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NCERT Solutions For Class 11 Chemistry Chapter 7

NCERT Solutions Class 11 Chemistry Equilibrium

NCERT solutions for class 11 Chemistry chapter 7 equilibrium are available in PDF format. Students can download these free NCERT solutions by visiting our website at BYJU’S. NCERT solutions for class 11 Chemistry chapter 7 includes solutions for all the questions provided in class 11 NCERT chemistry textbook. These solutions are solved by the subject experts and are framed according to the updated CBSE class 11 Chemistry syllabus. Apart from the regular textbook based questions, NCERT solutions for class 11 Chemistry chapter 7 pdf files include all the important questions from exam point of view.

Q.1. At a fixed temperature a liquid is in equilibrium with its vapour in a closed vessel. Suddenly, the volume of the vessel got increased.

I)What will be the final vapour pressure and  what will happen when equilibrium is restored finally?

II)Write down, how initially the rates of evaporation and condensation got changed?

III)Write down the effect observed when there was a change in vapour pressure.

Ans.

(I)Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

(II)On increasing the volume of the container, the rates of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.

(III)On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a large space.

Q.2.Find out Kc for the given reaction in equilibrium state

: [SO2]= 0.6 M, [O2] = 0.82 M and [SO3] = 1.9 M ?

2SO2(g)+O2(g)2SO3(g)$2SO_{2}(g)+O_{2}(g)\leftrightarrow 2SO_{3}(g)$

Ans.

As per the question,

2SO2(g)+O2(g)2SO3(g)$2SO_{2}(g)+O_{2}(g)\leftrightarrow 2SO_{3}(g)$ (Given)

Kc=[SO3]2[SO2]2[O2]=(1.9)2M2(0.6)2(0.82)M3=12.229M1$K_{c}=\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}\\ \\ =\frac{(1.9)^{2}M^{2}}{(0.6)^{2}(0.82)M^{3}}\\ \\ =12.229M^{-1}$(approximately)

Hence, K for the equilibrium is 12.229 M–1.

Q.3. At a definite temperature and a total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms

I2(g)2I(g)$I_{2} (g) \leftrightarrow 2I (g)$

Find Kp for the equilibrium.

Ans.

Partial pressure of Iodine atoms (I)

pI=40100×ptotal=40100×105=4×104Pa$p_{I}=\frac{40}{100}\times p_{total}\\ \\ =\frac{40}{100}\times 10^{5}\\ \\ =4\times 10^{4}Pa$

Partial pressure of I2 molecules,

pI=60100×ptotal=60100×105=6×104Pa$p_{I}=\frac{60}{100}\times p_{total}\\ \\ =\frac{60}{100}\times 10^{5}\\ \\ =6\times 10^{4}Pa$

Now, for the given reaction,

Kp=(pI)2pI2=(4×104)2Pa26×104Pa=2.67×104Pa$K_{p}=\frac{(p_{I})^{2}}{p_{I_{2}}} =\frac{(4\times 10^{4})^{2}Pa^{2}}{6\times 10^{4}Pa}\\ \\ =2.67\times 10^{4}Pa$

Q.4. For the given reaction , find expression for the equilibrium constant

(i)2NOCl(g)2NO(g)+Cl2(g)$2NOCl(g)\leftrightarrow 2NO(g)+Cl_{2}(g)$

(ii)2Cu(NO3)2(s)2CuO(s)+4NO2(g)+O2(g)$2Cu(NO_{3})_{2}(s)\leftrightarrow 2CuO(s)+4NO_{2}(g)+O_{2}(g)$

(iii)CH3COOC2H5(aq)+H2O(1)CH3COOH(aq)+C2H5OH(aq)$CH_{3}COOC_{2}H_{5}(aq)+H_{2}O(1)\leftrightarrow CH_{3}COOH(aq)+C_{2}H_{5}OH(aq)$

(iv)Fe3+(aq)+3OH(aq)Fe(OH)3(s)$Fe^{3+}(aq)+3OH^{-}(aq)\leftrightarrow Fe(OH)_{3}(s)$

(v)I2(s)+5F22IF5$I_{2}(s)+5F_{2}\leftrightarrow 2IF_{5}$

Ans.

KC=[NOg]2[Cl2(g)][NOCl(g)]2$K_{C}=\frac{[NO_{g}]^{2}[Cl_{2_{(g)}}]}{[NOCl_{(g)}]^{2}}$

(ii)KC=[CuO(s)]2[NO2(g)]4[O2(g)][Cu(NO3)2(g)]2=[NO2(g)]4[O2(g)]$K_{C}=\frac{[CuO_{(s)}]^{2}[NO_{2_{(g)}}]^{4}[O_{2_{(g)}}]}{[Cu(NO_{3})_{2(g)}]^{2}}\\ \\ =[NO_{2(g)}]^{4}[O_{2(g)}]$

(iii)KC=CH3COOH(aq)[C2H5OH(aq)][CH3COOC2H5(aq)][H2O(l)]=CH3COOH(aq)[C2H5OH(aq)][CH3COOC2H5(aq)]$K_{C}=\frac{CH_{3}COOH_{(aq)}[C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}][H_{2}O_{(l)}]}\\ \\ =\frac{CH_{3}COOH_{(aq)}[C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}]}$

(iv)KC=Fe(OH)3(s)[Fe3+(aq)][OH(aq)]3=1[Fe3+(aq)][OH(aq)]3$K_{C}=\frac{Fe(OH)_{3(s)}}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}\\ \\ =\frac{1}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}$

(v)KC=[IF5]2[I2(s)][F2]5=[IF5]2[F2]5$K_{C}=\frac{[IF_{5}]^{2}}{[I_{2(s)}][F_{2}]^{5}}\\ \\ =\frac{[IF_{5}]^{2}}{[F_{2}]^{5}}$

Q.5.Find the value of , Kc for each of the following equilibria from the given value of Kp:

(i)2NOCl(g)2NO(g)+Cl2(g);Kp=1.8×102at500K$2NOCl(g)\leftrightarrow 2NO(g)+Cl_{2}(g);\: \: \: K_{p}=1.8\times 10^{-2}\: \: at\: 500K$

(ii)CaCO3(s)CaO(s)+CO2(g);Kp=167at1073K$CaCO_{3}(s)\leftrightarrow CaO(s)+CO_{2}(g);\: \: \: K_{p}=167\: \: at\: 1073K$

Ans.

The relation between Kp and Kc is given as:

Kp=Kc(RT)Δn$K_{p}=K_{c}(RT)^{\Delta n}$

(a) Given,

R = 0.0831 barLmol–1K–1

Δn=32=1$\Delta n = 3-2 =1$

T = 500 K

Kp=1.8×102$1.8\times 10^{-2}$

Now,

Kp = Kc (RT) ∆n

1.8×102=Kc(0.0831×500)1Kc=1.8×1020.0831×500=4.33×104(approximately)$\Rightarrow 1.8\times 10^{-2}=K_{c}(0.0831\times 500)^{1}\\ \\ \Rightarrow K_{c}=\frac{1.8\times 10^{-2}}{0.0831\times 500}\\ \\ =4.33\times 10^{-4}(approximately)$

(b) Here,

∆n =2 – 1 = 1

R = 0.0831 barLmol–1K–1

T = 1073 K

Kp= 167

Now,

Kp = Kc (RT) ∆n

167=Kc(0.0831×1073)ΔnKc=1670.0831×1073=1.87(approximately)$\Rightarrow 167=K_{c}{(0.0831\times 1073)^{\Delta n}}\\ \\ \Rightarrow K_{c}=\frac{167}{0.0831\times 1073}\\ \\ =1.87(approximately)$

Q.6. For the following equilibrium, Kc=6.3×1014at1000KNO(g)+O3(g)NO2(g)+O2(g)$K_{c}=6.3\times 10^{14}\: \: \: at\: 1000K\\ \\ NO(g)+O_{3(g)}\leftrightarrow NO_{2(g)}+O_{2(g)}$

Both the reverse and forward reactions in the equilibrium are elementary bimolecular reactions. Calculate Kc, for the reverse reaction?

Ans.

For the reverse reaction, Kc=1Kc=16.3×1014=1.59×1015$K_{c}=\frac{1}{K_{c }}\\ \\ =\frac{1}{6.3\times 10^{14}}\\ \\ =1.59\times 10^{-15}$

Q.7. Explain why solids and pure liquids can be ignored while writing the equilibrium constant expression?

Ans.

This is because molar concentration of a pure solid or liquid is independent of the amount present.

Mole concentration= NumberofmolesVolumeMass/molecularmassVolume=MassVolume×Molecularmass=DensityMolecularmass$\frac{Number\: of\: moles}{Volume}\\ \\ \frac{Mass/molecular\:mass}{Volume}\\ \\ =\frac{Mass}{Volume\,\times\, Molecular \:mass}\\ \\ =\frac{Density}{Molecular\: mass}$

Though density of solid and pure liquid is fixed and molar mass is also fixed .

$∴$Molar concentration are constatnt.

Q.8. When  oxygen and nitrogen react with each other, then the following reaction takes place:

2N2(g)+O22N2O(g)$2N_{2}(g)+O_{2}\leftrightarrow 2N_{2}O(g)$

If a solution of 0.933 mol of oxygen and 0.482 mol of nitrogen is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc =2.0×1037$2.0\times 10^{-37}$, determine the composition of equilibrium solution.

Ans.

Let the concentration of N2O at equilibrium be x.

The given reaction is:

2N2(g)             +               O2(g)                               2N2O(g)

Initial conc.     0.482 mol                     0.933 mol                                  0

At equilibrium(0.482-x)mol            (1.933-x)mol                              x mol

[N2]=0.482x10[O2]=0.933x210,[N2O]=x10$[N_{2}]=\frac{0.482-x}{10}\cdot [O_{2}]=\frac{0.933-\frac{x}{2}}{10},[N_{2}O]=\frac{x}{10}$

The value of equilibrium constant  is extremely small. This means that only small amounts . Then,

[N2]=0.48210=0.0482molL1and[O2]=0.93310=0.0933molL1$[N_{2}]=\frac{0.482}{10}=0.0482molL^{-1} \: \:and\:\: [O_{2}]=\frac{0.933}{10}=0.0933molL^{-1}$

Now,

Kc=[N2O(g)]2[N2(g)][O2(g)]2.0×1037=(x10)2(0.0482)2(0.0933)x2100=2.0×1037×(0.0482)2×(0.0933)x2=43.35×1040x=6.6×1020$K_{c}=\frac{[N_{2}O_{(g)}]^{2}}{[N_{2(g)}][O_{2(g)}]}\\ \\ \Rightarrow 2.0\times 10^{-37}=\frac{(\frac{x}{10})^{2}}{(0.0482)^{2}(0.0933)}\\ \\ \Rightarrow \frac{x^{2}}{100}=2.0 \times 10^{-37}\times (0.0482)^{2}\times (0.0933)\\ \\ \Rightarrow x^{2}=43.35\times 10^{-40}\\ \\ \Rightarrow x=6.6\times 10^{-20}$ [N2O]=x10=6.6×102010=6.6×1021$[N_{2}O]=\frac{x}{10}=\frac{6.6\times 10^{-20}}{10}\\ \\ =6.6\times 10^{-21}$

Q.9. Nitric oxide reacts with bromine and gives nitrosyl bromide as per reaction is given below:

2NO(g)+Br2(g) ⇒ 2NOBr(g)

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.

Ans.

The given reaction is:

2NO(g)+Br2(g)                              2NOBr(g)

2mol       1mol                                2mol

Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.

Again, 2 mol of NOBr are formed from 1 mol of Br.

Therefore, 0.0518 mol of NOBr are formed from 0.05182$\frac{0.0518}{2}$ mol or Br, or 0.0259 mol of NO.

The amount of NO and Br present initially is as follows:

[NO] = 0.087 mol [Br2] = 0.0437 mol

Therefore, the amount of NO present at equilibrium is:

[NO] = 0.087 – 0.0518 = 0.0352 mol

And, the amount of Br present at equilibrium is:

[Br2] = 0.0437 – 0.0259 = 0.0178 mol

Q.10. At 450 K, Kp= 2.0×1010$2.0 \times 10^{10}$/bar for the given reaction at equilibrium.

2SO2(g)+O2(g) ⇒ 2SO3(g)

What is Kc at this temperature?

Ans.)

For the given reaction,

∆n = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K–1 mol–1

Kp=2.0×1010bar1$2.0 \times 10^{10}bar^{-1}$

We know that,

Kp=Kc(RT)Δn2.0×1010bar1=Kc(0.0831LbarK1mol1×450K)1Kc=2.0×1010bar1(0.0831LbarK1mol1×450K)1=(2.0×1010bar1)(0.0831LbarK1mol1×450K)=74.79×1010Lmol1=7.48×1011Lmol1=7.48×1011M1$K_{p}=K_{c}(RT)\Delta n\\ \\ \Rightarrow 2.0\times 10^{10}bar^{-1}=K_{c}(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)^{-1}\\ \\ \Rightarrow K_{c}=\frac{2.0\times 10^{10}bar^{-1}}{(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)^{-1}}\\ \\ =(2.0\times 10^{10}bar^{-1})(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)\\ \\ =74.79\times 10^{10}L\, mol^{-1}\\ \\ =7.48\times 10^{11}L\, mol^{-1}\\ \\ =7.48\times 10^{11}\, M^{-1}$

Q.11. A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?

2HI(g) ⇒ H2(g)+I2(g)

Ans.

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm.

Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:

2HI(g)                                     H2(g)       +      I2(g)

Initial conc.               0.2 atm                                        0                     0

At equilibrium        0.4 atm                                          0.16               2.15

2                      2

=0.08atm         =0.08atm

Therefore,

Kp=pH2×pI2p2HI=0.08×0.08(0.04)2=0.00640.0016=4.0$K_{p=}\frac{p_{H_{2}}\times p_{I_{2}}}{p^{2}_{HI}}\\ \\ =\frac{0.08\times 0.08}{(0.04)^{2}}\\ \\ =\frac{0.0064}{0.0016}\\ \\ =4.0$

Hence, the value of Kp for the given equilibrium is 4.0.

Q.12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction

N2(g)+3H2(g) ⇒ 2NH3(g) is 1.7×102$1.7\times 10^{2}$

Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Ans.

The given reaction is:

N2(g)+3H2(g) 2NH3(g)

The given concentration of various species is

[N2]= 1.5720molL1$\frac{1.57}{20}mol\, L^{-1}$

[H2]= 1.9220molL1$\frac{1.92}{20}mol\, L^{-1}$

[NH3]= 8.3120molL1$\frac{8.31}{20}mol\, L^{-1}$

Now, reaction quotient Qc is:

Q=[NH3]2[N2][H2]3=((8.13)20)2(1.5720)(1.9220)3=2.4×103$Q=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\ =\frac{(\frac{(8.13)}{20})^{2}}{(\frac{1.57}{20})(\frac{1.92}{20})^{3}}\\ \\ =2.4\times 10^{3}$

Since, 𝑄𝑐 ≠ 𝐾𝑐, the reaction mixture is not at equilibrium.

Again, 𝑄𝑐 > 𝐾𝑐. Hence, the reaction will proceed in the reverse direction.

Q.13. The equilibrium constant expression for a gas reaction is,

Kc=[NH3]4[O2]5[NO]4[H2O]6$K_{c}=\frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}$

Write the balanced chemical equation corresponding to this expression.

Ans.

The balanced chemical equation corresponding to the given expression can be written as:

4NO(g)+6H2o(g)4NH3(g)+5O2(g)

Q.14. One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 60% of water (by mass) reacts with CO according to the equation,

H2O(g)  +  CO(g) ⇒ H2(g)   +  CO2(g)

Calculate the equilibrium constant for the reaction.

Ans.

The given reaction is:

H2O(g)  +  CO(g)                            H2(g)   +  CO2(g)

Initial conc.                    1     M             1     M                            0               0

10                    10

At  equilibrium   1-0.6  M        1-0.6  M                                0.6  M               0.6  M

10                  10                                       10                       10

=0.04 M           =0.04M                                =0.06 M            =0.06M

Therefore, the equilibrium constant for the reaction,

Kc=Kc=[H2][CO2][H2O][CO]=0.06×0.060.04×0.04=0.00360.0016=2.25(approximately)$K_{c}=\frac{[H_{2}][CO_{2}]}{[H_{2}O][CO]}\\ \\ =\frac{0.06\times 0.06}{0.04\times 0.04}\\ \\ =\frac{0.0036}{0.0016}\\ \\ =2.25(approximately)$

Q.15. At 700 K, equilibrium constant for the reaction

H2(g)+I2(g) ⇒ 2HI(g)

is 54.8. If 0.5 molL–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

Ans.

It is given that equilibrium constant 𝐾𝑐 for the reaction

H2(g)+I2(g)                              2HI(g)   is  54.8.

Therefore, at equilibrium, the equilibrium constant Kc for the reaction

H2(g)+I2(g)                              2HI(g)

[HI]=0.5 molL-1 will be 1/54.8.

Let the concentrations of hydrogen and iodine at equilibrium be x molL–1

[H2]=[I2]=x mol L-1

Therefore, [H2][I2][HI]2=Kcx×x(0.5)2=154.8x2=0.2554.8x=0.06754x=0.068molL1(approximately)$\frac{[H_{2}][I_{2}]}{[HI]^{2}}=K^{‘}_{c}\\ \\ \Rightarrow \frac{x\times x}{(0.5)^{2}}=\frac{1}{54.8}\\ \\ \Rightarrow x^{2}=\frac{0.25}{54.8}\\ \\ \Rightarrow x=0.06754\\ \\ x=0.068molL^{-1}(approximately)$

Hence, at equilibrium, [H2]=[I2]=0.068 mol L-1.

Q.16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2ICl(g) ⇒ I2(g)+Cl2(g) ;  Kc =0.14

Ans.

The given reaction is:

2ICl(g)                                           I2(g)      +   Cl2(g)

Initial conc.                 0.78 M                                           0             0

At equilibrium      (0.78-2x) M                                       x M           x M

Now, we can write, [I2][Cl2][IC]2=Kcx×x(0.782x)2=0.14x2(0.782x)2=0.14x0.782x=0.374x=0.2920.748x1.748x=0.292x=0.167$\frac{[I_{2}][Cl_{2}]}{[IC]^{2}}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{(0.78-2x)^{2}}=0.14\\ \\ \Rightarrow \frac{x^{2}}{(0.78-2x)^{2}}=0.14\\ \\ \Rightarrow \frac{x}{0.78-2x}=0.374\\ \\ \Rightarrow x=0.292-0.748x\\ \\ \Rightarrow 1.748x=0.292\\ \\ \Rightarrow x=0.167$

Hence, at equilibrium,

[H2]=[I2]=0.167 M

[HI]= (0.782×0.167)M=0.446M$(0.78-2\times 0.167)M\\ \\ =0.446M$

Q.17. Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C2H6(g) ⇒ C2H4(g) +H2(g)

Ans.

Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now, according to the reaction,

C2H6(g)                                         C2H4(g)    +     H2(g)

Initial conc.                 4.0 M                                           0                      0

At equilibrium          (4.0-p)                                            p                     p

We can write,

pc2H4×pH2pc2H6=KPp×p40p=0.04p2=0.160.04pp2+0.04p0.16=0$\frac{p_{c_{2}H_{4}}\times p_{H_{2}}}{p_{c_{2}H_{6}}}=K_{P}\\ \\ \Rightarrow \frac{p\times p}{40-p}=0.04\\ \\ \Rightarrow p^{2}=0.16-0.04p\\ \\ \Rightarrow p^{2}+0.04p-0.16=0$

Now,

p=0.04±(0.04)24×1×(0.16)2×1=0.04±0.802=0.762(Takingpositivevalue)=0.38$p=\frac{-0.04\pm \sqrt{(0.04)^{2}-4\times 1\times (-0.16)}}{2\times 1}\\ \\ =\frac{-0.04\pm 0.80}{2}\\ \\ =\frac{0.76}{2}\: \: \: \: \: \: \: \: \: (Taking \:positive\: value)\\ \\ =0.38$

Hence, at equilibrium,

[C2H6] – 4 – p = 4 – 0.38

= 3.62 atm

Q.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

CH3COOH(l)+c2H5OH(l) ⇒ CH3COOC2H5(l)+H2O(l)

(i)Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)

(ii)At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

(iii)Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Ans.

(i)Reaction quotient,

Qc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]$Q_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}$

(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.

The given reaction is:

CH3COOH(l)+c2H5OH(l)                          CH3COOC2H5(l)+H2O(l)

Initial conc.                                                                        0                        0

At equilibrium

=            =

Therefore, equilibrium constant for the given reaction is:

Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]=0.171V×0.171V0.829V×0.009V=3.919=3.92(approximately)$K_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}\\ \\ =\frac{\frac{0.171}{V}\times \frac{0.171}{V}}{\frac{0.829}{V}\times \frac{0.009}{V}}=3.919\\ \\ =3.92(approximately)$

(iii)Let the volume of the reaction mixture be V.

CH3COOH(l)+c2H5OH(l)CH3COOC2H5(l)+H2O(l)

Initial conc.                                                                        0                        0

At equilibrium

=            =

Therefore, the reaction quotient is,

Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]=0.214V×0.214V0.786V×0.286V=0.2037=0.204(approximately)$K_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}\\ \\ =\frac{\frac{0.214}{V}\times \frac{0.214}{V}}{\frac{0.786}{V}\times \frac{0.286}{V}}=0.2037\\ \\ =0.204(approximately)$

Since Qc<Kc , equilibrium has not been reached.

Q.19. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5×101$0.5\times 10^{-1}$mol L–1. If value of Kc is 8.3×103$8.3\times 10^{-3}$, what are the concentrations of PCl3 and Cl2 at equilibrium?

PCl5(g) ⇒ PCl3(g)   +   Cl2(g)

Ans.

Consider the conc. Of both PCl3 and Cl2 at equilibrium be x molL–1. The given reaction is:

PCl5(g)PCl3(g      +      Cl2(g)

At equilibrium 0.5×1010molL1$0.5\times 10^{-10}molL^{-1}$         x mol L-1      x mol L-1

It is given that the value of equilibrium constant , Kc is 8.3×1010molL3$8.3\times 10^{-10}molL^{-3}$

Now we can write the expression for equilibrium as:

[PCl2][Cl2][PCl3]=Kcx×x0.5×1010=8.3×103x2=4.15×104x=2.04×102=0.0204=0.02(approximately)$\frac{[PCl_{2}][Cl_{2}]}{[PCl_{3}]}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{0.5\times 10^{-10}}=8.3\times 10^{-3}\\ \\ \Rightarrow x^{2}=4.15\times 10^{-4}\\ \\ \Rightarrow x=2.04\times 10^{-2}\\ \\ =0.0204\\ \\ =0.02(approximately)$

Therefore, at equilibrium,

[PCl3]=[Cl2]=0.02mol L-1

Q.20. One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.

FeO (s) + CO (g) ⇒ Fe (s) + CO2 (g); Kp= 0.265 at 1050 K.

What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO = 1.4 atm and 𝑝𝐶𝑂2= 0.80 atm?

Ans.

For the given reaction,

FeO(g)            +       CO(g)                             Fe(s)              +           CO2(g)

Initialy,              1.4 atm                                                      0.80 atm

Qp=pCO2pCO=0.801.4=0.571$\frac{p_{CO_{2}}}{p_{CO}}\\ \\ =\frac{0.80}{1.4}\\ \\ =0.571$

Since Qp > Kp , the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of CO2 will decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO2 be p. Then, we can write,

Kp=pCO2pCO0.265=0.80p1.4+p0.371+0.265p=0.80p1.265p=0.429p=0.339atm$K_{p}=\frac{p_{CO_{2}}}{p_{CO}}\\ \\ \Rightarrow 0.265=\frac{0.80-p}{1.4+p}\\ \\ \Rightarrow 0.371+0.265p=0.80-p\\ \\ \Rightarrow 1.265p=0.429\\ \\ \Rightarrow p=0.339 atm$

Therefore, equilibrium partial of CO2,pCO=0.80-0.339=0.461 atm

And, equilibrium partial pressure of CO,pCO=1.4+0.339=1.739 atm

Q.21. A reaction is given:

N2(g)+3H2(g) ⇒ 2NH3

For the above equation, Equilibrium constant = 0.061 at 500 K

At a specific time, from the analysis we can conclude that composition of the reaction mixture is, 2.0 mol L–1 H2 , 3.0 mol L –1 N2 and 0.5 mol L–1 NH3. Find out whether the reaction is at equilibrium or not? Find in which direction the reaction proceeds to reach equilibrium.

Ans.

N2(g)        +        3H2(g)                                  2NH3

At a particular time:  3.0 mol L-1          2.0mol L-1                              0.5 mol L-1

So,

Qc=[NH3]2[N2][H2]3=(0.5)2(3.0)(2.0)3=0.0104$\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\ =\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}\\ \\ =0.0104$

It is given that Kc=0.061

$∵$   QcKc$Qc \neq K_{c}$, the reaction is not at equilibrium.

$∵$    Qc<Kc$Qc < K_{c}$, the reaction preceeds in the forward direction to reach at equilibrium.

Q.22.Bromine monochloride(BrCl) decays into bromine and chlorine and reaches the equilibrium:

2BrCl(g) ⇒ Br2(g)   +   Cl2(g)

For which Kc= 42 at 600 K.

If initially pure BrCl is present at a concentration of 5.5×105$5.5 \times 10^{-5}$molL-1 , what is its molar concentration in the mixture at equilibrium?

Ans.)

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

2BrCl(g)                                               Br2(g)      +        Cl2(g)

Initial conc. 5.5×105$5.5\times 10^{-5}$                   0                         0

At equilibrium 5.5×1052x$5.5\times 10^{-5}-2x$         x                          x

Now, we can write,

[Br2][Cl2][BrCl]2=Kcx×x(5.5×1052x)2=42x5.5×1052x=6.48x=35.64×10512.96x13.96x=35.64×105x=35.6413.96×105=2.55×105$\frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{(5.5\times 10^{-5}-2x)^{2}}=42\\ \\ \Rightarrow \frac{x}{5.5\times 10^{-5}-2x}=6.48\\ \\ \Rightarrow x=35.64\times 10^{-5}-12.96x\\ \\ \Rightarrow 13.96x=35.64\times 10^{-5}\\ \\ \Rightarrow x=\frac{35.64}{13.96}\times 10^{-5}=2.55\times 10^{-5}$

So,$,$at equilibrium

[BrCl]=5.5×105(2×2.55×105)=5.5×1055.1×105=0.4×105=4.0×106molL1$[BrCl]=5.5\times 10^{-5}-(2\times 2.55\times 10^{-5})\\ \\ =5.5\times 10^{-5}-5.1\times 10^{-5}\\ \\ =0.4\times 10^{-5}\\ \\ =4.0\times 10^{-6}molL^{-1}$

Q.23. Find out Kc for the given reaction at temperature 1127K where the pressure is 1 atm. A solution of CO and CO2 is in equilibrium with carbon(solid). It has 93.55% CO by mass.

C(s)+CO2 (g) ⇒ 2CO (g)

Ans.)

Let us assume that the solution is of 100g in total.

Given, mass of CO = 93.55 g

Now, the mass of CO2 =(100 – 93.55)=6.45 g

Now, number of moles of CO, nCO=93.528=3.34mol$n_{CO}=\frac{93.5}{28}=3.34\: mol$

Number of moles of CO2, nCO2=6.4544=0.146mol$n_{CO_{2}}=\frac{6.45}{44}=0.146\: mol$

Partial pressure of CO,

PCO= nCOnCO+nCO2×ptotal=3.343.34+0.146×1=0.958atm$\frac{n_{CO}}{n_{CO}+n_{CO_{2}}}\times p_{total}=\frac{3.34}{3.34+0.146}\times 1=0.958\, atm$

Partial pressure of CO2, PCO2=nCO2nCO+nCO2×ptotal=0.1463.34+0.146×1=atm$P_{CO_{2}}=\frac{n_{CO_{2}}}{n_{CO}+n_{CO_{2}}}\times p_{total}=\frac{0.146}{3.34+0.146}\times 1=\, atm$

Therefore, Kp= [CO]2[CO2]=(0.938)20.062=14.19$\frac{[CO]^{2}}{[CO_{2}]}\\ \\ =\frac{(0.938)^{2}}{0.062}\\ \\ =14.19$

For the given reaction,

∆n = 2 – 1 = 1

We know that,

Kp=Kc(RT) Δn$\Delta n$

14.19=Kc(0.082×1127)1Kc=14.190.082×1127=0.154(approximately)$\Rightarrow 14.19=K_{c}(0.082\times 1127)^{1}\\ \\ \Rightarrow K_{c}=\frac{14.19}{0.082\times 1127}\\ \\ =0.154(approximately)$

Q.24.Find out

(I)The equilibrium constant for the formation of NO2 from NO and O2 at 298 K and

(II) ∆G°

NO(g)+12O2(g)NO2(g)$NO(g)+\frac{1}{2}O_{2}(g)\leftrightarrow NO_{2}(g)$

Where;

fG° (NO2) = 52.0 kJ/mol

∆fG° (NO) = 87.0 kJ/mol

∆fG° (O2) = 0 kJ/mol

Ans.)

(I)We know that,

∆G° = RT log Kc

∆G° = 2.303 RT log Kc

Kc=35.0×1032.303×8.314×298=6.134Kc=antilog(6.134)=1.36×106$K_{c}=\frac{-35.0\times 10^{-3}}{-2.303\times 8.314\times 298}\\ \\ =6.134\\ \\ ∴ K_{c}=antilog(6.134)\\ \\ =1.36\times 10^{6}$

Therefore, the equilibrium constant for the given reaction Kc is 1.36×106$1.36\times 10^{6}$

Q.25. When each of the following equilibria is subjected to a decrease in pressure by increasing the volume, does the number of moles of reaction products increase, decrease or remain same?

(I) PCl5(g) ⇒ PCl3 +Cl2 (g)

(II) Cao(s) + CO2 (g) ⇒ CaCO3(s)

(III) 3Fe (s) + 4H2O (g) ⇒ Fe3O4 (s) +4H2 (g)

Ans.

(I) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

(II) The number of moles of reaction products will decrease.

(III) The number of moles of reaction products remains the same.

Q.26. Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.

(I)COCl2 (g) ⇒ CO (g) +Cl2 (g)

(II)CH4 (g) +2S2 (g) ⇒ CS2 (g) + 2H2S (g)

(III)CO2 (g) +C (s) ⇒ 2CO (g)

(IV)2H2 (g) +CO  (g) ⇒ CH3OH(g)

(V)CaCO3 (s) ⇒ Cao (s) + CO2 (g)

(VI)4NH3 (g) +5O2 (g) ⇒ 4NO (g) + 6H2O (g)

Ans.)

When pressure is increased:

The reactions given in (i), (iii), (iv), (v), and (vi) will get affected.

Since, the number of moles of gaseous reactants is more than that of gaseous products; the reaction given in (iv) will proceed in the forward direction

Since, the number of moles of gaseous reactants is less than that of gaseous products, the reactions given in (i), (iii), (v), and (vi) will shift in the backward direction

Q.27. The equilibrium constant for the following reaction is 1.6×105$1.6 \times 10^{5}$at 1024 K.

H2 (g) + Br2 (g) ⇒ 2HBr (g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Ans.

Given, 𝐾𝑝 for the reaction i.e., H2 (g) + Br2 (g)2HBr (g) is 1.6×105$1.6 \times 10^{5}$.

Therefore, for the reaction   2HBr (g)H2 (g) + Br2 (g)  the equilibrium constant will be,

Kp= 1Kp=11.6×105=6.25×106$\frac{1}{K_{p}}\\ \\ =\frac{1}{1.6\times 10^{5}}\\ \\ =6.25\times 10^{-6}$

Now, let p be the pressure of both H2 and Br2 at equilibrium.

2HBr (g)                                         H2 (g)     +     Br2 (g)

Initial conc.                   10                                                         0                         0

At equilibrium           10-2p                                                       p                         p

Now, we can write,

pHBr×p2p2HBr=Kpp×p(102p)2=6.25×106p102p=2.5×103p=2.5×102(5.0×103)pp+(5.0×103)p=2.5×102(1005×103)=2.5×102p=2.49×102bar=2.5×102bar(approximately)$\frac{p_{HBr}\times p_{2}}{p^{2}_{HBr}}=K^{‘}_{p}\\ \\ \frac{p\times p}{(10-2p)^{2}}=6.25\times 10^{-6}\\ \\ \frac{p}{10-2p}=2.5\times 10^{-3}\\ \\ p=2.5\times 10^{-2}-(5.0\times 10^{-3})p\\ \\ p+(5.0\times 10^{-3})p=2.5\times 10^{-2}\\ \\ (1005\times 10^{-3})=2.5\times 10^{-2}\\ \\ p=2.49\times 10^{-2}bar=2.5\times 10^{-2}bar (approximately)$

Therefore, at equilibrium,

[H2]=[Br2]= 2.49×102bar$2.49\times 10^{-2}bar$

[HBr]= 102×(2.49×102)bar=9.95bar=10bar(approximately)$10-2\times (2.49\times 10^{-2})bar\\ \\ =9.95 bar=10 bar(approximately)$

Q.28.Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

CH4(g)+H2O(g)CO(g)+3H2(g)$CH_{4}(g)+H_{2}O(g)\leftrightarrow CO(g)+3H_{2}(g)$

(I)Write as expression for Kp for the above reaction.

(II) How will the values of Kp and composition of equilibrium mixture be affected by

(i)Increasing the pressure

(ii)Increasing the temperature

(iii)Using a catalyst?

Ans.)

(I)For the given reaction,

Kp=pCO×p3H2pCH4×pH2O$\frac{p_{CO}\times p_{H_{2}}^{3}}{p_{CH_{4}}\times p_{H_{2}O}}$

(II) (i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction.

(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.

(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

Q.29. Describe the effect of:

I) Removal of CO

III) Removal of CH3OH on the equilibrium of the reaction:

2H2 (g)+CO (g) ⇒ CH3OH (g)

Ans.)

(I) On removing CO, the equilibrium will shift in the backward direction.

(II) According to Le Chatelier’s principle, on addition of H2, the equilibrium of the given reaction will shift in the forward direction.

(III) On removing CH3OH, the equilibrium will shift in the forward direction.

(IV) On addition of CH3OH, the equilibrium will shift in the backward direction.

Q.30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3×103$8.3 \times 10^{-3}$. If decomposition is depicted as,

PCl5(g)PCl3(g)+Cl2(g)$PCl_{5}(g)\leftrightarrow PCl_{3}(g)+Cl_{2}(g)$

rH° = 124.0 kJmol–1

a) Write an expression for Kc for the reaction.

b) What is the value of Kc for the reverse reaction at the same temperature?

c) What would be the effect on Kc if

(ii) pressure is increased?

(iii) The temperature is increased?

Ans.)

(a)Kc=[PCl3(g)][Cl2(g)][PCl3(g)]$K_{c}=\frac{[PCl_{3}(g)][Cl_{2}(g)]}{[PCl_{3}(g)]}$

(b)Value of Kc for the reverse reaction at the same temperature is:

Kc=1Kc=18.3×103=1.2048×102=120.48$K_{c}^{‘} = \frac{1}{K_{c}}\\ \\ =\frac{1}{8.3 \times 10^{-3}} = 1.2048 \times 10^{2}\\ \\ =120.48$

(c)(i)Kc would remain the same because in this case, the temperature remains the same.

(ii)Kc is constant at constant temperature. Thus, in this case, Kc would not change. (iii)In an endothermic reaction, the value of Kc increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc will increase if the temperature is increased.

Q.31. Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

CO(g)+H2O(g)CO2(g)+H2(g)$CO(g)+H_{2}O(g)\leftrightarrow CO_{2}(g)+H_{2}(g)$

If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that Pco=PH2O  = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C

Ans.)

Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:

CO(g)       +    H2O CO2(g)      +     H2(g)

Initial conc.           4.0 bar          4.0 bar                            0                     0

At equilibrium      4.0-p             4.0-p                                p                     p

Given Kp = 10.1

PCO2×PH2PCO×PH2O=KPp×p(4.0p)(4.0p)=10.1p4.0p=3.178p=12.7123.178p4.178p=12.712p=12.7124.178p=3.04$\frac{P_{CO_{2}}\times P_{H_{2}}}{P_{CO}\times P_{H_{2}O}} = K_{P}\\ \\ \Rightarrow \frac{p\times p}{(4.0-p)(4.0-p)}=10.1\\ \\ \Rightarrow \frac{p}{4.0-p} =3.178\\ \\ \Rightarrow p =12.712-3.178p\\ \\ 4.178p = 12.712\\ \\ p =\frac{12.712}{4.178}\\ \\ p=3.04$

So, partial pressure of H2 is 3.04 bar at equilibrium.
Q.32. Predict which of the following reaction will have appreciable concentration of reactants and products:

(a)Cl2(g)2Cl(g);Kc=5×1039$Cl_{2}(g)\leftrightarrow 2Cl(g);K_{c}=5 \times 10^{-39}$

(b)Cl2(g)+2NO(g)2NOCl(g);Kc=3.7×108$Cl_{2}(g)+2NO(g)\leftrightarrow 2NOCl(g);K_{c}=3.7 \times 10^{8}$

(c)Cl2(g)+2NO2(g)2NO2Cl(g);Kc=1.8$Cl_{2}(g)+2NO_{2}(g)\leftrightarrow 2NO_{2}Cl(g);K_{c}=1.8$

Ans.)

If the value of Kc lies between 10–3 and 103 , a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

Q.33. The value of Kc for the reaction 3O2 (g) ⇒ 2O3 (g) is 2.0×1050$2.0\times 10^{-50}$ at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6×102$1.6 \times 10^{-2}$ , what is the concentration of O3?

Ans.)

Given,

3O2 (g)2O3 (g)

Then, Kc = [O3(g)]2[O2(g)]3$\frac{ [O_{3}(g)]^{2}}{[O_{2}(g)]^{3}}$

Given that Kc = 2.0×1050$2.0 \times 10^{-50}$ and [O2(g)] = 1.6×102$1.6 \times 10^{-2}$

Then,

2.0×1050=[O3(g)]2[1.6×102]3[O3(g)]2=2.0×1050×(1.6×102)3[O3(g)]2=8.192×1056[O3(g)]2=2.86×1028M$2.0 \times 10^{-50} = \frac { [O_{3}(g)]^{2} }{ [1.6 \times 10^{ -2 }]^{3}}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 2.0 \times 10^{-50} \times (1.6 \times 10^{-2})^{3}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 8.192 \times 10^{-56}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 2.86 \times 10^{-28} M$

So, the conc. of O3 is 2.86×1028M$2.86 \times 10^{-28}M$.

Q.34. The reaction, CO(g)+3H2(g)C


text:
SCIENTIA SINICA Informationis, Volume 48, Issue 4: 419-432(2018) https://doi.org/10.1360/N112017-00228

## Bayesian method for intent prediction in pervasive computing environments

Xin YI1,3,4, Chun YU1,2,3,4,*, Yuanchun SHI1,2,3,4
• AcceptedJan 29, 2018
• PublishedApr 10, 2018
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### Abstract

This paper describes the principle and examples of predicting users' intent using the Bayesian method. In natural interfaces, users use continuous, undetermined multi-modal data to express their interaction intention rather than using discrete, determined actions. In order to interpret their interaction intend, we can either use “black-box” machine-learning methods or use “white-box” user-behavior-modelling methods. The essence of the latter is to computationally model the users' interaction ability, which is crucial to understanding the users and exploring the computation principle of natural interaction. This paper summarizes the popular intelligent algorithms used in recent HCI researches, discusses the difference between these methods, and illustrates the methods of user modelling and the Bayesian method using some works from our laboratory.

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• Figure 1

Motion interaction interface

• Figure 2

Structure of a basic motion

• Figure 3

(Color online) Illustration of a touch model

• Figure 4

(Color online) When tapping the middle finger, the index finger moves along it due to the correlation movement between fingers

• Figure 5

(Color online) The amplitude ratio (AR) of correlated finger movement

• Figure 6

(Color online) Augmented Bayes model

Citations

• #### 0

Altmetric

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text:
Persistence Mayer Vietoris spectral sequence

PerMaViss

Welcome to PerMaViss! This is a Python3 implementation of the Persistence Mayer Vietoris spectral sequence. For full documentation, visit this page. For a mathematical description of the procedure, see Distributing Persistent Homology via Spectral Sequences.

In a nutshell, this library is intended to be a proof of concept for persistence homology parallelization. That is, one can divide a point cloud into covering regions, compute persistent homology on each part, and combine all results to obtain the global persistent homology again. This is done by means of the Persistence Mayer Vietoris spectral sequence. Here we present two examples, the torus and random point clouds in three dimensions. Both of these are divided into 8 mutually overlapping regions, and the spectral sequence is computed with respect to this cover. The resulting barcodes coincide with that which would be obtained by computing persistent homology directly.

This implementation is more of a prototype than a finished program. As such, it still needs to be optimized. Also, it would be great to have more examples for different covers. Additionally, it would be interesting to also have an implementation for cubical, alpha, and other complexes. Any collaboration or suggestion will be welcome!

Dependencies

PerMaViss requires:

• Python3

• NumPy

• Scipy

Optional for examples and notebooks:

• Matplotlib

• mpl_toolkits

Installation

Permaviss is built on Python 3, and relies only on NumPy and Scipy.

Additionally, Matplotlib and mpl_toolkits are used for the tutorials.

To install using pip3:

$pip3 install permaviss If you prefer to install from source, clone from GitHub repository: $ git clone https://github.com/atorras1618/PerMaViss
$cd PerMaViss$ pip3 install -e .

Quickstart

The main function which we use is permaviss.spectral_sequence.MV_spectral_seq.create_MV_ss. We start by taking 100 points in a noisy circle of radius 1

>>> from permaviss.sample_point_clouds.examples import random_circle
>>> point_cloud = random_circle(100, 1, epsilon=0.2)

Now we set the parameters for spectral sequence. These are

• a prime number p,

• the maximum dimension of the Rips Complex max_dim,

• the maximum radius of filtration max_r,

• the number of divisions max_div along the maximum range in point_cloud,

• and the overlap between different covering regions.

In our case, we set the parameters to cover our circle with 9 covering regions. Notice that in order for the algorithm to give the correct result we need overlap > max_r.

>>> p = 3
>>> max_dim = 3
>>> max_r = 0.2
>>> max_div = 3
>>> overlap = max_r * 1.01

Then, we compute the spectral sequence, notice that the method prints the successive page ranks.

>>> from permaviss.spectral_sequence.MV_spectral_seq import create_MV_ss
>>> MV_ss = create_MV_ss(point_cloud, max_r, max_dim, max_div, overlap, p)
PAGE: 1
[[  0   0   0   0   0]
[  7   0   0   0   0]
[133  33   0   0   0]]
PAGE: 2
[[  0   0   0   0   0]
[  7   0   0   0   0]
[100   0   0   0   0]]
PAGE: 3
[[  0   0   0   0   0]
[  7   0   0   0   0]
[100   0   0   0   0]]
PAGE: 4
[[  0   0   0   0   0]
[  7   0   0   0   0]
[100   0   0   0   0]]

We can inspect the obtained barcodes on the 1st dimension.

>>> MV_ss.persistent_homology[1].barcode
array([[ 0.08218822,  0.09287436],
[ 0.0874977 ,  0.11781674],
[ 0.10459203,  0.12520266],
[ 0.14999507,  0.18220508],
[ 0.15036084,  0.15760192],
[ 0.16260913,  0.1695936 ],
[ 0.16462541,  0.16942819]])

Notice that in this case, there was no need to solve the extension problem. See the examples folder for nontrivial extensions.

DISCLAIMER

The main purpose of this library is to explore how the Persistent Mayer Vietoris spectral sequence can be used for computing persistent homology.

This does not pretend to be an optimal library. Also, it does not parallelize the computations of persistent homology after the first page. Thus, this is slower than most other persistent homology computations.

This library is still on development and is still highly undertested. If you notice any issues, please email TorrasCasasA@cardiff.ac.uk

This library is published under the standard MIT licence. Thus: THE SOFTWARE IS PROVIDED “AS IS”, WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.

How to cite

Álvaro Torras Casas. (2020, January 20). PerMaViss: Persistence Mayer Vietoris spectral sequence (Version v0.0.2). Zenodo. http://doi.org/10.5281/zenodo.3613870

Reference

This module is written using the algorithm in Distributing Persistent Homology via Spectral Sequences.

Project details

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text:
TITLE

# Bulk universality for generalized Wigner matrices

AUTHOR(S)
Erdős, László; Yau, Horng-Tzer; Yin, Jun
PUB. DATE
October 2012
SOURCE
Probability Theory & Related Fields;Oct2012, Vol. 154 Issue 1/2, p341
SOURCE TYPE
DOC. TYPE
Article
ABSTRACT
Consider N Ã— N Hermitian or symmetric random matrices H where the distribution of the ( i, j) matrix element is given by a probability measure Î½ with a subexponential decay. Let $${\sigma_{ij}^2}$$ be the variance for the probability measure Î½ with the normalization property that $${\sum_{i} \sigma^2_{ij} = 1}$$ for all j. Under essentially the only condition that $${c\le N \sigma_{ij}^2 \le c^{-1}}$$ for some constant c > 0, we prove that, in the limit N â†’ âˆž, the eigenvalue spacing statistics of H in the bulk of the spectrum coincide with those of the Gaussian unitary or orthogonal ensemble (GUE or GOE). We also show that for band matrices with bandwidth M the local semicircle law holds to the energy scale M.
ACCESSION #
82301031

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text:
# Quick Start Guide¶

## Install¶

With Anaconda or Miniconda:

conda install h5py

If there are wheels for your platform (mac, linux, windows on x86) and you do not need MPI you can install h5py via pip:

pip install h5py

With Enthought Canopy, use the GUI package manager or:

enpkg h5py

To install from source see Installation.

## Core concepts¶

An HDF5 file is a container for two kinds of objects: datasets, which are array-like collections of data, and groups, which are folder-like containers that hold datasets and other groups. The most fundamental thing to remember when using h5py is:

Groups work like dictionaries, and datasets work like NumPy arrays

Suppose someone has sent you a HDF5 file, mytestfile.hdf5. (To create this file, read Appendix: Creating a file.) The very first thing you’ll need to do is to open the file for reading:

>>> import h5py
>>> f = h5py.File('mytestfile.hdf5', 'r')

The File object is your starting point. What is stored in this file? Remember h5py.File acts like a Python dictionary, thus we can check the keys,

>>> list(f.keys())
['mydataset']

Based on our observation, there is one data set, mydataset in the file. Let us examine the data set as a Dataset object

>>> dset = f['mydataset']

The object we obtained isn’t an array, but an HDF5 dataset. Like NumPy arrays, datasets have both a shape and a data type:

>>> dset.shape
(100,)
>>> dset.dtype
dtype('int32')

They also support array-style slicing. This is how you read and write data from a dataset in the file:

>>> dset[...] = np.arange(100)
>>> dset[0]
0
>>> dset[10]
10
>>> dset[0:100:10]
array([ 0, 10, 20, 30, 40, 50, 60, 70, 80, 90])

For more, see File Objects and Datasets.

### Appendix: Creating a file¶

At this point, you may wonder how mytestdata.hdf5 is created. We can create a file by setting the mode to w when the File object is initialized. Some other modes are a (for read/write/create access), and r+ (for read/write access). A full list of file access modes and their meanings is at File Objects.

>>> import h5py
>>> import numpy as np
>>> f = h5py.File("mytestfile.hdf5", "w")

The File object has a couple of methods which look interesting. One of them is create_dataset, which as the name suggests, creates a data set of given shape and dtype

>>> dset = f.create_dataset("mydataset", (100,), dtype='i')

The File object is a context manager; so the following code works too

>>> import h5py
>>> import numpy as np
>>> with h5py.File("mytestfile.hdf5", "w") as f:
>>>     dset = f.create_dataset("mydataset", (100,), dtype='i')

## Groups and hierarchical organization¶

“HDF” stands for “Hierarchical Data Format”. Every object in an HDF5 file has a name, and they’re arranged in a POSIX-style hierarchy with /-separators:

>>> dset.name
'/mydataset'

The “folders” in this system are called groups. The File object we created is itself a group, in this case the root group, named /:

>>> f.name
'/'

Creating a subgroup is accomplished via the aptly-named create_group. But we need to open the file in the “append” mode first (Read/write if exists, create otherwise)

>>> f = h5py.File('mydataset.hdf5', 'a')
>>> grp = f.create_group("subgroup")

All Group objects also have the create_* methods like File:

>>> dset2 = grp.create_dataset("another_dataset", (50,), dtype='f')
>>> dset2.name
'/subgroup/another_dataset'

By the way, you don’t have to create all the intermediate groups manually. Specifying a full path works just fine:

>>> dset3 = f.create_dataset('subgroup2/dataset_three', (10,), dtype='i')
>>> dset3.name
'/subgroup2/dataset_three'

Groups support most of the Python dictionary-style interface. You retrieve objects in the file using the item-retrieval syntax:

>>> dataset_three = f['subgroup2/dataset_three']

Iterating over a group provides the names of its members:

>>> for name in f:
...     print(name)
mydataset
subgroup
subgroup2

Membership testing also uses names:

>>> "mydataset" in f
True
>>> "somethingelse" in f
False

You can even use full path names:

>>> "subgroup/another_dataset" in f
True

There are also the familiar keys(), values(), items() and iter() methods, as well as get().

Since iterating over a group only yields its directly-attached members, iterating over an entire file is accomplished with the Group methods visit() and visititems(), which take a callable:

>>> def printname(name):
...     print(name)
>>> f.visit(printname)
mydataset
subgroup
subgroup/another_dataset
subgroup2
subgroup2/dataset_three

For more, see Groups.

## Attributes¶

One of the best features of HDF5 is that you can store metadata right next to the data it describes. All groups and datasets support attached named bits of data called attributes.

Attributes are accessed through the attrs proxy object, which again implements the dictionary interface:

>>> dset.attrs['temperature'] = 99.5
>>> dset.attrs['temperature']
99.5
>>> 'temperature' in dset.attrs
True

For more, see Attributes.


text:
Gauge Invariance (QED)

1. Mar 8, 2015

PhyAmateur

My book says that in this case $$e^+e^- \rightarrow \gamma \gamma$$ gauge invariance requires that $$k_{1\nu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0=k_{2\mu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})$$ Please see attachment. My question is how does this statement hold?

Last edited: Mar 8, 2015
2. Mar 9, 2015

vanhees71

This is a Ward-Takahashi identity. It comes from current conservation, $\partial_{\mu} j^{\mu}=0$, which is a necessary condition for gauge invariance.

3. Mar 9, 2015

PhyAmateur

But the author insisted that these conditions are met although the quantities in the equations k1ν(Aμν+A~μν)=0=k2μ(Aμν+A~μν) each separately are all different from zero. Why would he say that if it is already a consequence of ward identity? @vanhees71

4. Mar 10, 2015

vanhees71

The Ward-Takahashi identities refer to the proper vertex function and is valid order by order in $\hbar$ (number of loops), i.e., it's valid only for the complete set of diagrams of a given order.

5. Mar 10, 2015

PhyAmateur

How does this have to do with my question? I can't relate.

6. Mar 10, 2015

vanhees71

It has all to do with your last question, why it doesn't apply to a single diagram but only to the sum of the two diagrams relevant for the N-point function at the given loop order (which here is tree-level).

7. Mar 10, 2015

Avodyne

If you write out the expressions for $A^{\mu\nu}$ and $\tilde{A}^{\mu\nu}$ that you get from computing these diagrams, you will find that $k_{1\nu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0$ and $k_{2\mu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0$. So, while this can be predicted from gauge invariance, it is also the result of doing the explicit calculation.

8. Mar 10, 2015

PhyAmateur

Are you telling me that Ward Takashi holds for the amplitude and not necessarily for each of the Feynman diagrams whose sum is the amplitude? @vanhees71

9. Mar 11, 2015

vanhees71

Yes, that's it. The same is true for other symmetries, like charge conjugation symmetry, which implies that n-photon vertices with an odd number of photons must vanish. This also holds true only for the sum at a given loop order. Take, e.g., the one-loop triangle diagrams. You need to add both contibutions (which are different only by the orientation of the electron-positron loop making up the triangle).


text:
# Bijection of a segment in $\mathbb{R}^n$ to an interval. Path connectedness.

So this question is probably trivial.

It seems to me that I have a problem understanding how $\mathbb{R}^n$ generally looks like, and proving things with it.

Context: I started from the fact that I want to prove that any open connected subset of $\mathbb{R}^n$ is path connected. During the proof I use that between any 2 points in an open ball in $\mathbb{R}^n$ you can draw a line segment included in the ball.

So from this I want to show that the ball is path connected. Fix x and y in B. If there is a bijection from such a ball (call it $B$) such that f(x) = a and f(y) = b.then exists $f\colon B \rightarrow[a,b]$ bijective and continuous, and then $f^{-1}\colon[a,b]\rightarrow B$ will be bijective (here, the fact that it is also continuous is not self implied) with $f(a) = x$ and $f(b) = y$. There exists a bijection from g\colon [0,1] \rightarrow [a,b] with g(0)= a, and g(1) = b, so $f^{-1}. g\colon [0,1] \rightarrow B$ such that $(f^{-1}. g)(0) = x$ and $(f^{-1}. g)(1) = y$.

The issue is that I have to find $f$ because $f^{-1}$ is not guranteed to be continuous.

So what came to mind is projecting the line segment from $x$ to $y$ one of of the axis of the space. This should be a segment, but it is not necessarily bijective.

So what I need is a continuous bijective function from an interval $[a,b]$ to two points connected by a segment in $\mathbb{R}^n$ (ie. any $2$ points in $\mathbb{R}^n$).

Thank you! (I dumped a lot of my train thought here, but I think that might for someone who wants to explain this to me)

-
I don't think there will exist a continuous bijection from a ball in $\Bbb R^n$ to an interval in $\Bbb R$ in general. My suggestion to prove this: pick a point $x_0\in U$, where $U$ is your given open, connected set, and consider the set of points that can be connected by a path to $x_0$. If you can show this set is open and closed in $U$, you'll be done since $U$ is connected. – Clayton May 4 '13 at 16:49
That is exactly what I am doing. But in order to show that, I need to show that there exists a path between any 2 points in an open ball in $R^n$. I do not claim that there exists a bijection from a ball in $R^n$ to a closed interval in $R$, but from a segment in a ball in $R^n$ to a closed interval (a closed segment) in $R$. – elaRosca May 4 '13 at 16:53
From my question:"So what I need is a continuous bijective function from an interval [a,b] to two points connected by a segment in Rn (ie. any 2 points in R^n)." – elaRosca May 4 '13 at 16:54
Also from your question: "If there is a bijection from such a ball (call it B) such that $f(x) = a$ and $f(y) = b$, then exists $f:B−>[a,b]$ bijective and continuous," – Clayton May 4 '13 at 16:58

Without loss of generality you can assume that your ball is the one described by $x_1^2+\ldots+x_n^2<1$, otherwise use a translation (to bring the center of the ball to the origin) and a homothety to make the ball of the appropriate size. Your two points, $a,b$ will satisfy then $|a|<1,|b|<1$.
Consider the following path $C:[0,1]\to\mathbb{R}^n,\quad C(t)=ta+(1-t)b$, what we need to prove to conclude that $B$ is path connected is to show that $|C(t)|<1$ for all $t$, but this is very simple: $|C(t)|=|ta+(1-t)b|\leq t|a|+(1-t)|b|<t+(1-t)=1$
The first inequality is due to the fact that $\mathbb{R}^n$ is metric vector space and that $t>0, t-1>0$.


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# Given a rhombus (square) shaped grid, find which cell a point is in

Sorry for this simple question but I can't find a way to solve it in a simple way. I've found "point location" solutions, but there must be a much simpler way to do this, given that it's a regular grid and not arbitrary polygons.

As seen in the attached picture, I have a rhombus shaped grid. They have 90 degree angles, so they're squares, but I think the word rhombus is more appropriate here because everyone assumes something else when thinking of a square grid. The diagonal of a rhombus can be of any size (although if it makes the solution simpler (unlikely), my diagonals are always an even number). In this picture it's 6.

The center of a rhombus cell is indicated by a green square.

Given a random point on the graph, I need to get the coordinates of the center of the rhombus cell that the point is in. If the point is on the edge of 2 cells, it can favor either of them. 1 http://puu.sh/7Ej6V.png

• Rotate 45 degrees; then it's easy. – bubba Mar 21 '14 at 22:53
• Rotate the point 45 degrees, get the cell in a square grid, then rotate the center of the cell back 45 degrees? Thanks, it was indeed simple. – Wa. Mar 21 '14 at 23:08

Let $L$ be the "diagonal" width of a rhombus, by which I mean the distance between its two opposite corners. Note that by replacing $x$ with $\frac{x+y}{\sqrt{2}}$ and replacing $y$ with $\frac{x-y}{\sqrt{2}}$ we rotate everything by $45$ degrees around the origin.

Thus, given a point $a,b$, set

$$a' = \frac{a+b}{\sqrt{2}}, b' = \frac{a-b}{\sqrt{2}}.$$

This is like rotating 45 degrees around the origin. Then we want to align the square grid with the normal unit grid (I'm assuming the origin was in the center of one of the rhombuses), so we set

$$a'' = \frac{a' + \frac{L}{2}}{L}, b'' = \frac{b + \frac{L}{2}}{L}$$

Now we round down to the nearest integer to move each point to the bottom left of its square, and shift it by $\frac{1}{2}$ to move it to the center of that square:

$$a''' = \lfloor a''\rfloor + \frac{1}{2}, b''' = \lfloor b''\rfloor + \frac{1}{2}$$

Finally, we can transform these back to our rotated grid by multiplying by $L$ and rotating backwards:

$$a'''' = \frac{\sqrt{2}\left(La''' + Lb'''\right)}{2}, b'''' = \frac{\sqrt{2}\left(La''' - Lb'''\right)}{2}$$

Note that here we're just saying we want $La''' = \frac{a'''' + b''''}{\sqrt{2}}$ and $Lb''' = \frac{a'''' - b''''}{\sqrt{2}}$ and solving for $a''''$ and $b''''$, which will be the center of the rhombus you want.

The diagonal lines have equations $x+y = r + 2ar$ or $x-y= r + 2br$, where $r$ is half the length of the diagonal. Let $a = \left \lfloor \frac{x+y - r}{2r}\right\rfloor$ and $b = \left \lfloor \frac{x-y - r}{2r}\right\rfloor$. Then the point lies in the rhombus with center $\left((a+b+2)r, (a-b)r\right)$.


text:
# The Effect of Mitochondrial DNA Half-Life on Deletion Mutation Proliferation in Long Lived Cells

The Effect of Mitochondrial DNA Half-Life on Deletion Mutation Proliferation in Long Lived Cells The proliferation of mitochondrial DNA (mtDNA) with deletion mutations has been linked to aging and age related neurodegenerative conditions. In this study we model the effect of mtDNA half-life on mtDNA competition and selection. It has been proposed that mutation deletions (mtDNAdel\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{del}$$\end{document}) have a replicative advantage over wild-type (mtDNAwild\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{wild}$$\end{document}) and that this is detrimental to the host cell, especially in post-mitotic cells. An individual cell can be viewed as forming a closed ecosystem containing a large population of independently replicating mtDNA. Within this enclosed environment a selfishly replicating mtDNAdel\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{del}$$\end{document} would compete with the mtDNAwild\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{wild}$$\end{document} for space and resources to the detriment of the host cell. In this paper, we use a computer simulation to model cell survival in an environment where mtDNAwild\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{wild}$$\end{document} compete with mtDNAdel\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{del}$$\end{document} such that the cell expires upon mtDNAwild\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{wild}$$\end{document} extinction. We focus on the survival time for long lived post-mitotic cells, such as neurons. We confirm previous observations that mtDNAdel\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{del}$$\end{document} do have a replicative advantage over mtDNAwild\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{wild}$$\end{document}. As expected, cell survival times diminished with increased mutation probabilities, however, the relationship between survival time and mutation rate was non-linear, that is, a ten-fold increase in mutation probability only halved the survival time. The results of our model also showed that a modest increase in half-life had a profound affect on extending cell survival time, thereby, mitigating the replicative advantage of mtDNAdel\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{del}$$\end{document}. Given the relevance of mitochondrial dysfunction to various neurodegenerative conditions, we propose that therapies to increase mtDNA half-life could significantly delay their onset. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Acta Biotheoretica Springer Journals

# The Effect of Mitochondrial DNA Half-Life on Deletion Mutation Proliferation in Long Lived Cells

, Volume 69 (4) – Dec 1, 2021
25 pages

/lp/springer-journals/the-effect-of-mitochondrial-dna-half-life-on-deletion-mutation-rR1ifhzKEB
Publisher
Springer Journals
ISSN
0001-5342
eISSN
1572-8358
DOI
10.1007/s10441-021-09417-z
Publisher site
See Article on Publisher Site

### Abstract

The proliferation of mitochondrial DNA (mtDNA) with deletion mutations has been linked to aging and age related neurodegenerative conditions. In this study we model the effect of mtDNA half-life on mtDNA competition and selection. It has been proposed that mutation deletions (mtDNAdel\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{del}$$\end{document}) have a replicative advantage over wild-type (mtDNAwild\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{wild}$$\end{document}) and that this is detrimental to the host cell, especially in post-mitotic cells. An individual cell can be viewed as forming a closed ecosystem containing a large population of independently replicating mtDNA. Within this enclosed environment a selfishly replicating mtDNAdel\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{del}$$\end{document} would compete with the mtDNAwild\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{wild}$$\end{document} for space and resources to the detriment of the host cell. In this paper, we use a computer simulation to model cell survival in an environment where mtDNAwild\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{wild}$$\end{document} compete with mtDNAdel\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{del}$$\end{document} such that the cell expires upon mtDNAwild\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{wild}$$\end{document} extinction. We focus on the survival time for long lived post-mitotic cells, such as neurons. We confirm previous observations that mtDNAdel\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{del}$$\end{document} do have a replicative advantage over mtDNAwild\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{wild}$$\end{document}. As expected, cell survival times diminished with increased mutation probabilities, however, the relationship between survival time and mutation rate was non-linear, that is, a ten-fold increase in mutation probability only halved the survival time. The results of our model also showed that a modest increase in half-life had a profound affect on extending cell survival time, thereby, mitigating the replicative advantage of mtDNAdel\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\text {mtDNA}_{del}$$\end{document}. Given the relevance of mitochondrial dysfunction to various neurodegenerative conditions, we propose that therapies to increase mtDNA half-life could significantly delay their onset.

### Journal

Acta BiotheoreticaSpringer Journals

Published: Dec 1, 2021

Keywords: Mitochondrial DNA; Mitochondrial mutation; Aging; Computer simulation; Half-life; Neurodegenerative diseases

### References

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Roy's identity

Roy's identity (named for French economist René Roy) is a major result in microeconomics having applications in consumer choice and the theory of the firm. The lemma relates the ordinary (Marshallian) demand function to the derivatives of the indirect utility function. Specifically, denoting the indirect utility function as ${\displaystyle v(p,w),}$ the Marshallian demand function for good ${\displaystyle i}$ can be calculated as

${\displaystyle x_{i}^{m}=-{\frac {\frac {\partial v}{\partial p_{i}}}{\frac {\partial v}{\partial w}}}}$

where ${\displaystyle p}$ is the price vector of goods and ${\displaystyle w}$ is income.[1]

Derivation of Roy's identity

Roy's identity reformulates Shephard's lemma in order to get a Marshallian demand function for an individual and a good (${\displaystyle i}$) from some indirect utility function.

The first step is to consider the trivial identity obtained by substituting the expenditure function for wealth or income ${\displaystyle w}$ in the indirect utility function ${\displaystyle v(p,w)}$, at a utility of ${\displaystyle u}$:

${\displaystyle v(p,e(p,u))=u}$

This says that the indirect utility function evaluated in such a way that minimizes the cost for achieving a certain utility given a set of prices (a vector ${\displaystyle p}$) is equal to that utility when evaluated at those prices.

Taking the derivative of both sides of this equation with respect to the price of a single good ${\displaystyle p_{i}}$ (with the utility level held constant) gives:

${\displaystyle {\frac {\partial v[p,e(p,u)]}{\partial w}}{\frac {\partial e(p,u)}{\partial p_{i}}}+{\frac {\partial v[p,e(p,u)]}{\partial p_{i}}}=0}$.

Rearranging gives the desired result:

${\displaystyle -{\frac {\frac {\partial v[p,e(p,u)]}{\partial p_{i}}}{\frac {\partial v[p,e(p,u)]}{\partial w}}}={\frac {\partial e(p,u)}{\partial p_{i}}}=h_{i}(p,u)=x_{i}(p,e(p,u))}$

with the second-to-last equality following from Shephard's lemma and the last equality from a basic property of Hicksian demand.

Alternative proof using the envelope theorem

For expositional ease, consider the two-goods case. The indirect utility function ${\displaystyle v(p_{1},p_{2},w)}$ is the value function of the constrained optimization problem characterized by the following Lagrangian[2]:

${\displaystyle {\mathcal {L}}=u(x_{1},x_{2})+\lambda (w-p_{1}x_{1}-p_{2}x_{2})}$

By the envelope theorem, the derivatives of the value function ${\displaystyle v(p_{1},p_{2},w)}$ with respect to the parameters are:

${\displaystyle {\frac {\partial v}{\partial p_{1}}}=-\lambda x_{1}^{m}}$
${\displaystyle {\frac {\partial v}{\partial w}}=\lambda }$

where ${\displaystyle x_{1}^{m}}$ is the maximizer (i.e. the Marshallian demand function for good 1). Hence:

${\displaystyle -{\frac {\frac {\partial v}{\partial p_{1}}}{\frac {\partial v}{\partial w}}}=-{\frac {-\lambda x_{1}^{m}}{\lambda }}=x_{1}^{m}}$

Application

This gives a method of deriving the Marshallian demand function of a good for some consumer from the indirect utility function of that consumer. It is also fundamental in deriving the Slutsky equation.

References

1. ^ Varian, Hal (1992). Microeconomic Analysis (Third ed.). New York: Norton. pp. 106–108.
2. ^ Cornes, Richard (1992). Duality and Modern Economics. New York: Cambridge University Press. pp. 45–47. ISBN 0-521-33291-5.


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# Calculate $\int \log(1+\log(x))x^ndx$

How to calculate the following integral? $$\int \log(1+\log(x))x^ndx,$$ $n$ is an integer $\in N$.

• Mathematica produces $$\frac{x^{n+1} \log (\log (x)+1)-e^{-n-1} \text{Ei}((n+1) (\log (x)+1))}{n+1}.$$ – user64494 Aug 2 '14 at 16:15
• Well, for $|\log(x)| < 1$ you can use the Taylor series of $\log(1+x)$. – abnry Aug 2 '14 at 16:28

$\textbf{small hint}$ This was a direct result of the Mathematica output as shown by user* in the comments above. $$\int \log(1+\log(x))x^ndx = \int \log(\log(ex))x^ndx$$

substitute $t = ex$ we find $$\begin{eqnarray} \int \log(1+\log(x))x^ndx &=& \frac{1}{e}\int \log(\log(t))\left(\frac{t}{e}\right)^ndt \\ &=& e^{-(n+1)}\int\log(\log(t))t^ndt \end{eqnarray}$$ make another sub $\log t = u$ we find $$e^{-(n+1)}\int \log u \mathrm{e}^{(n+1)u}du$$

using the fact $$\int \mathrm{e}^{cx}\ln x = \frac{1}{c}\left(\mathrm{e}^{cx}\ln |x| - \mathrm{Ei}(cx)\right)$$ we find that with $c=n+1$ and $x = u$ $$e^{-(n+1)}\int \log u \mathrm{e}^{(n+1)u}du = e^{-(n+1)}\frac{1}{n+1}\left(\mathrm{e}^{(n+1)u}\ln |u| - \mathrm{Ei}[(n-1)u]\right)$$ and you should be able to finish off with subbing in $u = 1+ \log(x)$

• No I was just seeing that myself..I have messed up that exponential. I will fix it now. Thank you. – Chinny84 Aug 2 '14 at 16:32
• @Michael hopefully it is corrected now. Thank you again. – Chinny84 Aug 2 '14 at 16:39
• I think that the difficult part is $$\int \mathrm{e}^{cx}\ln x = \frac{1}{c}\left(\mathrm{e}^{cx}\ln |x| - \mathrm{Ei}(cx)\right)$$ Do you agree ? – Claude Leibovici Aug 2 '14 at 16:43
• How do you mean? is it with regards to "$\textbf{small hint}$"?. Or was it the wrong approach? – Chinny84 Aug 2 '14 at 16:49
• Not at all; your work is really good. I was just commenting the part I felt the most difficult (to me !!). Thanks for your answer. Cheers :-) – Claude Leibovici Aug 2 '14 at 16:58

I think it can be solved by this way $$\int \ln(1+\ln x)x^n dx$$ using integration by parts $$u=\ln(1+\ln x)\Rightarrow du=\frac{\frac{1}{x}}{1+\ln x}dx=\frac{dx}{x(1+\ln x)}\\ dv=x^ndx\Rightarrow v=\frac{x^{n+1}}{n+1}\\ \int \ln(1+\ln x)x^n dx=\ln(1+\ln x)\frac{x^{n+1}}{n+1}-\frac{1}{n+1}\int\frac{x^n}{1+\ln x}dx=\\ \frac{1}{n+1}\left[\ln(1+\ln x)x^{n+1}-\int\frac{x^n}{1+\ln x}dx\right]=\\ \frac{1}{n+1}\left[\ln(1+\ln x)x^{n+1}-\int\frac{x^n}{\ln ex}dx\right]$$ then $$\int\frac{x^n}{\ln ex}dx\\ u=\ln ex\\ e^u=ex\Rightarrow x=e^{u-1}\\ du=\frac{dx}{x}\Rightarrow dx=xdu=e^{u-1}du\\ \int\frac{x^n}{\ln ex}dx=\int\frac{(e^{u-1})^n}{u}e^{u-1}du=\int\frac{e^{nu-n}e^{u-1}}{u}du=\\ \int\frac{e^{(n+1)u}e^{-(n+1)}}{u}du=e^{-(n+1)}\int\frac{e^{(n+1)u}}{u}du=\\ e^{-(n+1)}\text{Ei}[(n+1)u]=e^{-(n+1)}\text{Ei}[(n+1)\ln ex]=e^{-(n+1)}\text{Ei}[(n+1)(1+\ln x)]$$ which gives $$\frac{\ln(1+\ln x)x^{n+1}-e^{-(n+1)}\text{Ei}[(n+1)(1+\ln x)]}{n+1}$$

• I like the fact you showed all the steps :) +1 – Chinny84 Aug 3 '14 at 10:06


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# American Institute of Mathematical Sciences

January  2011, 29(1): 169-192. doi: 10.3934/dcds.2011.29.169

## Periodic, subharmonic, and quasi-periodic oscillations under the action of a central force

 1 Dipartimento di Matematica e Geoscienze, Università di Trieste, P. le Europa 1, I-34127 Trieste, Italy 2 Departamento de Matematica Aplicada, Facultad de Ciencias, Universidad de Granada, E-18071, Granada, Spain

Received  November 2009 Revised  February 2010 Published  September 2010

We consider planar systems driven by a central force which depends periodically on time. If the force is sublinear and attractive, then there is a connected set of subharmonic and quasi-periodic solutions rotating around the origin at different speeds; moreover, this connected set stretches from zero to infinity. The result still holds allowing the force to be attractive only in average provided that an uniformity condition is satisfied and there are no periodic oscillations with zero angular momentum. We provide examples showing that these assumptions cannot be skipped.
Citation: Alessandro Fonda, Antonio J. Ureña. Periodic, subharmonic, and quasi-periodic oscillations under the action of a central force. Discrete & Continuous Dynamical Systems - A, 2011, 29 (1) : 169-192. doi: 10.3934/dcds.2011.29.169
##### References:
 [1] A. Ambrosetti and V. Coti Zelati, "Periodic Solutions of Singular Lagrangian Systems, Progress in Nonlinear Differential Equations and their Applications,", Birkhäuser Boston, (1993).   Google Scholar [2] J.C. Alexander, A primer on connectivity, Fixed point theory (Sherbrooke, Que., 1980), pp. 455-483,, Lecture Notes in Math. 886, (1981).   Google Scholar [3] V. I. Arnold, "Mathematical Methods of Classical Mechanics," 2nd edition,, Graduate Texts in Mathematics 60, (1978).   Google Scholar [4] E. N. Dancer, On the use of asymptotics in nonlinear boundary value problems,, Ann. Mat. Pura Appl. (4), 131 (1982), 167.  doi: doi:10.1007/BF01765151.  Google Scholar [5] K. Deimling, "Nonlinear Functional Analysis,", Springer-Verlag, (1985).   Google Scholar [6] A. Fonda and R. Toader, Periodic orbits of radially symmetric Keplerian-like systems: A topological degree approach,, J. Differential Equations, 244 (2008), 3235.  doi: doi:10.1016/j.jde.2007.11.005.  Google Scholar [7] P. Habets and L. Sanchez, Periodic solutions of dissipative dynamical systems with singular potentials,, Differential Integral Equations, 3 (1990), 1139.   Google Scholar [8] J. Leray and J. Schauder, Topologie et équations fonctionnelles (French),, Ann. Sci. École Norm. Sup. (3), 51 (1934), 45.   Google Scholar [9] J. Mawhin, Leray-Schauder continuation theorems in the absence of a priori bounds,, Topol. Methods Nonlinear Anal., 9 (1997), 179.   Google Scholar [10] J. Mawhin, C. Rebelo and F. Zanolin, Continuation theorems for Ambrosetti-Prodi type periodic problems,, Communications in Contemporary Mathematics, 2 (2000), 87.   Google Scholar [11] W. Magnus and S. Winkler, "Hill's Equation,", Dover Publ., (1979).   Google Scholar [12] I. Newton, "Principes Mathématiques de la Philosophie Naturelle," 2nd section,, Livre Premier, (1759).   Google Scholar

show all references

##### References:
 [1] A. Ambrosetti and V. Coti Zelati, "Periodic Solutions of Singular Lagrangian Systems, Progress in Nonlinear Differential Equations and their Applications,", Birkhäuser Boston, (1993).   Google Scholar [2] J.C. Alexander, A primer on connectivity, Fixed point theory (Sherbrooke, Que., 1980), pp. 455-483,, Lecture Notes in Math. 886, (1981).   Google Scholar [3] V. I. Arnold, "Mathematical Methods of Classical Mechanics," 2nd edition,, Graduate Texts in Mathematics 60, (1978).   Google Scholar [4] E. N. Dancer, On the use of asymptotics in nonlinear boundary value problems,, Ann. Mat. Pura Appl. (4), 131 (1982), 167.  doi: doi:10.1007/BF01765151.  Google Scholar [5] K. Deimling, "Nonlinear Functional Analysis,", Springer-Verlag, (1985).   Google Scholar [6] A. Fonda and R. Toader, Periodic orbits of radially symmetric Keplerian-like systems: A topological degree approach,, J. Differential Equations, 244 (2008), 3235.  doi: doi:10.1016/j.jde.2007.11.005.  Google Scholar [7] P. Habets and L. Sanchez, Periodic solutions of dissipative dynamical systems with singular potentials,, Differential Integral Equations, 3 (1990), 1139.   Google Scholar [8] J. Leray and J. Schauder, Topologie et équations fonctionnelles (French),, Ann. Sci. École Norm. Sup. (3), 51 (1934), 45.   Google Scholar [9] J. Mawhin, Leray-Schauder continuation theorems in the absence of a priori bounds,, Topol. Methods Nonlinear Anal., 9 (1997), 179.   Google Scholar [10] J. Mawhin, C. Rebelo and F. Zanolin, Continuation theorems for Ambrosetti-Prodi type periodic problems,, Communications in Contemporary Mathematics, 2 (2000), 87.   Google Scholar [11] W. Magnus and S. Winkler, "Hill's Equation,", Dover Publ., (1979).   Google Scholar [12] I. Newton, "Principes Mathématiques de la Philosophie Naturelle," 2nd section,, Livre Premier, (1759).   Google Scholar
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# How to derive Balck Scholes from the Binomial Model?

The book gives the following recipe, but no further details:

1. Do a Taylor series expansion of $$V = V(S,t)$$
2. Do a Taylor series expansion of $$V^{+} = V(u \cdot S, t + dt) \hspace{5mm}:\hspace{5 mm} u = 1 + \sigma \cdot \sqrt{dt}$$
3. Do a Taylor series expansion of $$V^{-} = V(d \cdot S, t + dt) \hspace{5mm}:\hspace{5 mm} d = 1 - \sigma \cdot \sqrt{dt}$$
4. Stick the three expansions into: $$V = \frac{1}{k} \cdot \bigg(p' \cdot V^{+} + (1-p') \cdot V^{-}\bigg) \hspace{5mm}:\hspace{5 mm} k \text{ is some discounting factor, } p'=\frac{1}{2} + \frac{r \cdot \sqrt{dt}}{2 \cdot \sigma} \text{ (r -- risk free rate)}$$.

However a number of things are impeding me doing the Taylor expansion:

1. The three equations given have no flesh, i.e. there is nothing to take a derivative on.
2. Even if we assume that $$V()$$ is the same as the one where we will be plugging back our expansions it is not clear to me where exactly $$S$$ and $$t$$ inputs go.
3. It is unclear which point to do the Taylor expansion around.

## 1 Answer

Check out Approximation of CRR as Black Scholes PDE. I show the derivation in my post there