diameter of the iron ring (which is now smaller!). You are given the coefficient of linear expansion of bronze so you can calcuate how much the linear dimension (in particular, the diameter) of the brass plug increases with each degree and so what the diameter is at any temperature. Now find the temperature at which those two diameters are the same.

3. Apr 25, 2005

### OlderDan

Diameter of brass plug as a function of the temperature change is

$$D_b = D_{b0}(1+10*10^{-6}/C^o*{\Delta}T)$$

Diameter of iron ring as a function of the temperature change is

$$D_i = D_{i0}(1+12*10^{-6}/C^o*{\Delta}T)$$

Set the two diameters equal and solve for the temperature change. Add the change to the starting temperature.

4. Apr 25, 2005

### OlderDan

The diameter of the inside of the ring actually gets larger. If it were a disk, all the linear dimensions would increase. The diameter of the "missing" part of the disk that makes it a ring will expand in proportion to the outer diameter.

5. Apr 25, 2005

### Jayhawk1

Thank you but...

Hahaha... I'm sorry, but now I'm just really confused. I tried making the two equations equal, but I got a very very small number. Can someone please clarify for me? I appreciate all your help so far.

6. Apr 25, 2005

### OlderDan

$$D_b = 8.755cm(1+10*10^{-6}/C^o*{\Delta}T)$$

$$D_i = 8.745cm(1+12*10^{-6}/C^o*{\Delta}T)$$

$$8.755cm(1+10*10^{-6}/C^o*{\Delta}T) = 8.745cm(1+12*10^{-6}/C^o*{\Delta}T)$$

$$8.755cm-8.745cm = 8.745cm(12*10^{-6}/C^o*{\Delta}T) - 8.755cm(10*10^{-6}/C^o*{\Delta}T)$$

$$(8.755cm-8.745cm)*10^6C^o = (8.745cm*12 - 8.755cm*10){\Delta}T$$

Can you finish this?


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# How do you find the product (k-m)(k+m)(k-m)?

May 2, 2017

See the entire solution process below:

#### Explanation:

First, we can multiply the two terms on the left of the expression using this rule and substituting $k$ for $a$ and $m$ for $b$:

$\left(a + b\right) \left(a - b\right) = {a}^{a} - {b}^{2}$

(color(red)((k - m))color(blue)((k + m))(k - m) => (k^2 - m^2)(k - m)

Next, we can multiply these two remaining terms by multiplying each term in the parenthesis on the left by each term in the parenthesis on the right:

$\textcolor{red}{\left({k}^{2} - {m}^{2}\right)} \textcolor{b l u e}{\left(k - m\right)} \implies$

$\left(\textcolor{red}{{k}^{2}} \cdot \textcolor{b l u e}{k}\right) - \left(\textcolor{red}{{k}^{2}} \cdot \textcolor{b l u e}{m}\right) - \left(\textcolor{red}{{m}^{2}} \cdot \textcolor{b l u e}{k}\right) + \left(\textcolor{red}{{m}^{2}} \cdot \textcolor{b l u e}{m}\right) \implies$

${k}^{3} - {k}^{2} m - k {m}^{2} + {m}^{3}$