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Thermodynamics is an important branch of Classical Physics, and its application to chemical systems plays a decisive role in the understanding of their behaviour. This thread contains a selection of problems collected or adapted from several sources: Thermodynamics books, university Physical Chemistry exams, several Internet resources (including this forum), and a few elaborated by me. The subjects covered by these problems are the following: basic definitions in Thermodynamics; Zeroth, First, Second, and Third Laws of Thermodynamics; Phase Equilibrium; Reaction Equilibrium; and Solutions. The material here can be used both by students preparing for Chemistry Olympiads and by university students. Note that the problems do not follow the order of the subjects covered as presented above: this because it is my opinion that a given problem is more challenging, and there is a better chance for the student to develop its understanding, if the problem is not explicitly linked to a given subject. I hope you find this material useful. Also, feel free to make any comments or suggestions.

[b]Important note:[/b] in order to prevent this thread to become a mess, please [i]do not post here discussions or solutions relative to these problems[/i] – use instead [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185861]this thread[/url] to do that. However, feel free to post here any interesting problems you might have from other sources.
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[size=150][b]Problem 1[/b][/size]. Consider the reversible expansion of an ideal gas from $ V_1$ to $ V_2$.

(a) Deduce the expression for the work done by the system on the surroundings if the process is isothermic.

(b) If the process is performed in an adiabatic way, will the final pressure of the system be greater, equal, or lower compared with (a)? Justify your answer.

[size=150][b]Problem 2[/b][/size]. For a fixed amount of a perfect gas, which of the following statements must be true?

(a) U and H each depend only on T.
(b) $ C_P$ is a constant.
(c) $ P\,dV \equal{} nR\,dT$ for every infinitesimal process.
(d) $ C_P \minus{} C_V \equal{} nR$
(e) $ dU \equal{} C_V\,dT$ for a reversible process.

[size=150][b]Problem 3[/b][/size]. For each of the following processes deduce whether each of the quantities $ q$, $ w$, $ \Delta U$, $ \Delta H$, $ \Delta S$, and $ \Delta S_{univ}$ is positive, zero, or negative:

(a) Reversible melting of solid benzene at 1 atm and the normal melting point.
(b) Reversible melting of ice at 1 atm and 0ºC.
(c) Reversible adiabatic expansion of a perfect gas.
(d) Reversible isothermal expansion of a perfect gas.
(e) Adiabatic expansion of a perfect gas into a vacuum.
(f) Reversible heating of a perfect gas at constant P.
(g) Reversible cooling of a perfect gas at constant V.
(h) Joule-Thomson adiabatic throttling of a perfect gas.
(i) Combustion of benzene in a sealed container with rigid, adiabatic walls.
(j) Adiabatic expansion of a nonideal gas into vacuum.

[size=150][b]Problem 4[/b][/size]. A perfect gas undergoes an expansion process at constant pressure. Does its internal energy increase or decrease? Justify your answer.

[size=150][b]Problem 5[/b][/size]. A perfect gas with $ C_{V,m} \equal{} 3R$ independent of T expands adiabatically into a vacuum, thereby doubling its volume. Two students present the following conflicting analyses. John uses the expression $ TV^{\gamma \minus{} 1} \equal{} constant$ to write $ T_2/T_1 \equal{} (V_1/2V_1)^{1/3} \equal{}> T_2 \equal{} T_1/\sqrt[3]{2}$. Wendy writes $ \Delta U \equal{} q \plus{} w \equal{} 0 \plus{} 0 \equal{} 0$ and $ \Delta U \equal{} C_V \Delta T$, so $ \Delta T \equal{} 0$ and $ T_2 \equal{} T_1$. Which student is correct? What error did the other student make?

[size=150][b]Problem 6[/b][/size]. Classify each of the following sentences as true or false. Always justify your choice.

(1)	A closed system cannot interact with its surroundings.

(2)	The Avogadro constant has no units.

(3)	For every cyclic process, the final state of the system is the same as the initial state.

(4)	$ \Delta T \equal{} 0$ for every isothermal process.

(5)	Every closed system is isolated.

(6)	Doubling the absolute temperature of an ideal gas at fixed volume and amount of gas will double its pressure.

(7)	Density is an intensive property.

(8)	A thermodynamic process is defined by the final state and the initial state.

(9)	$ \Delta U \equal{} 0$ for every cyclic process.

(10)	The Atlantic Ocean is an open system.

(11)	For a fixed amount of an ideal gas, the product PV remains constant during any process.

(12)	The P-V work in a mechanically reversible process in a closed system always equals $ \minus{}P \Delta V$.

(13)	The ratio PV/mT is the same for all gases in the limit of zero pressure.

(14)	For every process, $ \Delta E_{syst} \equal{} \minus{} \Delta E_{surr}$.

(15)	A homogeneous system must be a pure substance.

(16)	$ \Delta U \equal{} q \plus{} w$ for every thermodynamic system at rest in the absence of external fields.

(17)	A system containing only one substance must be homogeneous.

(18)	The pressure of a nonideal gas mixture is equal to the sum of the partial pressures defined by $ P_i \equal{} x_i P$.

(19)	A change of state from state 1 to state 2 produces a greater increase in entropy when carried out irreversibly than when done reversibly.

(20)	Mole fractions are intensive properties.

(21)	$ \Delta U \equal{} 0$ for a reversible phase change at constant T and P.

(22)	For every cyclic process, the final state of the surroundings is the same as the initial state of the surroundings.

(23)	On the Celsius scale, the boiling point of water is slightly less than 100.00ºC.

(24)	The value of the work in a reversible process in a closed system can be found so long as we know the initial state and the final state of the system.

(25)	Every isolated system is closed.

(26)	Ideal gas isotherms further away from the axes of a PV plot correspond to higher temperatures.

(27)	All ideal gases have the same density at 25ºC and 1 bar.

(28)	Increasing the temperature of the hot reservoir of a Carnot cycle engine must increase the efficiency of the engine.

(29)	The infinitesimal P-V work in a mechanically reversible process in a closed system always equals $ \minus{}P dV$.

(30)	For a closed system at rest with no fields present, the sum $ q \plus{} w$ has the same value for every process that goes from a given state 1 to a given state 2.

(31)	The ratio PV/nT is the same for all gases in the limit of zero pressure.

(32)	The value of the integral $ \int_1^2 P\,dV$ is fixed once the initial and final states 1 and 2 and the equation of state P = P(T,V) are known.

(33)	Every process for which $ \Delta T \equal{} 0$ is an isothermal process.

(34)	$ C_P$ is an extensive property and a state function.

(35)	We have $ \int_1^2 P\,dV \equal{} \int_1^2 nR\,dT$ for every reversible process in an ideal gas.

(36)	If systems A and B each consist of pure liquid water at 1 bar pressure and if $ T_A > T_B$, then the internal energy of system A must be greater than that of B.

(37)	The equation $ w_{rev} \equal{} \minus{} \int_1^2 P\,dV$ applies only to constant-pressure processes.

(38)	The quantities $ H$, $ U$, $ PV$, $ \Delta H$, and $ P \Delta V$ all have the same dimensions.

(39)	The heat $ q$ for an irreversible change of state from state 1 to state 2 might differ from the heat for the same change of state carried out reversibly.

(40)	$ q$ must be zero for an isothermal process.

(41)	$ \Delta H$ is a state function.

(42)	For a closed system at rest in the absence of external fields, $ U \equal{} q \plus{} w$.

(43)	The value of the work done by a system in a given process can only be calculated if the process is reversible.

(44)	$ \Delta T \equal{} 0$ for every adiabatic process in a closed system.

(45)	Decreasing the temperature of the cold reservoir of a Carnot cycle engine must increase the efficiency of the engine.

(46)	$ \Delta G \equal{} 0$ always for a reversible process in a closed system capable of P-V work only.

(47)	$ C_V$ is independent of T for every perfect gas.

(48)	The higher the absolute temperature of a system, the smaller the increase in its entropy produced by a given positive amount $ \delta q_{rev}$ of reversible heat flow.

(49)	If a closed system undergoes a reversible process for which $ \Delta V \equal{} 0$, then the P-V work done on the system in this process must be zero.

(50)	A closed system process that has $ \Delta T \equal{} 0$, must have $ \Delta U \equal{} 0$.

(51)	$ \Delta H$ is defined only for a constant-pressure process.

(52)	For an adiabatic process in a closed system, $ \Delta S$ must be zero.

(53)	The entropy of 20 g of $ H_2O(l)$ at 300 K and 1 bar is twice the entropy of 10 g of $ H_2O(l)$ at the same T and P.

(54)	U remains constant in every isothermal process in a closed system.

(55)	The relation $ \Delta G \equal{} \Delta H \minus{} T \Delta S$ is valid for all processes.

(56)	A Carnot cycle is, by definition, a reversible cycle.

(57)	For a constant-volume process in a closed system, $ \Delta U \equal{} \Delta H$.

(58)	An adiabatic process cannot decrease the entropy of a closed system.

(59)	$ q \equal{} 0$ for every cyclic process.

(60)	The molar entropy of 20 g of $ H_2O(l)$ at 300 K and 1 bar is equal to the molar entropy of 10 g of $ H_2O(l)$ at the same T and P.

(61)	For every closed system in thermal and mechanical equilibrium and capable of P-V work only, the state function G is minimized when material equilibrium is reached.

(62)	If a closed system at rest in the absence of external fields undergoes an adiabatic process that has $ w \equal{} 0$, then the system’s temperature must remain constant.

(63)	The quantities SdT, TdS, VdP, and $ \int_1^2 V\,dP$ all have dimensions of energy.

(64)	The chemical potential $ \mu_i$ is a state function and an intensive property.

(65)	For a reversible isothermal process in a closed system, $ \Delta S$ must be zero.

(66)	$ \Delta G$ is undefined for a process in which T changes.

(67)	P-V work is usually negligible for solids and liquids.

(68)	Since a Carnot cycle is a cyclic process, the work done in a Carnot cycle is zero.

(69)	The Gibbs energy of 12 g of ice at 0ºC and 1 atm is less than the Gibbs energy of 12 g of liquid water at 0ºC and 1 atm.

(70)	If neither heat nor matter can enter or leave a system, that system must be isolated.

(71)	$ \Delta G \equal{} 0$ for a reversible phase change at constant T and P.

(72)	We always have $ \int_1^2 \frac{C_V}{T}\,dT \equal{} C_V \ln \frac{T_2}{T_1}$.

(73)	$ C_P \minus{} C_V \equal{} nR$ for all gases.

(74)	For a closed system with P-V work only, a constant-pressure process that has $ q > 0$ must have $ \Delta T > 0$.

(75)	For an adiabatic process in a closed system, $ \Delta S_{syst}$ must be zero.

(76)	For a closed system, $ \Delta S$ can never be negative.

(77)	For a reversible process in a closed system, $ \Delta S$ must be zero.

(78)	For a closed system, equilibrium has been reached when S has been maximized.

(79)	Thermodynamics cannot calculate $ \Delta S$ for an irreversible process.

(80)	For a reversible process in a closed system, $ \delta q \equal{} T\,dS$.

(81)	For every process in an isolated system, $ \Delta T \equal{} 0$.

(82)	For every process in an isolated system that has no macroscopic kinetic or potential energy, $ \Delta U \equal{} 0$.

(83)	For an adiabatic process in a closed system, $ \Delta S$ cannot be negative.

(84)	For every process in an isolated system, $ \Delta S \equal{} 0$.

(85)	$ \Delta S \equal{} 0$ for every adiabatic process in a closed system.

(86)	For every reversible process in a closed system, $ \Delta S \equal{} \Delta H/T$.

(87)	For a process in an isolated system, $ \Delta S$ cannot be negative.

(88)	For every isothermal process in a closed system, $ \Delta S \equal{} \Delta H/T$.

(89)	$ q \equal{} 0$ for every isothermal process in a closed system.

(90)	For the conversion of 1 mol of $ N_2(g)$ from 25ºC and 10 L to 25ºC and 20 L, $ \Delta S$ is the same independently of the conversion being reversible or irreversible.

(91)	For a reversible process in a closed system, $ \Delta S_{univ}$ must be zero.

(92)	$ C_P \minus{} C_V \equal{} TV \alpha ^2/ \kappa$ for any substance.

(93)	For an irreversible, isothermal, isobaric process in a closed system with P-V work only, $ \Delta G$ must be negative.

(94)	$ \Delta S$ is positive for every irreversible process.

(95)	The work done by a closed system can exceed the decrease in the system’s internal energy.

(96)	The Gibbs energy of a closed system with P-V work only is always minimized at equilibrium.

(97)	If a system remains in thermal and mechanical equilibrium during a process, then its T and P are constant during the process.

(98)	The entropy S of a closed system with P-V work only is always maximized at equilibrium.

(99)	$ \Delta G \equal{} 0$ for the freezing of supercooled water at -10ºC and 1 atm.

(100)	$ \Delta S_{syst} \plus{} \Delta S_{surr} > 0$ for every irreversible process.

(101)	$ G_{syst} \plus{} G_{surr}$ is constant for any process.

(102)	If a system absorbs heat, then its temperature will always increase.

(103)	If the temperature of a system increases, then it must have absorbed heat.

(104)	By no means can a given quantity of heat be completely converted into work, according to the second law.

(105)	For every spontaneous process $ \Delta S > 0$.

(106)	For every isochoric process in a closed system $ \Delta U \equal{} q_V$, where $ q_V$ is the heat exchanged at constant volume.

(107)	If the temperature of a system increases then its volume must also increase.

(108)	$ \Delta G < 0$ for every spontaneous process.

(109)	For every isobaric process in a closed system $ \Delta H \equal{} q_P$, where $ q_P$ is the heat exchanged at constant pressure.

(110)	On isothermal conditions, if the pressure of a system increases then its volume must decrease.

[size=150][b]Problem 7[/b][/size]. Because the heat capacities per unit volume of gases are small, accurate measurement of $ C_P$ or $ C_V$ for gases is not easy. However, accurate measurement of the adiabatic coefficient $ \gamma$ of a gas (for example, by measurement of the speed of sound in the gas) is easy. For gaseous $ CCl_4$ at 0.1 bar and 20ºC, experiment gives $ \gamma \equal{} 1.13$. Find $ C_{P,m}$ and $ C_{V,m}$ for $ CCl_4(g)$ at 20ºC and 1 bar.

[size=150][b]Problem 8[/b][/size]. Which of the following cyclic integrals must vanish for a closed system with P-V work only?

(a) $ \oint P\,dV$

(b) $ \oint (P\,dV \plus{} V\,dP)$

(c) $ \oint V\,dV$

(d) $ \oint \frac{\delta q_{rev}}{T}$

(e) $ \oint H\,dT$

(f) $ \oint dU$

(g) $ \oint \delta q_{rev}$

(h) $ \oint \delta q_P$

(i) $ \oint \delta w_{rev}$

(j) $ \oint \frac{\delta w_{rev}}{P}$

[size=150][b]Problem 9[/b][/size]. The molar heat capacity of oxygen at constant pressure for temperatures in the range 300 to 400 K and for low or moderate pressures can be approximated as $ C_{P,m} \equal{} a \plus{} bT$, where $ a \equal{} 6.15\,\,cal\,K^{\minus{}1}\,mol^{\minus{}1}$ and $ b \equal{} 0.00310\,\,cal\,K^{\minus{}2}\,mol^{\minus{}1}$.

(a) Calculate $ q$, $ w$, $ \Delta U$, and $ \Delta H$ when 2.00 mol of $ O_2$ are reversibly heated from 27ºC to 127ºC with P held fixed at 1.00 atm. Assume perfect-gas behaviour.

(b) Calculate $ q$, $ w$, $ \Delta U$, and $ \Delta H$ when 2.00 mol of $ O_2$ initially at 1.00 atm are reversibly heated from 27ºC to 127ºC with V held fixed.

(c) Find $ \Delta S$ when 2.00 mol of $ O_2$ are heated from 27ºC to 127ºC with P held fixed.

[size=150][b]Problem 10[/b][/size]. The molar heat of vaporization of Ar at its normal boiling point 87.3 K is 1.56 kcal/mol.

(a) Compute $ \Delta S$ for the vaporization of 1.00 mol of Ar at 87.3 K and 1 atm.

(b) Find $ \Delta S$ when 5.00 g of Ar gas condenses to liquid at 87.3 K and 1 atm.

[size=150][b]Problem 11[/b][/size]. Explain how liquid water can go from 25ºC and 1 atm to 30ºC and 1 atm in a process for which $ q < 0$.

[size=150][b]Problem 12[/b][/size]. One mole of liquid water at 30ºC is adiabatically compressed, with P increasing from 1.00 to 10.00 atm. Since liquids and solids are rather incompressible, it is a fairly good approximation to take V as unchanged for this process. With this approximation calculate $ q$, $ \Delta U$, and $ \Delta H$ for this process.

[size=150][b]Problem 13[/b][/size]. Calculate $ \Delta U$ and $ \Delta H$ for each of the following changes in state of 2.50 mol of a perfect monoatomic gas with $ C_{V,m} \equal{} 1.5R$ for all temperatures:

(a) (1.50 atm, 400 K) -----> (3.00 atm, 600 K)

(b) (2.50 atm, 20.0 L) -----> (2.00 atm, 30.0 L)

(c) (28.5 L, 400 K) -----> (42.0 L, 400 K)

[size=150][b]Problem 14[/b][/size]. Can $ q$ and $ w$ be calculated for the processes of the previous problem? If the answer is yes, calculate them for each process.

[size=150][b]Problem 15[/b][/size]. For each of the following sets of quantities, all the quantities except one have something in common. State what they have in common and state which quantity does not belong with to each set (in some cases, more than one answer for the property in common might be possible).

(a) $ H$, $ U$, $ q$, $ S$, $ T$.

(b) $ T$, $ \Delta S$, $ q$, $ w$, $ \Delta H$.

(c) $ q$, $ w$, $ U$, $ \Delta U$, $ V$, $ H$.

(d) $ C_V$, $ C_P$, $ U$, $ T$, $ S$, $ G$, $ A$, $ V$.

(e) $ \rho$, $ S_m$, $ M$, $ V$.

(f) $ \Delta H$, $ \Delta S$, $ dV$, $ \Delta P$.

(g) $ H$, $ U$, $ G$, $ S$, $ A$.

(h) $ U$, $ V$, $ \Delta H$, $ S$, $ T$.

[size=150][b]Problem 16[/b][/size]. For each of the following statements, tell which state function(s) is (are) being described.

(a) It enables one to find the rates of change of $ H$ and of $ S$ with respect to T at constant P.

(b) They determine whether substance $ i$ in phase $ \alpha$ is in phase equilibrium with $ i$ in phase $ \beta$.

(c) It enables one to find the rates of change of $ U$ and of $ S$ with respect to T at constant V.

(d) It is maximized when an isolated system reaches equilibrium.

(e) It is maximized when a system reaches equilibrium.

(f) It is minimized when a closed system capable of P-V work only and held at constant T and P reaches equilibrium.

[size=150][b]Problem 17[/b][/size]. Show that it is impossible to attain the absolute zero of temperature.

[size=150][b]Problem 18[/b][/size]. Derive expressions for $ \left(\frac{\partial S}{\partial T}\right)_P$, $ \left(\frac{\partial S}{\partial T}\right)_V$, and $ \left(\frac{\partial H}{\partial P}\right)_T$.

[size=150][b]Problem 19[/b][/size]. Show that $ \left(\frac{\partial U}{\partial P}\right)_T \equal{} PV \kappa \minus{} TV \alpha$.

[size=150][b]Problem 20[/b][/size]. Show that $ \left(\frac{\partial U}{\partial T}\right)_P \equal{} C_P \minus{} PV \alpha$.

[size=150][b]Problem 21[/b][/size]. Show that $ \left(\frac{\partial H}{\partial V}\right)_T \equal{} \frac{\alpha T \minus{} 1}{\kappa}$.

[size=150][b]Problem 22[/b][/size]. Show that $ \mu_J \equal{} \frac{P \minus{} \alpha T / \kappa}{C_V}$, where $ \mu_J$ is the Joule coefficient.

[size=150][b]Problem 23[/b][/size]. Show that $ \mu_{JT} \equal{} \frac{V(\alpha T \minus{} 1)}{C_P}$, where $ \mu_{JT}$ is the Joule-Thomson coefficient.

[size=150][b]Problem 24[/b][/size]. Prove the [i]Gibbs-Helmholtz equation[/i], $ \left[\frac {\partial (G/T)}{\partial T}\right]_P \equal{} \minus{} \frac {H}{T^2}$.

[size=150][b]Problem 25[/b][/size]. Prove [i]Mayer’s Relation[/i], $ C_P \minus{} C_V \equal{} \frac {TV \alpha ^2}{\kappa}$.

[size=150][b]Problem 26[/b][/size]. Show that $ \left(\frac {\partial C_P}{\partial P}\right)_T \equal{} \minus{} T \left(\frac {\partial ^2V}{\partial T^2}\right)_P$.  The volumes of substances increase approximately linearly with T, so $ \frac {\partial ^2V}{\partial T^2}$ is usually quite small. Therefore, the pressure dependence of $ C_P$ can usually be neglected unless one is dealing with high pressures.

[size=150][b]Problem 27[/b][/size]. Show that $ \left(\frac {\partial C_V}{\partial V}\right)_T \equal{} T \left(\frac {\partial ^2P}{\partial T^2}\right)_V$.

[size=150][b]Problem 28[/b][/size]. A certain gas obeys the equation of state $ PV \equal{} nRT(1 \plus{} bP)$, where $ b$ is a constant. For this gas, find expressions for (a) $ \left(\frac {\partial U}{\partial V}\right)_T$, (b) $ C_P \minus{} C_V$, and (c) $ \mu_{JT}$.

[size=150][b]Problem 29[/b][/size]. A certain author states that a given solid obeys the two equations of state shown bellow, where $ V_o$, $ a$, $ b$, and $ c$ are constants. Show that this statement is wrong, because both equations cannot describe the same substance.

$ V \equal{} V_o \minus{} aP \plus{} bT$ and $ U \equal{} cT \minus{} bPT$

[size=150][b]Problem 30[/b][/size]. Show that $ \frac{\left(\frac{\partial P}{\partial V}\right)_S}{\left(\frac{\partial P}{\partial V}\right)_T} \equal{} \gamma$ (this is [i]Reech’s Relation[/i]).

[size=150][b]Problem 31[/b][/size]. Show that

$ \left(C_P \minus{} C_V \right) \left( \frac{\partial^2 T}{\partial P \partial V} \right) \plus{} \left( \frac{\partial C_P}{\partial P} \right)_V \left( \frac{\partial T}{\partial V} \right)_P \minus{} \left( \frac{\partial C_V}{\partial V} \right)_P \left( \frac{\partial T}{\partial P} \right)_V \equal{} 1$.

[size=150][b]Problem 32[/b][/size]. The behaviour of a given gas is described by the thermal equation of state $ PV \equal{} kT^3$ and by the energetic equation of state $ U \equal{} bT^3 \ln\left(\frac{V}{V_o}\right)$, where $ k$ and $ V_o$ are known constants.

(a) Express the constant $ b$ as a function of $ k$ and $ V_o$.

(b) Is $ C_V$ for this gas compatible with the Third Law of thermodynamics?

[size=150][b]Problem 33[/b][/size]. An ideal gas goes through a reversible adiabatic process from the initial state $ (P_1, V_1, T_1)$ to the final state $ (P_2, V_2, T_2)$. You may assume the heat capacities to be constant.

(a) Show that for such a process Poisson’s equation $ PV^{\gamma} \equal{} const.$ holds.

(b) Derive an expression for the work done by (or on) the gas.

(c) Show that, if a quantity of heat equivalent to the work done by the gas in the previous process is added to the gas in the final state $ (P_2, V_2, T_2)$ while keeping the volume constant, the temperature of the gas will return to $ T_1$.

[size=150][b]Problem 34[/b][/size]. Show that the internal energy of a system whose equation of state has the form $ P \equal{} T \cdot f(V)$ (where $ f(V)$ is a function of V only) is independent of volume.

[size=150][b]Problem 35[/b][/size]. The internal energy per unit volume of a gas, $ u$, is a function of T only, and the equation of state for the gas is given by $ P \equal{} \frac {1}{3} u(T)$. Is the heat capacity at constant volume for this gas compatible with the Third Law of thermodynamics?

[size=150][b]Problem 36[/b][/size]. Show that the Hessian determinant $ H$ for the internal energy $ U \equal{} U(S,V)$ is given by $ H[U(S,V)] \equal{} \minus{} \frac{T}{C_V} \left(\frac{\partial P}{\partial V}\right)_T$.

[size=150][b]Problem 37[/b][/size]. (a) Provide a deduction of [i]Clapeyron equation [/i] (for a bonus, try to do this in at least two different ways).

(b) Use Clapeyron equation to estimate the slope of the equilibrium curve solid-liquid of water, knowing that the enthalpy of fusion of water is $ \Delta H_{fus} \equal{} 6.008\,kJ/mol$, the densities of ice and liquid water at 0ºC are $ \rho_{ice} \equal{} 0.91671\,g/mL$ and $ \rho_{lw} \equal{} 0.99984\,g/mL$, and the molar mass of water is 18 g/mol.

[size=150][b]Problem 38[/b][/size]. The expansion and pressure coefficients for a given system are given by

$ \alpha \equal{} \frac {b}{V_o \minus{} aP \plus{} bT}$ and $ \alpha _P \equal{} \frac {b}{aP}$

where $ a,\,b, \,V_o$ are constants. Derive the thermal state equation for this system.

[size=150][b]Problem 39[/b][/size]. The heat capacity at constant volume for a system is given by

$ C_V \equal{} A \frac {e^{ \minus{} k/T}}{T^3}$,

where [i]A[/i] and [i]k[/i] are constants. Is this expression compatible with the Third Law of Thermodynamics?

[size=150][b]Problem 40[/b][/size]. Is the following state equation for a gas, $ PV \equal{} nRT \left(1\minus{}\frac{c}{V}\right)$, where $ c$ is a positive constant, acceptable for all values of pressure and volume?

[size=150][b]Problem 41[/b][/size]. Show that $ \left(\frac {\partial H}{\partial P}\right)_T \equal{} V \minus{} T \left(\frac {\partial V}{\partial T}\right)_P$.

[size=150][b]Problem 42[/b][/size]. Prove the equation $ \left(\frac {\partial U}{\partial V}\right)_T \equal{} T \left(\frac {\partial P}{\partial T}\right)_V \minus{} P$ by at least three different ways. This is a very important expression, since it relates the thermal and energetic equations of state.

The Joule-Thomson or “throttling” process is an experimental procedure of great practical importance that is commonly used to cool and liquefy gases. It was originally devised in 1853 by James Joule and William Thomson (in later life, Lord Kelvin). In this process (refer to figure I below), a gas is made to pass through a porous barrier from a region maintained at the higher pressure $ P_1$ to a region maintained at a lower pressure $ P_2$. Note that all the walls of the system (including the pistons and porous plug) are adiabatic. The throttling barrier is porous but not greatly so, which allows a gas to pass slowly from one chamber to the other, therefore maintaining pressure equilibrium on both sides. The porous wall was originally made of a wad of cotton, but nowadays it can be made of glass fibbers or a porous ceramic material, or the porous plug can be replaced by a narrow opening (a needle valve). The last two options are used in industrial applications, where the pressure difference is maintained continuously by a pump.
To treat thermodynamically this process, start by noting that all the pressure drop occurs at the porous wall, therefore the gas on the left is maintained at constant pressure $ P_1$ and the gas that emerges on the right is at constant pressure $ P_2$. As there is no transfer of energy to/from the gas as heat, we can write for $ \Delta U \equal{} W_T$, where $ W_T$ is total amount of work done on/by the gas. If one mole of the gas occupying a volume $ V_1$ is made to pass through the barrier by the left piston, the work done [i]on[/i] the gas is $ W_{left} \equal{} P_1V_1$. As this amount of gas emerges on the right, it will occupy a new volume $ V_2$ (since the pressure is different on the right) and it will push the right piston, therefore the work done [i]by[/i] the gas is $ W_{right} \equal{} \minus{} P_2V_2$. Hence,

$ \Delta U \equal{} W_T \equal{} W_{left} \plus{} W_{right} \equal{} > U_2 \minus{} U_1 \equal{} P_1V_1 \minus{} P_2V_2 \equal{} > U_1 \plus{} P_1V_1 \equal{} U_2 \plus{} P_2V_2$

Remembering the definition of enthalpy, $ H \equiv U \plus{} PV$, the equation above can be written as

$ H_1 \equal{} H_2$

and so the Joule-Thomson process occurs with constant enthalpy. As we are usually interested in the change of temperature that occurs with the forced change in pressure, this process allow us to measure the partial derivative

$ \left(\frac {\partial T}{\partial P}\right)_H$

(considering small changes in pressure), which is commonly called the [i]Joule-Thomson coefficient[/i] and symbolized by $ \mu _{JT}$. Now we are to apply the mathematical formalism of thermodynamics in order to relate this partial derivative with readily measured quantities. We have

$ \left(\frac {\partial T}{\partial P}\right)_H \left(\frac {\partial P}{\partial H}\right)_T \left(\frac {\partial H}{\partial T}\right)_P \equal{} \minus{} 1 \equal{} > \left(\frac {\partial T}{\partial P}\right)_H \equal{} \minus{} \frac {1}{\left(\frac {\partial P}{\partial H}\right)_T \left(\frac {\partial H}{\partial T}\right)_P }$

$ \equal{} > \left(\frac {\partial T}{\partial P}\right)_H \equal{} \minus{} \frac {\left(\frac {\partial H}{\partial P}\right)_T}{\left(\frac {\partial H}{\partial T}\right)_P}$

But $ \left(\frac {\partial H}{\partial T}\right)_P \equal{} C_P$, and from the definition of enthalpy, differentiating, and using $ dU \equal{} TdS \minus{} PdV$ we have

$ dH \equal{} TdS \plus{} VdP$

Now considering a constant T process and dividing the last equation by $ dP$ we get

$ \left(\frac {\partial H}{\partial P}\right)_T \equal{} T \left(\frac {\partial S}{\partial P}\right)_T \plus{} V$

But from $ dG \equal{} \minus{} SdT \plus{} VdP$ we can write

$ \minus{} \left(\frac {\partial S}{\partial P}\right)_T \equal{} \left(\frac {\partial V}{\partial T}\right)_P \equal{} > \left(\frac {\partial S}{\partial P}\right)_T \equal{} \minus{} \alpha V$, since by definition $ \alpha \equal{} \frac {1}{V} \left(\frac {\partial V}{\partial T}\right)_P$.

Therefore,

$ \left(\frac {\partial T}{\partial P}\right)_H \equal{} \minus{} \frac { \minus{} \alpha VT \plus{} V}{C_P} \equal{} > \left(\frac {\partial T}{\partial P}\right)_H \equal{} \frac {V(\alpha T \minus{} 1)}{C_P}$.

In the throttling experiment we always have $ dP < 0$. Also, the stability of thermodynamic systems requires that $ C_P > 0$. Therefore, from the last expression it is easy to see that $ \alpha T > 1$ (where T is the initial temperature of the gas) implies $ dT < 0$, and the throttling process cools the gas. If $ \alpha T < 1$ then $ dT > 0$ and the gas is heated. For $ \alpha T \equal{} 1$, the partial derivative vanishes, and an infinitesimal pressure change neither heats nor cools the gas. Since the condition $ \alpha T \equal{} 1$ defines the turn point from an heating process to a cooling process, we therefore define the [i]inversion temperature[/i] $ T_{inv}$ of a gas by

$ T_{inv} \equal{} \frac {1}{\alpha}$

Note that the inversion temperature of a gas [i]is not[/i] a constant, since the coefficient of thermal expansion $ \alpha$ is an intensive state function and is, therefore, a function of temperature and pressure.
Experimentally, it is found that the T versus P curve for a series of throttling experiments with the same initial state (defined by its T and P) in common has the form shown in figure II: note that the extremum corresponding to the inversion condition $ \mu _{JT} \equal{} 0$ is a maximum. As we have shown above, to the right of point [i]a[/i] $ \mu _{JT} < 0$ and a gas is heated in that region; to left of [i]a[/i], $ \mu _{JT} > 0$ and a gas is cooled. Note also that the curve shown connects points with the same enthalpy, and is therefore an [i]isenthalpic[/i] curve. From the considerations given until now, one may conclude that a gas must be first pre-cooled below its inversion temperature (therefore assuring that $ \mu _{JT} > 0$) in order for it to be cooled through a Joule-Thomson experiment. Otherwise, heating occur instead.
For most gases at atmospheric pressure the inversion temperature is quite higher than room temperature - for example, 621 K (348ºC) for $ \ce{N2}$ and 764 K (491ºC) for $ \ce{O2}$ -, and so these gases can be cooled at room temperature by isenthalpic expansion. On the other hand, at atmospheric pressure the gases $ \ce{He}$ and $ \ce{H2}$ have very low inversion temperatures (-222ºC for Helium), and so these two gases will warm by a throttling process at room temperature.

[size=150][b]Problem 43[/b][/size]. In the text above it was shown that the Joule-Thomson process occurs at constant enthalpy. However, this statement is not completely correct. Why? What better statement can we make?

[size=150][b]Problem 44[/b][/size]. Is $ \mu _{JT}$ a state function? If yes, then is it extensive or intensive?

[size=150][b]Problem 45[/b][/size]. Compute $ \mu _{JT}$ for an ideal gas, and comment your result.

[size=150][b]Problem 46[/b][/size]. For air at temperatures near 25ºC and pressures in the range 0 to 50 bar, the $ \mu _{JT}$ values are all reasonably close to 0.2ºC/bar. Estimate the final temperature of the gas if 58 g of air at 25ºC and 50 bar undergoes a Joule-Thomson throttling to a final pressure of 1 bar.

[size=150][b]Problem 47[/b][/size]. Find an expression for the inversion temperature of real gases, assuming them to be described by the Van der Waals equation of state.

[size=150][b]Problem 48[/b][/size]. A gas has the following equations of state, $ U \equal{} PV$ and $ T \equal{} 3B \sqrt[3]{\frac{U^2}{NV}}$, where B is a positive constant, and the system obeys the Nernst postulate. The gas, at an initial temperature $ T_1$ and initial pressure $ P_1$, goes through a Joule-Thomson experiment, with the final pressure being $ P_2$. Find the final temperature $ T_2$ of the gas.

[size=150][b]Problem 49[/b][/size]. For a particular gaseous system it has been determined that the energy is given by $ U \equal{} \frac{5}{2}PV \plus{} B$, where B is a constant. Find the equation of the adiabats in the P-V plane for this gas.

[size=150][b]Problem 50[/b][/size]. The energy of a particular system, of one mole, is given by $ U \equal{} AP^2V$, where A is a positive constant. Find the equation of the adiabats in the P-V plane.

[size=150][b]Problem 51[/b][/size]. Let [i]n[/i] moles  of a perfect gas undergo an adiabatic free expansion into a vacuum (the Joule experiment). Express $ \Delta S$ in terms of the initial and final temperatures and volumes.

[size=150][b]Problem 52[/b][/size]. Calculate $ \Delta S$ for the conversion of 1 mole of supercooled liquid water at -10ºC and 1 atm to 1 mole of ice at -10ºC and 1 atm.

[size=150][b]Problem 53[/b][/size]. The specific heat capacity $ c_P$ of water is nearly constant in the temperature range 25ºC to 75ºC at 1 atm, having the value of 1.00 cal/g/ºC.

(a) Find $ \Delta S$ when 100 g of water are reversibly heated from 25ºC to 50ºC at 1 atm.

(b) Without doing the calculation, state whether $ \Delta S$ for heating 100 g of water from 50ºC to 75ºC at 1 atm will be greater than, equal to, or less than $ \Delta S$ for the previous 25ºC to 50ºC heating. Justify your answer.

(c) Find $ \Delta S$ when 100 g of water are reversibly heated from 50ºC to 75ºC at 1 atm. Does your result confirm the answer given to part (b)?


Yes I'm an idiot...I would know this stuff, but I haven't been able to attend class lately.

Help and/or answers would be GREEEEEEEEEEATLY appreciated w/ factoring the following problems:

1.)   X^2-X-12 (X^2 means squared just in case nobody knew that)

2.) X^2-4X-12

3.) X^2-X-6

4.) X^2-5X+6

5.) X^2-7X+12

Again Thanks a lot if you can help me.

1:
$x^2-x-12=(x-4)(x+3)$

2:
$x^2-4x-12=(x-6)(x+2)$

3:
$x^2-x-6=(x-3)(x+2)$

4:
$x^2-5x+6=(x-2)(x-3)$

5:
$x^2-7x+12=(x-3)(x-4)$

6:
This is not suitable for Algebra Unsolved Problems

7:
This thread should be posted at the Getting Started forum.

My bad...Thanks for your help though.

I'm gonna go post it over there but is there any way you can show me how you arrived at your conclusions? Thanks again.

or u can do it with the hard way and efficient way  :?
 :ninja: Discriminant :ninja:


[b][u]Question #1[/u][/b]
How can i make an expression appear in the middle of the screen ?
Also is there a way to make it appear anywhere we want ??

$\int f(x) dx$

[b][u]Question #2[/u][/b]
Why there is a space in the first line of the post below and is there is a way to avoid it ???

$(a+b)^2 =(a+b)(a+b) = \\ a(a+b)+b(a+b)=a^2+ab+ba+b^2 =\\ a^2+2ab+b^2$ .

[b][u]Question #3[/u][/b]

For the expression above : Why the [b]dot[/b] at the end does not appear correctly ?
I dont use correctly this  ->  \\  ?

[b][u]Question #4[/u][/b]

When i write a lengthy expression i have to change lines because it appears
something like this...

$\frac{a}{b+2c+d}+\frac{b}{c+2d+a}+\frac{c}{d+2a+b}+\frac{d}{a+2b+c}= \frac{a}{S-(a-c)}+\frac{b}{S-(b-d)}+\frac{c}{S+(a-c)}+\frac{d}{S+(b-d)}$

That is for now  :D  . Thanks  a lot   :)

Q1: [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#spacing]Try here[/url] You type \begin{center} [i] text [/i] \end{center}
Q2: [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php#eqnarray]This is really useful too[/url] Use the \begin{eqnarray*} tool


function
limit
integration
logarithms
calculus
inequalities
real analysis

Given that $f:[0,1]\rightarrow\mathbb{R}$ is a continuous function, compute the following limit:

\[ \lim_{n\rightarrow\infty} \int_0^1 \frac{x^{\frac{1}{n}}-1}{\frac{1}{n}} \cdot f(x)\ dx \]

The pointwise limit of $\frac{x^{\frac1n}-1}{\frac1n} = \ln x$. Furthermore (as can be shown by using Taylor's Theorem on $x^{\frac1n}=e^{\frac{\ln x}n}$), each term of this sequence is bounded by $|\ln x|$. Since $f$ is continuous on [0, 1], there is some $M$ such that $|f(x)|\le M$, so the integrands are dominated by the integrable function $M|\ln x|$. By the Dominated Convergence Theorem, the limit of the integral is the integral of the limit, namely $\int_0^1 f(x)\ln x dx$. Of course, this could have been done by more elementary means without quoting the DCT, but the result is the same.

I'm sorry, can you post a more rigorous and elementary proof? This was intented for students aged 18-19 who do not posess superior knowledge in the integration theory area. Thank you very much!

OK, I'll give it a try. Interchanging limits of sequences of functions with integrals isn't normally dealt with in the first calculus sequence.

The key inequality is that for $y\le 0, 1+y\le e^y\le 1+y+\frac{y^2}2$.
So, $1+\frac{\ln x}n\le e^{\ln x/n}\le 1 + \frac{\ln x}n+\frac{(\ln x)^2}{2n^2}$
$\frac{\ln x}n\le x^{1/n}-1\le \frac{\ln x}n+\frac{(\ln x)^2}{2n^2}$
$\ln x\le \frac{x^{1/n}-1}{1/n}\le \ln x + \frac{(\ln x)^2}{2n}$
$0\le \frac{x^{1/n}-1}{1/n}-\ln x\le \frac{(\ln x)^2}{2n}$
$\left|\frac{x^{1/n}-1}{1/n}-\ln x\right|\le\frac{(\ln x)^2}{2n}$
Since $f$ is continuous, there exists $M$ such that $|f(x)|\le M$.
Thus, $\left|\int_0^1\frac{x^{1/n}-1}{1/n}\cdot f(x)dx - \int_0^1\ln x\cdot f(x)dx\right|\le\int_0^1\left|\frac{x^{1/n}-1}{1/n}-\ln x\right||f(x|dx$
$\le \int_0^1 \frac{M(\ln x)^2}{2n}dx = \frac{M}n \to 0$ as $n\to\infty$.

[quote="Kent Merryfield"]A somewhat related question, but with a different answer. Again, assume that $f$ is continuous on [0, 1]. Compute
$\lim_{n\to\infty} n\int_0^1 f(x)(1-x)^{n-1}dx$.[/quote]

As for this one, I'll assume $f(x)\ge 0$, because otherwise we should employ some boring $|\cdot|$ :). Moreover, we can assume $f(x)=0$, because, since $\int^1_0n(1-x)^{n-1}dx=1$, we can replace $f$ with $g(x)=f(x)-f(0)$. Now we write the integral as $\int^{a_k}_0+\int^1_{a_k}$. The second one is $\le M(1-a_k)^n$, while the first one is $\le M_k(1-(1-a_k)^n)$, where $M,M_k$ are bounds for $f$ on [$0,1$] and [$0,a_k$] respectively. When $n\to \infty$, the second integral goes to $0$, while the first integral becomes $\le M_k+\varepsilon$ with $\varepsilon$ as small as we want. If we make $a_k\to 0$, using $f(0)=0$, we get the result $n\int^1_0f(x)(1-x)^{n-1}\to 0$.

For general $f$ the limit is thus $f(0)$.

That's it, grobber, clean and simple enough. An old Fourier analyst like me can't resist a good "approximation to the identity".


function
calculus
derivative
limit
real analysis
real analysis unsolved

Suppose that $f(x)=\sum_{k\geq 1}{a_k e^{ikx}}$ is a continuous function (attention, the convergence is only simple). Then the $a_k$ are its Fourier coefficients. I have a very beautiful proof for this one, but I'd like to see many approaches.

"simple" meaning "pointwise", i.e., the assumption being that the series converges to $f(x)$ for every $x$? :?

Yes, sorry, I was too influenced by the french vocabulary.  :blush:

please can you give a hint to your proof ...it's very strange , my teacher had never talk about this result even if it's very interesting

the trick is to introduce $F(x) = \sum_{k \geq 1}\frac{a_{k}}{k^{2}}e^{ikx}$, hopping that it will looks like a function such that $F''=-f$. This time everything converges normally because the $(a_{k})$ are bounded, so one knows the fourier coefficient of $F$, and when one computes $DF(x)$, the pseudo second derivative, one can see that $DF(x)=-f(x)$, where $Dg(x) = \lim_{h \to 0}\frac{g(x+h)+g(x-h)-2g(x)}{h^{2}}$. So if $G \in C^{2}(\mathbb{R})$ satisfies $G''(x)=f(x)$, one can see that $D(F+G)=0$, which means that $F+G$ is an affine function. From here, the conclusion follows.


Gauss
limit
trigonometry
floor function
real analysis
real analysis solved

Congratulations on Mathlink's recovery! :)  :)

I am very glad to see this forum again.Thank you for your taking troubles to whom it may concern.

Evaluate \[\lim_{n\to\infty}\sin \left( 2\pi\sqrt{n^2+\left[\frac{n}{3}\right]}\right )\]

$\lfloor\frac{n}3\rfloor=\frac{n}3+O(1)$

$1+\frac1{n^2}\lfloor\frac{n}3\rfloor= 1+\frac1{3n}+O\left(\frac1{n^2}\right)$

$\sqrt{n^2+\lfloor\frac{n}3\rfloor}= n\sqrt{1+\frac1{n^2}\lfloor\frac{n}3\rfloor}= n\left(1+\frac1{6n}+O\left(\frac1{n^2}\right)\right)$

$2\pi\sqrt{n^2+\lfloor\frac{n}3\rfloor}= 2\pi n+\frac{2\pi}6+O\left(\frac1n\right)$

So the limit is $\sin\left(\frac{\pi}3\right)=\frac{\sqrt{3}}2.$

Newton's binomial series is a powerful thing.

[quote="kunny"]Congratulations on Mathlink's recovery! :)  :)

I am very glad to see this forum.Thank you for your taking troubles to whom it may concern.
[/quote] Altough it's a little bit off-topic here, you're welcome.  ;)

@kunny. For whom this problem was proposed? I mean what level of students was being assumed?

Here is my solution.

$\frac{n}{3}-1<[\frac{n}{3}]\leqq\frac{n}{3}$ ,so we have $\frac{1}{3}-\frac{1}{n}<\frac{1}{n}[\frac{n}{3}]\leqq\frac{1}{3}$ by Archimedes'axiom we have $\lim_{n\to\infty}\frac{1}{n}[\frac{n}{3}]=\frac{1}{3}$.
 Analogously $\lim_{n\to\infty}\frac{1}{n^2}[\frac{n}{3}]=0$

Then $\lim_{n\to\infty}(\sqrt{n^2+[\frac{n}{3}]}-n)=\lim_{n\to\infty}\frac{[\frac{n}{3}]}{\sqrt{n^2+[\frac{n}{3}}]+n}=\lim_{n\to\infty}\frac{\frac{1}{n}[\frac{n}{3}]}{\sqrt{1+\frac{1}{n^2}[\frac{n}{3}]}+1}=\frac{1}{6}$

Thus $\lim_{n\to\infty}\sin (2\pi\sqrt{n^2+[\frac{n}{3}]})=\lim_{n\to\infty}\sin (2\pi(\sqrt{n^2+[\frac{n}{3}]}-n))$.

By The continuity of $\sin x$, the desired limit is $\sin \frac{2\pi}{6}=\frac{\sqrt{3}}{2}$


The convex hexagon $ABCDEF$ satisfies $\angle A + \angle C + \angle E = 360^{\cdot}$ and $AB \cdot CD \cdot EF = BC \cdot DE \cdot FA$. Show that $AB \cdot FD \cdot EC = BF \cdot DE \cdot CA$.

At least posted before:

http://www.mathlinks.ro/Forum/viewtopic.php?t=48100
http://www.mathlinks.ro/Forum/viewtopic.php?t=1117
http://www.mathlinks.ro/Forum/viewtopic.php?t=19479

  darij


graph theory
combinatorics unsolved
combinatorics

For what positive integers $ n$ is it possible to divide the complete graph on $ n$ points into triangles so that each edge is used exactly once?


Let $ a,b,c\geq\ 0$ such that $ a \plus{} b \plus{} c \equal{} 3$.Find $ max$
[size=150]$ D \equal{} (a^\frac {7}{5} b \plus{} a^\frac {3}{2} b \plus{} 1)( b^\frac {7}{5} c \plus{} b^\frac {3}{2} c \plus{} 1)(c^\frac {7}{5} a \plus{} c^\frac {3}{2} a \plus{} 1)$[/size]
It is my sorry to can_hang2007

The maximum is $ 27$, use the following
\[ a^{3/2}b \plus{}b^{3/2}c\plus{}c^{3/2}a \le 3\]
:)

[quote="Allnames"][quote="can_hang2007"]The maximum is $ 27$, use the following
\[ a^{3/2}b \plus{} b^{3/2}c \plus{} c^{3/2}a \leq 3
\]
:)[/quote]
it is not enough,how about $ a^{7/5}b \plus{} b^{7/5}c \plus{} c^{7/5}a$,can you detail it,i think it is not easy[/quote]
We also have the inequality
\[ a^{7/5}b\plus{}b^{7/5}c\plus{}c^{7/5}a \le 3\]
and it follows from the inequality
\[ a^{3/2}b\plus{}b^{3/2}c\plus{}c^{3/2}a \le 3\]
;)


category theory
abstract algebra
invariant
group theory
superior algebra
superior algebra unsolved

Hello

Consider R-modules

let  F be a covariant functor to the abelian group category such that each short exact sequence

$0\rightarrow A\buildrel{f}\over{ \rightarrow}B\buildrel{g}\over{\rightarrow}C\rightarrow 0$

is transformed into a new short exact sequence

$0\rightarrow FA\buildrel{F(f)}\over{\rightarrow}FB\buildrel{F(g)}\over{\rightarrow}FC\rightarrow 0$

That is our definition of exact functor.

Prove now that each long exact sequence is also transformed into a long exact sequence.
It's not as easy as it looks!  I got stuck... :(

It's obvious after showing the following important equivalence:

$F$ is exact  $\Longleftrightarrow$  for all exact sequences $\bullet \buildrel{f}\over{\rightarrow} \bullet \buildrel{g}\over{\rightarrow} \bullet$ the sequence  $\bullet \buildrel{F(f)}\over{\rightarrow} \bullet\buildrel{F(g)}\over{\rightarrow} \bullet$ is also exact.

where $F$ is an additive functors between abelian categories.  The important stuff are the arrows, so I don't denote the objects.

$\Longleftarrow$ :

Let $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{f}\over{ \rightarrow} \bullet \buildrel{g}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$ be a short exact sequence. Then $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{f}\over{ \rightarrow} \bullet$ is exact, hence $F(0) \buildrel{0} \over{ \rightarrow} \bullet \buildrel{F(f)}\over{ \rightarrow} \bullet$ by assumption. It follows $0 = im(0) = ker F(f)$, so that $F(f)$ is a monomorphism. Dually, we see that $F(g)$ is an epimorphism. So, $0 \buildrel{0} \over{ \rightarrow} \bullet \buildrel{F(f)}\over{ \rightarrow} \bullet$ and $\bullet \buildrel{F(g)}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$ are exact. Besides, the exactness of $\bullet \buildrel{f}\over{ \rightarrow} \bullet \buildrel{g}\over{ \rightarrow} \bullet$ implies the exactness of $\bullet \buildrel{F(f)}\over{ \rightarrow} \bullet \buildrel{F(g)}\over{ \rightarrow} \bullet$. Composing gives, that $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{F(f)}\over{ \rightarrow} \bullet \buildrel{F(g)}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$ is exact.

$\Longrightarrow$:

Let $f$ be an arbitrary epimorphism. Then $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{ker(f)}\over{ \rightarrow} \bullet \buildrel{f}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$ is exact, hence $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{F(ker f)}\over{ \rightarrow} \bullet \buildrel{F(f)}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$. Therefore, $F(f)$ is an epimorphism. Thus $F$ preserves epimorphisms, and dually we derive that monomorphisms are preserved.

Not, let $f$ be an arbitrary morphism. It can be uniquely (up to isomorphism) factorized by an epimorphism $coim(f)$ and a monomorphism $im(f)$. Then $F(f) = F(im(f) \circ coim(f))=F(im(f)) \circ F(coim(f))$. We've seen that $F(im(f))$ is a monomorphism and that $F(coim(f))$ is an epimorphism. So, $F(im(f))=im(F(f))$ and $F(coim(f))=coim(F(f))$. Besides, the exactness of $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{ker(f)}\over{ \rightarrow} \bullet \buildrel{coim(f)}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$ yields the exactness of $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{F(ker(f))}\over{ \rightarrow} \bullet \buildrel{F(coim(f))}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$. Thus, $F(Ker(f))=ker(F(coim(f))=ker(coim(F(f))=ker(F(f))$. Dually it follows $F(coker(f))=coker(F(f))$.

Now, we have established enough invariants to proof the assertion:

Let $\bullet \buildrel{f}\over{ \rightarrow} \bullet \buildrel{g}\over{ \rightarrow} \bullet$ be exact. This means, that $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{ker(g)}\over{ \rightarrow} \bullet \buildrel{coker(f)}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$  is exact, hence also $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{F(ker(g))}\over{ \rightarrow} \bullet \buildrel{F(coker(f))}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$ by assumption. This sequence is isomorphic to $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{ker(F(g))}\over{ \rightarrow} \bullet \buildrel{coker(F(f))}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$. So it is also exact, hence also $\bullet \buildrel{F(f)}\over{ \rightarrow} \bullet \buildrel{F(g)}\over{ \rightarrow} \bullet$.

Danke, du bist wirklich sehr sehr gut! :D
Aber :

I have some questions.  First of all, I 'should' be able to deduce that the decomposition in an abelian category of a random morphism into an epic and then a monic morphism is 'essentially unique', right?  I had a go at it, and I needed to consider a lot of things, is it possible you explicitly need for instance the assumption that every monic morphism is a kernel of its own cokernel (as far as I know that is an explicit axiom of abelian category)

You said  : this sequence is isomorphic to, what exactly do you mean by isomorphic sequence?

Then finally, what I initially asked for was : prove that  a covariant exact functor from $R-mod$ to $Ab$ maps lojng exact sequence to long exact sequence.
Now I think I could at least change $Ab$ into $S-mod$

You however, worked with abelian categories (which is nicer as it is more general) but you assumed additivity of your functor F.  That is 'bad' for me as it in conflict with my assumptions in the original problem, but .. where do you use it in your proof?

[quote="fredbel6"]Danke, du bist wirklich sehr sehr gut! :D [/quote]
Und du kannst sehr sehr gut Deutsch, lass uns doch so weiterschreiben? ;-)

[quote="fredbel6"]is it possible you explicitly need for instance the assumption that every monic morphism is a kernel of its own cokernel [/quote]
Yes - this is quite important.

[quote="fredbel6"]You said  : this sequence is isomorphic to, what exactly do you mean by isomorphic sequence?[/quote]
Sequences of the same "type" are isomorphic, when there are isomorphisms between the occuring objects (of the same position) so that the resulting diagram commutes. This should remind you of the category of chain complexes ;)

[quote="fredbel6"]Now I think I could at least change $Ab$ into $S-mod$[/quote]
Yes, of course. As I already said, the whole think works for arbitrary abelian categories.

[quote="fredbel6"]... you assumed additivity of your functor F ... where do you use it in your proof?[/quote]
Good question. :-) At the beginning, I use that the null morphism is mapped to the null morphism. I think that's it.

Nein, es tut mir leid aber ich glaube nicht das wir können weiterschreiben in Deutsch  :oops: .  Even if I would call it a language that I speak, it would be very low on the list, number four actually, and I cannot say difficult sentences in German, like this one!

But I must admit that is wrong  ;)  the functor had to be additive and covariant and from $R-mod$  to $Ab$
You used only abelian categories and additive covariant functor
So I guess nothing is lost!

Actually, I am quite ashamed that I only think of it now, but I have been working in$R-mod$ for so long, how exactly do you define exact sequence in a general abelian category.

My guess would be :

$\bullet \buildrel{f}\over{ \rightarrow} \bullet \buildrel{g}\over{ \rightarrow} \bullet$ is said to be exact if :

1.  $g\circ f=0$

2.  $coker(f)\circ ker(g)=0$

Or am I leaving out a crucial assumption?

The confusing this is that you use exactness of the last sequence I wrote to prove exactness of $\bullet \buildrel{f}\over{ \rightarrow} \bullet \buildrel{g}\over{ \rightarrow} \bullet$, while I use that sequence to define exactness! :ninja:

The definition of exactness is correct. But I don't understand your confusion ...

This is all so nice and interesting but stuff like that should be included in my syllabus in my opinion then  :|

[quote]Let $\bullet \buildrel{f}\over{ \rightarrow} \bullet \buildrel{g}\over{ \rightarrow} \bullet$ be exact. This means, that $0 \buildrel{0}\over{ \rightarrow} \bullet \buildrel{ker(g)}\over{ \rightarrow} \bullet \buildrel{coker(f)}\over{ \rightarrow} \bullet \buildrel{0}\over{ \rightarrow} 0$  is exact[/quote]

So how would you go about proving this then?
This is confusing at it involves kernels of cokernels

The definition of exactness can be replaced by im f = ker g, and he have Im ker g = ker coker ker g = ker g, ker coker f = im f ....


For getting into a top tier college (Stanford, Harvard, MIT, Princeton...), how important is it to be well-rounded. I'm a high school sophomore right now, and I swim, but that takes up a lot of time every day and my junior year is going to be packed. Is it worth it to keep swimming through high school? Or should I devoted more time to studying?

I'd say you shouldn't base your decision about swimming just on what would look good on your college applications. Do you like swimming? If not, do something else that you like more. That will probably make you look well-rounded still, and you'll also have more fun. If you do like swimming, then keep doing it! It's not hurting you.

In high school I was under a lot of pressure to do a sport, because if I didn't, I wouldn't be "well-rounded" and no top-tier college would want me.  I resisted, and ultimately proved the prevailing wisdom wrong.  So, I would take any advice along the lines of "you must do X in order to be a well-rounded person and get into a good school" with a grain of salt.

If you truly enjoy swimming, by all means, do it!  If you don't, find something else you like.  Don't worry about what other people think.  Feeling empowered to do activities that you enjoy, and that are meaningful to you-- to [b]you[/b], not to parents, teachers, or colleges-- is absolutely priceless, and will serve you well for the rest of your life.


limit
real analysis
real analysis unsolved

I know it's well-known but I couldn't find a proof:

Let $ (a_n)_{n\ge 1}$ be a sequence of real numbers with $ \sum \frac{a_n}{n}$ convergent. Prove that $ \lim_{n\to \infty}\frac{a_1\plus{}a_2\plus{}\dots \plus{}a_n}{n}\equal{} 0$

If $ b_n \geq 0$ and $ \sum b_n$ converges, then:

1) $ \lim b_n \equal{} 0$;

2) $ \lim n b_n \equal{} 0$.

In your case, we can conclude that $ \lim n \cdot \frac {a_n}{n} \equal{} \lim a_n \equal{} 0$. Now we need a theorem about sequences:

(3) If $ \lim a_n$ exists then $ \lim \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n}$ also exists and we have

$ \lim \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} \equal{} \lim a_n$.

Therefore, regarding your problem, we can then conclude that $ \lim \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} \equal{} 0$. I will leave the proofs of (1), (2), and (3) as exercises for you.

Don't underestimate this problem.

[quote="Carcul"]

2) $ \lim n b_n \equal{} 0$.

In your case, we can conclude that $ \lim n \cdot \frac {a_n}{n} \equal{} \lim a_n \equal{} 0$.
[/quote]

Not true. Consider $ a_n\equal{}(\minus{}1)^n$ so $ b_n\equal{}\frac{(\minus{}1)^n}{n}$ and the series $ \sum b_n$ converges conditionally. So it can not be solved with a simple application of the Cezaro-Stolz lemma.

You are right: I was considering sequences of non-negative real numbers in my post.

Let $ b_n \equal{} a_n / n$ so that $ a_n \equal{} n b_n$.
Then $ \sum b_n$ converges.
Let $ f(n)\equal{}\frac{b_1\plus{}2b_2\plus{}\ldots\plus{}nb_n}{n}$
We want to show that $ \lim_{n\to\infty}f(n)\equal{}0$
Let $ s_n\equal{}b_1\plus{}\ldots\plus{}b_n$.
Then $ f(n)\equal{}s_n\minus{}\frac{s_1\plus{}s_2\plus{}\ldots\plus{}s_{n\minus{}1}}{n}$ (just consider the number of times each $ b_i$ appears).

But $ \lim_{n\to\infty}s_n$ does exist (since the sum converges), so we can use Stolz-Cesaro on $ \frac{s_1\plus{}s_2\plus{}\ldots\plus{}s_{n\minus{}1}}{n}$, and the desired conclusion follows.

I think this works.


algebra
polynomial
number theory unsolved
number theory

$P(x)\in \mathbb{Z}[x],P(x)=a_{0}+\dots+a_{p-1}x^{p-1}$ and if we know that for prime $p>2$ following holds: if $p$ does not divide $a-b$ then $p$ does not divide $P(a)-P(b)$.prove that $p|a_{p-1}$

In $\mathbb F_{p}$ (the field with $p$ elements), we have relations of the form $P(x_{i})=y_{i},\ i=\overline{0,p-1}$, where both $\{x_{i}\}$ and $\{y_{i}\}$ are permutations of $\overline{0,p-1}$.

Since our polynomial has degree $\le p-1$, we can find it in terms of $a_{i}$ and $b_{i}$ through Lagrange interpolation, and writing down the expression for the interpolation polynomial shows the result (the main coefficient $a_{p-1}$ will turn out to be $0+1+\ldots+p-1=0$).


[size=167][b][color=red]Opera[/color] for Windows [ Win 95/98/ME/NT/2000/XP ][/b][/size]
[b]Surf the Internet in a safer, faster, and easier way[/b]

Visit [url=http://www.opera.com]Opera web site[/url].

The most full-featured Internet power tool on the market, Opera includes pop-up blocking, tabbed browsing, integrated searches, and advanced functions like Opera's groundbreaking E-mail program, RSS Newsfeeds and IRC chat. And because we know that our users have different needs, you can customize the look and content of your Opera browser with a few clicks of the mouse.

[size=125]Key features in Opera 8[/size]

:arrow: [b]Fresh new look[/b]
We've cleaned up our front yard. The Opera 8 interface is designed to make the advanced functions easy and effective to use. Menus, toolbars and other elements have undergone our "slim and clean"-routine. The licensed version has the largest browsing area in the industry.

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Stay away from spy-ware, viruses, and other malicious applications that silently attack your computer while you are surfing the Web. For example, Opera displays security information inside the address bar, located next to the padlock icon that indicates the level of security present on a site. Opera 8 also provides protection against phishing attacks and automatically checks for security updates.

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For further information on just how good our rewritten scripting engine is, go to [url=http://www.howtocreate.co.uk/browserSpeed.html](Browser speed comparison)[/url]

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Opera pioneered tabs and pop-up blocking ages ago. But they're still key features that more and more people start enjoying. With tabs you can open multiple pages within the same application window. Opera lets you control whether to block all pop-ups, or open only the ones that you have requested.

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Search your favorite sites, for example Google, eBay, or Amazon without having to go to their Web pages. Use the integrated search window or search directly in the address field using shortcuts (e.g. writing "g Opera" in the address field will search for "Opera" using Google).

:arrow: [b]Customization[/b]
Using the 'appearance' dialog you can make Opera look almost any way you want. Move buttons and search fields, add and remove toolbars, and so forth. Opera skins can change the look of your browser by giving it the icons and buttons of your choice.

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Opera's password manager, the Wand, remembers your usernames and passwords so you will not have to. The password manager can store more than one username and password combination for each page, and will let you choose which one to use if multiple combinations are stored.

:arrow: [b]Fit to window width[/b]
Opera's "fit to window width"-feature means no more horizontal scrolling and improved ability to print entire Web pages. In combination with Opera's Zoom function, you can magnify Web pages dramatically and still view them without having to scroll sideways.

:arrow: [b]Voice[/b]
Opera is the first browser to prepare for a future of Web sites offering interactive, voice-enabled shopping and booking systems. You can browse the Web using spoken commands, such as "Opera next link", "Opera back", or "Opera speak". The latter command will make Opera read Web page content and e-mail messages to you aloud ,adding usability as either a screen reader or advanced dictionary. Voice is currently offered in English and works on Windows 2000 and XP. A headset with a microphone is required to use Voice.

:arrow: [b]Additional features[/b]
 :!: Full support for Gmail
Support for Atom newsfeeds
First Web browser to natively support Scalable Vector Graphics (SVG)

I have used Opera for ages and I recommend it as the best ever tool for intenet surfing.

This is interesting, the site gives me permanent refused connection :what?:

Looks a lot nicer, and seems to be a bit faster. However I still don't like the ads too much :(

Don't listen to him!

Opera Software has special [b]FREE OF CHARGE[/b] offers for educational institutions.

[quote="fanaticsm"]Is there any way to configure the fonts in Opera? The posts on AoPS are not displaying properly ;)[/quote]As in any browser there must be ways to configure font sizes. Also there is a small switch in the upper right corner on the index page (if you are using the AoPS skin) which you might try tweaking at the same time with the browser font settings.

Mikhail, but how to obtain the add-free version?!  :?

[quote="fanaticsm"]The ad-free version can be obtained by purchasing the browser. More information can be found at [url=https://secure.bmtmicro.com/servlets/Orders.ShoppingCart?CID=310&PRODUCTID=3100024&AID=812930]this[/url] website.
I believe the cost is 39 USD..[/quote]Please, don't give a partial information.

Student license costs 20 USD.

Also visit [b]Opera Higher Education Program[/b] at http://www.opera.com/education/

[quote]Opera offers free site licenses to higher education institutions.[/quote]
[quote="stupidkid"]...but you can always use Mozilla Firefox, which is free and also very good.[/quote] This opinion comes only from lack of knowledges. Opera's features cover everything what you have with IE or Firefox, moreover it gives many unique great things, which help with internet surfing.
Opera contains built-in mail client M2, but I don't use it (here I think that Thunderbird is very good).

[quote="Valentin Vornicu"][quote="stupidkid"][quote="Myself"]Is there possible a way to obtain the ad-free version free and [i]legally[/i]?[/quote]
I doubt that you can, but you can always use Mozilla Firefox, which is free and also very good. :D[/quote]Not very good when it comes to java unfortunately :([/quote]
Really? I don't know, I use it all the time, both in Linux and Windows and usually Java is not a problem. Sometimes Flash is.

My Firefox (1.0.3) crashes about 2-3 times a day, especially when loading pages with JavaScript. Opera, IE and Netscape don't have such problems.

If only IE had Tabs  :(

Well, I don't know, there are things that I don't like about the Explorer, but mainly is because I like Linux and open source stuff. And yes, I love the tabs in FireFox. If my wife wouldn't hate Linux, it would be the only operative system in my computer (unfortunatelly she does :( ).

As I told, it is interesting, I have never had FireFox crashing down with Java Applets or JavaScript. But it never shows the Flash stuff, even when I have tried to configure the pluggin. My version is 1.0, released on November 7, 2004!

I know I have ages without using Opera (probably at least three or four years) but when I used long time ago, it was not that good as I remember. I suppose that they have corrected almost all the thousands of bugs that it had. Nevertheless, I prefer FireFox

It just crashed 1 minute ago when I was trying to post a reply. This was the 21st crash in the past week! :mad:

BTW, this topic is devoted to the best browser on the Earth! So I don't understand your discussion here. Indeed, Firefox and IE are not so good.
Sorry, but I even can't understand who seriously can use Firefox and IE for work. :?

[size=125][color=red][b]OPERA[/b][/color] forever! [/size]:first:

They're giving out free registration codes today only! Now's a great time to try it if you haven't already.
Link:
[url=http://my.opera.com/community/party/reg.dml]my.opera.com/community/party/reg.dml[/url]

Disclaimer: I'm still using Firefox because of certain invaluable extensions.  :)

I use FF too. David, you've GOT to get a newer version. (Flash works fine for me on Deer Park Alpha 1/2, but Java takes forever (if not a crash too))

And yes, although I don't use it much, I've gotten my free registration, and also :) posted it in F&G.

[quote="Myth"]
[quote="stupidkid"]...but you can always use Mozilla Firefox, which is free and also very good.[/quote] This opinion comes only from lack of knowledges. Opera's features cover everything what you have with IE or Firefox, moreover it gives many unique great things, which help with internet surfing.[/quote]
Firefox still has some features unavailable in Opera, most importantly extensions. (Does Opera support extensions? I don't think so...)

[quote="solafidefarms"]Does it support Greasemonkey?! Platypus?! Undo Close Tab for any tab up to one closed 25 tabs ago? Hm????[/quote]

Let me guess... No.


geometry
circumcircle
parallelogram
geometric transformation
homothety
ratio
geometry unsolved

Consider a triangle ABC and select a point M on the circumcircle of the given triangle. The feet of the perpendiculars from M to the sides of the triangle all lie on the same line, Simson's line (I can prove this). Prove that Simson's line bisects the segment MH, where H is the orthocenter of the triangle ABC. Also prove that the midpoint of the segment MH lie on the Euler's circle!

Please help me to prove this! Thanks

[hide="solution"]before proving let's have some terminologies
$ ABC$ is the triangle
$ P$ is a point on the circumcircle($ \Gamma$)
$ U$ is the other point of intersection of simson's line with $ \Gamma$
$ AH$ extended meets $ BC$ at $ D$ and $ \Gamma$ at $ D^{'}$
$ PD^{'}$ meets $ BC$ at $ Q$ .
$ HQ$ meets $ PU$ at $ V$
$ A_{1},B_{1},C_{1}$ are the feet of perpendiculars from $ P$ onto $ BC,AC,CB$,
[u][b][color=red]Lemma[/color][/b][/u]:
$ AU$ is parallel to $ A_{1}B_{1}$
[u][b]Proof of Lemma[/b][/u]
consider the cyclic quadrilaterals $ PAUC$ and $ PB_{1}A_{1}C$
angle $ PUA \equal{} PCA \equal{} PA_{1}B_{1}$

[u][b]proof[/b][/u]
since $ HD \equal{} DD^{'}$ we and that $ PV$ and $ AD$ are both parallel since they are perpendicular to the same line.
we have $ \Delta HQD^{'}$ and $ \Delta PQV$ are isosceles.
now angles $ D^{'}HV \equal{} PVH \equal{} D^{'}PU \equal{} DAU$
so $ HV$ is parallel to $ AU$ is  parallel to simson line of $ P$(by lemma)
also note that in $ \Delta PHV$ $ A_{1}B_{1}$ is parallel to $ HV$ and bisects $ PV$ hence it bisects $ PH$
thus we are done
[/hide]

[quote="pardesi"]
$ U$ is the other point of intersection of simson's line with $ \Gamma$
$ PD$ meets $ BC$ at $ Q$ .
[/quote]
I think you meant $ U$ is the other intersection of $ PA_{1}$ with $ \Gamma$, and $ PD'\cap BC \equal{}\{Q\}$.

A somewhat similar solution is the following:
Let $ AH$ meet the Simson line at $ A'$. Since $ A'H\|PA_{1}$, we have to prove $ PA'HA_{1}$ is a parallelogram, or $ PA' \equal{} A_{1}H$. Let $ D$ and $ D'$ be points of intersection of $ AH$ with $ BC$ and $ \Gamma$. Using similar angle chasing as in pardesi's solution, we obtain that $ PA'D'A_{1}$ is an isosceles trapezoid. Then $ PA' \equal{} A_{1}D' \equal{} A_{1}H$.

Now since a homothety of center $ H$ and ratio $ \frac{1}2$ sends $ \Gamma$ to Euler's circle $ \omega$, we see that the midpoint of $ MH$ lies on $ \omega$.


(a) Let $a,b>1$ be real numbers such that $a+b \leq 4$ find the minimum value of the
     expression:

   $F = \frac{a^4}{(b-1)^3} + \frac{b^4}{(a-1)^3}$

(b) Find the maximum value of the expression:

  $\frac{x+y}{1+z} + \frac{y+z}{1+x} + \frac{z+x}{1+y}$

   where x,y,z are real numbers in the interval $[\frac{1}{2} ; 1]$

[quote="Rust"]1. a=b=2,F=32,
2. x=y=z=1, max=3.[/quote]

how do you get that? :lol:


calculus
integration
logarithms
trigonometry
calculus computations

Evaluate $ \int_0^2 \frac{2x\plus{}1}{\sqrt{x^2\plus{}4}}dx$

Note that you aren't allowed to use the formula $ \int \frac{1}{\sqrt{x^2\plus{}a}}dx\equal{}...$

[hide="Solution"]Lets calculate $ I\equal{}\int\frac{2x\plus{}1}{\sqrt{x^2\plus{}4}}dx\equal{}\int\frac{2x}{\sqrt{x^2\plus{}4}}dx\plus{}\frac{1}{2}\int\frac{dx}{\sqrt{(\frac{x}{2})^2\plus{}1}}$

1st integral is ofc $ 2\sqrt{x^2\plus{}4}$, and now the 2nd:
$ \frac{1}{2}\int\frac{dx}{\sqrt{(\frac{x}{2})^2\plus{}1}}\equal{}\left\|\begin{array}{l} x\equal{}2\sinh t \\ dx\equal{}2\cosh t \end{array}\right\|\equal{}\int\frac{\cosh t}{\sqrt{\sinh^2t\plus{}1}}dt\equal{}\int\frac{\cosh t}{\cosh t}dt\equal{}\int dt\equal{}t\plus{}C\equal{}\mbox{arsinh}\frac{x}{2}\plus{}C\equal{}\ln\left(\frac{x}{2}\plus{}\sqrt{\frac{x^2}{4}\plus{}1}\right)\plus{}C$, hence:

$ \int_0^2\frac{2x\plus{}1}{\sqrt{x^2\plus{}4}}dx\equal{}\left[2\sqrt{x^2\plus{}4}\plus{}\ln\left(\frac{x}{2}\plus{}\sqrt{\frac{x^2}{4}\plus{}1}\right)\right]^2_0\equal{}4(\sqrt{2}\minus{}1)\plus{}\ln (1\plus{}\sqrt{2})$[/hide]

That's right.

To Japanese high school students.

Let $ x\equal{}2\tan \theta$, we have $ \int_0^2 \frac{1}{\sqrt{x^2\plus{}4}}dx\equal{}\int_0^{\frac{\pi}{4}} \frac{\cos \theta}{1\minus{}\sin ^ 2 \theta}d\theta$

$ \sin ht\equal{}\frac{e^t\minus{}e^{\minus{}t}}{2},\ \cos ht\equal{}\frac{e^t\plus{}e^{\minus{}t}}{2}$

For the last one you can also set $ u\equal{}x\plus{}\sqrt{x^2\plus{}4}.$


search
real analysis
real analysis unsolved

Riemann proved that for convergent series $ \sum x_{n}, x_{n}\in \mathbb{R}$, such that $ \sum |x_{n}|$ ,doesn't converges and for each $ \sigma \in \mathbb{\overline{R}}$ there exists bijection $ s: \mathbb{N}\rightarrow \mathbb{N}$ such that $ \sum_{n=1}^{\infty}x_{s(n)}=\sigma$.

Sorry didn't know it was posted already. Grobber's last post gives the solution to this problem, and it's more rigorous then my solution, so I wont post it.
Now try to solve the other problem that I gave  :P .


I 'm trying t insert a table in my page headers, but latex seems to put the header always wrong.

This is what i do to make the table:

\newpagestyle{teststyle}{
  \sethead{
{\footnotesize
\begin{tabular}{|p{2.6cm}|p{3.6cm}|p{3.6cm}|p{3.6cm}|}
\hline
\multicolumn{1}{|l|}{Kwaliteitshandboek} &
\multicolumn{3}{l|}{\thechapter.~\chaptertitle}\\
\multicolumn{1}{|l|}{} &
\multicolumn{3}{l|}{\hspace{3mm}\thesection ~\sectiontitle} \\
\hline
Opgesteld door:  & Goedgekeurd door: & Paraaf:  &
Datum Goedkeuring:\\
& & & 01-01-2005\\
Res. Avondrust &  \large Deswert Luk & . . . . . . . . . . . . . . .
. &
Datum toepassing:\\
& & & 01-01-2005\\
\hline
\end{tabular}
}
  }
          {}
          {}
}

The table is perfect, but it always apears on top of the text of my document. When i increase the headersize, the table just goos down as well...

These are my page settings:
\documentclass[a4paper,oneside,11pt]{book}

\usepackage[dvips]{graphicx}
\usepackage[dutch]{babel}
\usepackage{makeidx}
\usepackage{wrapfig}
\usepackage{url}
\usepackage{picins}
\usepackage[ansinew]{inputenc}
\usepackage[big, pagestyles]{titlesec}
\usepackage{eso-pic,graphicx}

\setlength{\topmargin}{1cm}
\setlength{\headheight}{10cm}

\setlength{\parskip}{1ex plus 0.5ex minus 0.2ex}
\setlength{\parindent}{0pt} \sloppy \setlength{\textwidth}{15cm}
\setlength{\oddsidemargin}{0.5in}
\addtolength{\textheight}{54,4pt}
\renewcommand{\baselinestretch}{1.2}

Please help me out here :)
tnx a lot!

Got it!
Just need to put [b] next to the table definition... :) GREAT


Triangle ABC is acute. P is satisfying $ \angle APB\equal{}\angle BPC\equal{}\angle CPA$. Show that three Euler lines of triangles $ APB,BPC,CPA$are concurrent.

Who can solve it?does anyone have some idea?

[quote="mr.danh"]Triangle ABC is acute. P is satisfying $ \angle APB \equal{} \angle BPC \equal{} \angle CPA$. Show that three Euler lines of triangles $ APB,BPC,CPA$are concurrent.[/quote]
It is very nice, but not hard. Here is my solution:
Construct the equilateral triangle BCE.
Easy to show that A,P,E are collinear and P,B,C,E lie on a circle, whose center is denoted by I.
K,G are the centroids of triangles BPC,ABC respectively. Hence, IK is the Euler line of triangle BPC.
Let's prove that I,K,G are collinear. Similarly, three Euler lines of triangles APB,BPC,CPA are concurrent at the centroid of triangle ABC.
Let M be the midpoint of side BC. we have $ \frac{MI}{ME}\equal{}\frac{MK}{MP}\equal{}\frac{MG}{MA}\equal{}\frac{1}{3}$, hence I,K,G lie on a line parallel to the line (AKE).
Done.


I found some AHSME problems on resource sections, but most of them are incomplete and I can't find solutions. Anyone knows where to find ASHME problems and solutions  (1970s and 1980s), maybe other websites?

You can purchase them from the AMC's website.

Purchase the AHSME VOl II CD from the AMC office, it has the complete set of contests, problems and solutions from 1975-2000, exactly as they appeared.  The Volume I CD has all the contests 1950-1974.  Details are at:
http://www.unl.edu/amc/d-publication/d1-pubarchive/20090515-PubRev.pdf

Alternatively, the Contest Problem Book Series from the MAA Bookstore has the contest problems and detailed solutions in book form.  You probably want Problem Books IV, V, VI for the 70's and 80's.

-- Steve Dunbar
MAA Director, American Mathematics Competitions

Are there any free ones available of the web, like amc10 and ahsme/amc12??

AoPS/MathLinks has thousands of problems from various old competitions. Go to the "Contests" or "Resources" button at the top of the page.

[quote="mathemagician1729"]Does anyone know where to find the ahsme's from like 1990-1995 for free?[/quote]

You have to buy them.
Long time ago, I found them through a friend, but if there's something I can tell you, it's that most of "good" problems in those years are found in AoPS book, especially the 2nd one.


$ x,y,z > 0$
$ xyz \geq xy \plus{} yz \plus{} zx$
Prove
$ xyz\geq3(x \plus{} y \plus{} z)$

India 2001

[quote="pardesi"]let $ \frac {1}{x} \equal{} a,\frac {1}{y} \equal{} b,\frac {1}{z} \equal{} c$
thus the given condition is $ \sum a \leq 1$
 $ \Rightarrow 3(\sum ab) \leq (\sum a)^{2} \leq 1$
done[/quote]

Its the solution I had.
Is there any other solution :D ?

$ xyz(zy\plus{}yz\plus{}xz)\ge (xy\plus{}yz\plus{}zx)^2\ge 3xyz(x\plus{}y\plus{}z)$ then $ xyz\ge xy\plus{}yz\plus{}zx\ge 3(x\plus{}y\plus{}z)$  :)

[quote=quangpbc]$ (xy\plus{}yz\plus{}zx)^2\ge 3xyz(x\plus{}y\plus{}z)$ [/quote]
can anyone pls tell how we got this?? :help:


Thank you.  I hope someone understands the question.


On a rectangular coordinate system, what is the number of units in the distance between $ (4,1)$ and $ (16,\minus{}4)$?

Using the distance formula gives: $ \sqrt{12^2\plus{}5^2}\equal{}\sqrt{169}\equal{}\boxed{13}$.


linear algebra
matrix
superior algebra
superior algebra unsolved

Let $A \in M_n(R)$ such that: $A=A^t$ and ${a_{ii}>\sum_{j=1;j \neq i}^n|a_ia_j}|$.
 Prove that $det(A- xI_n)$ has all positive roots

Do you mean $a_{ii}>\sum_{j=1;j \neq i}^n|a_{ij}|$?

It is a classical problem.

Let $\lambda$ be an eigenvalue of $A$. (Since $A$ is symmetric, all eigenvalues are real)
Suppose $A\alpha=\lambda\alpha$, where $\alpha = (x_1,\dots,x_n)^T\neq 0$
Let $|x_i| = \max \{|x_1|,\dots,|x_n|\}$, then $x_i\neq 0$
We have
$a_{i1}x_1+a_{i2}x_2+ \cdots + a_{in}x_n = \lambda x_i$
then
$a_{i1}\frac{x_1}{x_i} +  \cdots + a_{ii} + \cdots + a_{in}\frac{x_n}{x_i} = \lambda$

Notice that $a_{ii} > \sum_{j\neq i} |a_{ij}|$ and $|x_j/x_i| \leq 1$
we have
$\lambda = a_{i1}\frac{x_1}{x_i} +  \cdots + a_{ii} + \cdots + a_{in}\frac{x_n}{x_i} \geq a_{ii} - \sum \left|a_{ij}\frac{x_j}{x_i}\right| \geq a_{ii} - \sum |a_{ij}| > 0 $

There is also another approach with Gershgorin Circle Theorem:
as liyi said our matrix $A$ has only real eigenvalues, since $\displaystyle a_{ii}>\sum_{j \neq i}^n|a_{ij}|$, $a_{ii}>0$ and all the Gershgorin Circles have their centers at $a_{ii}$ on positive real axis and their radiuses $<a_{ii}$ from where the conclusion immediately follows

I believe this was posted by Mubinool before, and I posted another solution there.


Try this:

[url]http://216.239.59.104/search?q=cache:GGgReWIh8OcJ:www.win.tue.nl/~henkvt/debruijn-sequence.pdf+Debruijn+sequence&hl=en&ie=UTF-8[/url]

Even the address looks scary, doesn't it?  :D  :D  :D

You know, I have no idea what D... sequences mean. I just looked it up with Google... :D (I can't even remember the name if I don't peek at the topic review  :D )


search
combinatorics unsolved
combinatorics

How many subsets of the set {1,2,...,2005} have a sum congruent to 2006 modulo 2048?


function
inequalities
topology
real analysis
triangle inequality
calculus
calculus computations

[b]Problem:[/b] Let $X$ be a metric space with metric $d$, and let $x_0$ be a fixed point in $X$. Show that the real function $f_{x_0}$ defined on $X$ by $f_{x_0}(x)=d(x,x_0)$ is continuous. Is it uniformly continuous?

[b]My work:[/b] First of all, the reason for asking about this is that I'm a bit unsure about how to measure the distance between $d(x,x_0)$ and $d(y,x_0)$. These are clearly two real numbers and so it is natural (at least to me) to show that we are capable of finding a $\delta >0$ such that for any $\varepsilon >0$ we have $|d(x,x_0)-d(y,x_0)|<\varepsilon$ whenever $d(x,y)<\delta$.

From the triangle inequality we have
\[ d(x,x_0)\le d(x,y)+d(y,x_0), \]
and similarly $d(y,x_0)\le d(x,y)+d(x,x_0)$, so that we draw the conclusion that
\[ |d(x,x_0)-d(y,x_0)|\le d(x,y)<\delta : =\varepsilon , \]
whenever $d(x,y)<\delta$. This also implies that $f_{x_0}$ is uniformly continuous.

Am I correct? If not, should I generalize to some more arbitrary metric on $\mathbb{R}$?

[quote="fredbel6"]that is completely correct in my opinion
I don't think that it will work for every metric on the reals : consider the discrete metric
It will of course work for every equivalent metric, as it gives the same topology

If you want something else to work on, take a random subset A, nonempty in your metric space

and define $d(x,A)=inf\{d(x,a)|a\in A\}$

show that this function is also uniformly continuous.  When is it zero?[/quote]

Right, accepting your challange I get
\[ \inf _{a\in A}d(x,a)\le \inf_{a\in A}(d(x,y)+d(y,a))=d(x,y)+d(y,A) \]
and as above we have
\[ |d(x,A)-d(y,A)|\le d(x,y), \]
which proves uniform continuity. Also, it is zero for any $x\in \overline{A}$ (closure), I guess.

Thanks for your response!


linear algebra
matrix
linear algebra unsolved

Does anyone know of any quick ways of working out the determinant of a matrix of the form D-I, where I is the identity and D has determinant one?

There's really no shortcut, unless you know more about $D$ than that. Of course, finding determinants isn't very hard.

For $D$ diagonal, there's a simple rule saying that just like $det(D) = \prod(\lambda_i)$, we have that $det(D-I)=\prod(\lambda_i-1)$. Note that the diagonal elements on a diagonal matrix are the eigenvalues of that matrix, and that the product of the eigenvalues is always the determinant of a matrix :)


Before saying anything else, I want people not to get angry that I made both this poll and the one on invented vs discovered. I am just curious as to the difference in what a person says in regards to this question as opposed to math being invented or discovered. I want to see if I am understanding people views correctly and see people's opinions on this topic.

I hope this vote does not turn out negatively, I have no bad intentions. I will only say that I believe that 1+1 is always 2 because of the universal meaning of 1 and +.

What do you guys think? I encourage all opinions and the same to continue in the other thread, I apologize for saying everything is obvious.

Well to be technical, 1+1=2 in base 10. If it were in base 2, it would be 1+1=10

[quote="G-UNIT"][quote="10000th User"][quote="WindSlicer"]It depends how you define, 1, +, =, and 2. If we use our modern definitions, then yep.

But of course we can always define 1 to be 2 and then we have 2+2=2.[/quote]
That's not correct because if I replace '2' by $x$, I get $x+x=x$ and this is not the same as $1+1=2$. Maybe you mean something like $2+2=5$, with '2' replacing 1 and '5' replacing 2.[/quote]

  Thats assuming that "+" is defined using our "modern definitions".  As windlsicer said, "=" "+" and numbers can be defined differently, which means algebra which only holds for our modern definitons, will not hold for our new cases.[/quote]
I'm sorry but I don't understand what you mean "modern definitions". I understand WindSlicer 3rd sentence: "we can always define 1 to be 2 and then we have 2+2=2". He define "1 to be 2" and everything same but he put "2+2=2", and this is not correct proof.

[quote="forever"]i'm a little suprised that 75% of people vote "no".
is it bacause of the different bases?[/quote]
I know it is only different in base 2 because $1_2+1_2=10_2$ but $1_b+1_b=2_b$ in $b>2$

I know there is a $1+1=0$ or $1+1=1$ in engineering??? I don't know where its from but I know there are formulas like that :blush:

[quote="10000th User"]I know there is a $1+1=0$ or $1+1=1$ in engineering??? I don't know where its from but I know there are formulas like that :blush:[/quote]
Boolean ring?
Boolean Algebra?

I was thinking about Chemistry when I voted for no.
$50ml ( \text{spirit})+50ml(\text{water}) < 100ml$.

[quote="10000th User"][/quote][quote="G-UNIT"][quote="10000th User"][quote="WindSlicer"]It depends how you define, 1, +, =, and 2. If we use our modern definitions, then yep.

But of course we can always define 1 to be 2 and then we have 2+2=2.[/quote]
That's not correct because if I replace '2' by $x$, I get $x+x=x$ and this is not the same as $1+1=2$. Maybe you mean something like $2+2=5$, with '2' replacing 1 and '5' replacing 2.[/quote]

  Thats assuming that "+" is defined using our "modern definitions".  As windlsicer said, "=" "+" and numbers can be defined differently, which means algebra which only holds for our modern definitons, will not hold for our new cases.[/quote][quote="10000th User"]
I'm sorry but I don't understand what you mean "modern definitions". I understand WindSlicer 3rd sentence: "we can always define 1 to be 2 and then we have 2+2=2". He define "1 to be 2" and everything same but he put "2+2=2", and this is not correct proof.[/quote]

Yeah, I wasn't really paying attention to the end product, my focus was on the fact that you can define anything to be something else, therefore 1+1 may not always equal 2 depending how how you define the characters.

Yeah, technically you [i]can[/i] define the symbols to be anything.  You could redefine the '+' operator such that it can spit out different values for the same inputs.  But reasonably speaking, it is a common consensus that '+' means addition and that by using the most commonly held axioms 1+1=2...ALL THE TIME :D (well, mathematically speaking...as shobber points out it is a little different when dealing with chemistry and one could make other arguments in other fields I presume).

As posted earlier, if we do addition and subtraction on the set 0,1 and take it (mod 2), all of our "modern" definitions are satisifed (it is a field) and we get

1+1=0.

No real problem here.

I think he's trying to round down somehow.

Well, if you read the nutrition labels on food items, you will find that not only does $1+1\not=2$, but $1+0+0+0=3$, $4\cdot 2=0$, and $9\cdot k\equiv 0\pmod{10}$, for all $k\in\mathbb{Z}$.

[quote="frt"][quote="Arne"]Isn't 1 + 1 = 3 for very large values of 1? ;)[/quote]
Sorry, but what did you mean by "very large values of 1"?[/quote]

I was actually joking :) Don't worry about it.

from 1984:
-Two and two are four.
-Sometimes, Winston. Sometimes they are five. Sometimes they are three. Sometimes they are all of them at once. You must try harder.

so, 1+1 doesn't have to equal two, just like 2+2 doesn't have to equal 4.  :lol:

People, sorry I didnt mention in the original post, i mean the meaning of 1+1 always gives the same asnwer, knowing what "1" and "+" means, that is what I was refering to.

I am surprised that no accountant has yet attempted to answer the question. If (s)he (the accountant) is anyting like the one who does tax  returns  would make sure that all the windows are closed and would wispher:
[quote] How much you want it to be? [/quote]

(Yeah I know, an old joke...! :)
(Not to mention, if the calculator  you use is MS windows 3.1 or have  an old Pentium chip - your answer may vary)

if it was in base 2, it would be 10, not 2 :lol:

Suppose it was $1_{\infty}+1_{\infty}=2_{\infty}=2\cdot \infty=\infty$ or indetermined.

Sorry I did not say in my first post, but I meant that the meaning of 1+1 in our basic arithmetical answer produces the answer 2, u cant changed the meaning of 1 or + in what I was asking about in my question. How would the answer then change?


Peter and Darij play the game of the three knights. It is played on a chess board, initially three knights are placed on positions $h1,g2$ and $f3$. See the left image below. Alternately the players move one of the knights to an empty field whereas contrary to chess the knight may just perform the four moves shown in the right image. The goal is to move the knight to its home. A knight is at home when it is one of the hatched fields $a8,a7,b8$ or $b7$. The person who gets the third knight home will win the game. If there might be a situation before that no movement is possible anymore there will be a tie. Peter starts.

Is there a winning strategy for any of the players ? If so, what is the strategy and why does it force the victory ?

Hmm. Nobody there who wants to help Darij and Peter ?  :D

[quote="DPopov"]can the knights take each other..lol? and also are the hatching fields for each knight respectively or for all the knights?[/quote]

Obviously the first point is not true, otherwise you cannot move all knights in the upper left four squares. For the second point: Finally all the three knights will be in the previously mentioned squares whereas one of fours squares remains empty.  :)


function
analytic geometry
graphing lines
slope

A botanist measures the growth of a plant, in centimeters, every day. Connecting the dots placed by him on a graph, we get the picture beside. If you always keep this relationship between time and height, the plant will, in the 30 (thirtieth) day, a height equal to:

[img]http://img4.imageshack.us/img4/9803/cals.jpg[/img]

a) 5 cm
b) 6 cm
c) 3 cm
d) 15 cm
e) 30 cm

I'm very novice in this subject (Functions), can anyone explain me more clearly ?
Thanks.  :lol:

The key is to determine the equation of the graph.  It's pretty clear that it's a line so what is its equation?  We know that $ (10,2)$ and $ (5,1)$ are both points, so the slope of the line is $ \frac{2\minus{}1}{10\minus{}5}\equal{}\frac{1}{5}$.  So the line has equation $ y\equal{}\frac{x}{5}$.  So when $ x\equal{}30$ (30 days), $ y\equal{}\frac{30}{5}\equal{}6$.  So the height is 6cm.

Thanks, but is the equation  Y = Ax + B ?
And B, what's it value ?
A = x/5 ?

[quote="luiseduardo"]Ok, but how do you know that B = 0 ?[/quote]
Straight line from the origin

hum ...  :(
Thanks, but i think it's difficult to understand. It's my first time with exercises like that.

Ah ok  :lol:
No problem. I understand now. Thanks.

So can say that the line has equation $ y\equal{}\frac{x}{5}\plus{}b$ if you want to, but since $ (0,0)$ is a point, you get that $ 0\equal{}0\plus{}b$, so $ b\equal{}0$.

Basically b is the number's y intercept, and since it goes through the origin, its y intercept is 0.

actually, there is not even the need to find the equation of ther line.you can use the similarity between the triangles. :wink: (this is much more simple, i think)

It is b

It starts from the origin, and has slope of 1/5. So you just count up 6 and move over 30

:starwars:  :spam:  :nhl:

[quote="v235711"]actually, there is not even the need to find the equation of the line.you can use the similarity between the triangles. :wink:  (this is much more simple, i think)[/quote]
Not if you're trying to understand functions and equations of lines, which is what the OP is trying to do.

Thanks JRav.
I understand the problem now.  :lol:


calculus
derivative
trigonometry
calculus computations

Well I have this problem here, and so far the calculus isn't the problem, its the algebra. I just want to see if anyone can solve this for me, as I am stuck and its valentine's day and the combination of not having a girlfriend and doing f--king problems like these is making me want to find a tall building.

x(arc cosy) = (arc sinxy)

Perhaps I am missing a simpler way to do this, or maybe this is just one ugly mother.  I tried getting y on the left side, but then after taking the derivative (using multiple chain/product rules), there are some dy/dx's wrapped up in the right side.

Also, these inverses are NOT 1/sin (csc), 1/cos (sec), these are the inverse trig fxns (someone told me they are also knows as arc fxns)

I have the formulas for the arc fxns:
d/dx[arc sin(u)]=1/[sqr rt=1-(u^2)]
d/dx[arc cos(u)]=(-)1/[sqr rt=1-(u^2)]

Any help would be appreciated as it is surely not my time to leave this world, but every second i spend on this problem speeds it towards me.

I finally got dy/dx, if someone still wants to do this i would like to compare answers and methods.
-- Thanks

I started writing this up before your second post:

Let me just double-check: are you trying to find the derivative $\frac{dy}{dx}$ given the equation $x \cdot \cos^{-1}y = \sin^{-1}(xy)$?  That equation isn't going to solve for a nice expression of $y$ in terms of $x$, which suggests that you have to use implicit differentiation.  After you do that, it should be easy to solve for $\frac{dy}{dx}$ (although the resulting expression is pretty ugly).

Yes sir, thats the equation. I got an answer that checks out, and yea youre right i had to use implicit diff. i just had to work it out in my head before i messed with in on paper to make sure... thanks! like i said if you wanna try it, ill compare notes.


Solve for x:
\[ y \equal{} \frac{2x}{t\minus{}1}\]

whoops sorry, the t is supposed to be another x. so instead its:
\[ y \equal{} \frac{2x}{x\minus{}1}\]

For other problems like those, first cross-multiply to eliminate fractions. Then solve the equation to get the answer (which isabella already gave) \[ \frac{y(t \minus{} 1)}{2} \equal{} x\]

[quote="Lord Voldemort"]whoops sorry, the t is supposed to be another x. so instead its:
\[ y \equal{} \frac {2x}{x \minus{} 1}
\]
[/quote]
that's what i meant, sorry for the confusion.

Then substitute the $ t$ in the answer for $ x$. I don't think you can get any simpler than that  :) .

$ x\equal{}\frac{y}{y\minus{}2}$

I'll explain it better when I get home.

[hide="solution"]
$ y\equal{}\frac{2x}{x\minus{}1}$

$ yx\minus{}y\equal{}2x$

$ \minus{}y\equal{}2x\minus{}yx$

$ \minus{}y\equal{}x(2\minus{}y)$

$ \frac{\minus{}y}{2\minus{}y}\equal{}x$

$ \boxed{\frac{y}{y\minus{}2}\equal{}x}$

[/hide]
i guess i'll save markkram the time.

Minus one point for forgetting restrictions. Remember that y can't be 2.

Ok we have:  y=(2x)/(x-1)

WE need to isolate "x" or just get it on one side to show what it equals..

First:  we need to multiply each side by (x-1) to get rid of the denominator.

Now we have: xy-y=2x   <--- a lot simpler huh?

Now, our final step.  WE just need to divide each side by 2 to get rid of the coefficient and completely isolate x.

The answer:  [(xy-y)/2]=x

[quote]
$ x\equal{} \frac{xy\minus{}y}{2}$
[/quote]

but x isn't completely isolated. You need to bring xy to the other side, then factor out x.


i have posted this exercire in the High school intermediate topic but no one answer
so i post in this topic , please help me solve the equation
(sin2x)^2 + sin2x + sinx + 1 = 0

hello, simplifying your equation we get
$ 4\, \left( \sin \left( x \right)  \right) ^{2} \left( \cos \left( x
 \right)  \right) ^{2}\plus{}2\,\sin \left( x \right) \cos \left( x \right)
\plus{}\sin \left( x \right) \plus{}1\equal{}0$
converting this in $ \tan(\frac{x}{2})$ we get
$ \displaystyle{16\,{\frac { \left( \tan \left( 1/2\,x \right)  \right) ^{2} \left( 1\minus{}
 \left( \tan \left( 1/2\,x \right)  \right) ^{2} \right) ^{2}}{
 \left( 1\plus{} \left( \tan \left( 1/2\,x \right)  \right) ^{2} \right) ^{4
}}}\plus{}4\,{\frac {\tan \left( 1/2\,x \right)  \left( 1\minus{} \left( \tan
 \left( 1/2\,x \right)  \right) ^{2} \right) }{ \left( 1\plus{} \left( \tan
 \left( 1/2\,x \right)  \right) ^{2} \right) ^{2}}}\plus{}2\,{\frac {\tan
 \left( 1/2\,x \right) }{1\plus{} \left( \tan \left( 1/2\,x \right)
 \right) ^{2}}}\plus{}1\equal{}0}$
substituting $ \tan(\frac{x}{2})\equal{}t$ and factorring we have to solve
$ \left( t\plus{}1 \right)  \left( {t}^{7}\minus{}3\,{t}^{6}\plus{}23\,{t}^{5}\minus{}21\,{t}^{4}
\minus{}5\,{t}^{3}\plus{}15\,{t}^{2}\plus{}5\,t\plus{}1 \right)\equal{}0$.
Sonnhard.


Prove that for $a,b,c$ are nonegative real, we have:

$\frac{2a^2}{b^2+c^2}+\frac{2b^2}{a^2+c^2}+\frac{2c^2}{a^2+b^2} \ge \frac{(a+b+c)^2}{ab+bc+ca}$

Pham Kim Hung

I use  Vasc'c  inequality $a,b,c>0$  then  $\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\geq\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ and  BCS
$LHS\geq\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}=\frac{2a^2}{ab+ac}+\frac{2b^2}{bc+ba}+\frac{2c^2}{ca+cb}\geq\frac{2(a+b+c)^2}{2(ab+bc+ca)}=RHS$

Sorry  , i  cannot  find the link  with  Vasc's  inequality,,,,

Very nice solution and It's much simpler than I think at first.
Other hand, $2$ is the best constant for this.

[quote="hungkhtn"]
Other hand, $2$ is the best constant for this.[/quote]

What do you mean?


Let's say we have $ (x+y)^{2}$ . We'd automatically think it equals $ x^{2}+2xy+y^{2}$. However, this presents a problem when dealing with coefficients.

If we have, for instance $ (3a+4b)$, the only way this can be factored with the above formula is to do the following:

$ (3+4)^{2}= (3^{2}+2(3)(4)+4^{2}) = (9+24+16)$

$ (a+b)^{2}= (1)a^{2}+(2)ab+(1)b^{2}$

Ignoring the coefficients of the second equation and replacing them with the coefficients of the first equation, we have:

$ 9a^{2}+24ab+16b^{2}$ which is correct.

But instead of doing all that work, wouldn't it make more sense just to use a more thorough formula such as $ (cx+dy)^{2}= c^{2}x^{2}+2cdxy+d^{2}x^{2}$ where $ c,d$ are coefficients and $ x,y$ are variables? This way, we could get our answer instantly. Example:

$ (8x+9y)^{2}= 8^{2}x^{2}+2(8)(9)xy+9^{2}y^{2}= 64x^{2}+144xy+81y^{2}$

Sorry if the question seems really confusing, my main issue is whether to list factoring formulas with or without variables representing coefficients.

[quote="aidan"]I think when factoring formulas are listed, they usually omit the coefficients because we can assign $ a = 8x$ and $ b = 9y$, for example, in $ (a+b)^{2}$ and we get $ a^{2}+2ab+b^{2}= (8x)^{2}+2(8x)(9y)+(9y)^{2}$. So it's easy to substitute the coefficients into the formula and we don't have to explicitly represent coefficients in the formula?[/quote]

That pretty much answered my question. Thank you!

[quote="max_tm"]If I understand your question correctly, you're starting to touch on something called the binomial theorem. Do a search on the site for it.[/quote]

I've seen the binomial theorem before, but that's not what I'm asking about. Thanks anyway.


Prove or disprove, for any $n$

$ \displaystyle \sum^n_{k=3} \frac{1}{k^3}<\frac{1}{12}$

[quote="Elemennop"]Isn't it enough to notice that for $n=3$, $\frac{1}{k^3}=\frac{1}{27}$, which is less than $\frac{1}{12}$, and that for any $n>3$, $\displaystyle \sum^n_{k=3}k^3$ will be greater than when $n=3$, and thus $\displaystyle \displaystyle \sum^n_{k=3} \frac{1}{k^3}$ will be less than $\frac{1}{27}$ (and consenquently less than $\frac{1}{2}$?[/quote]
Yes, all of the individual terms will be less than $\frac{1}{12}$, but, how do you know that $\frac{1}{27}+\frac{1}{64}+\frac{1}{125}+\frac{1}{216} + ... + \frac{1}{n^3} < \frac{1}{12}$?

[quote="Elemennop"]Isn't it enough to notice that for $n=3$, $\frac{1}{k^3}=\frac{1}{27}$, which is less than $\frac{1}{12}$, and that for any $n>3$, $\displaystyle \sum^n_{k=3}k^3$ will be greater than when $n=3$, and thus $\displaystyle \displaystyle \sum^n_{k=3} \frac{1}{k^3}$ will be less than $\frac{1}{27}$ (and consenquently less than $\frac{1}{2}$?[/quote]
I think what you have proved is that as n increases each term will decrease, but this is not enough to say that the whole sum will decrease.
$\sum^n_{k=3}k^3 $is not the reciprocal of $\sum^n_{k=3} \frac{1}{k^3}$

Hmm.

I would say the way to prove it now would be show that if $n=3$, the expression is less than $\frac{1}{12}$, then find what it converges to as $n$ approaches $\infty$ (and show it is less than $\frac{1}{12}$), and then show that the expression is decreasing for each $n$ as $n$ approaches $\infty$.

i guess we can't express zeta(3) as an exact number yet.  i looked it up (http://64.233.187.104/search?q=cache:IkjY0zJIzu8J:mathworld.wolfram.com/AperysConstant.html+zeta(3)&hl=en), and all i found was an approximation ~1.2020569, so i dont know if thats the way to go.  telescoping it seems more likely.

Hum diddly dum.

$\sum_3 1/k^3<\sum_2 1/[k(k+1)(k+2)]=1/4-1/6=1/12$

I think there's some minor pedagogical value for you guys to figure out why what I wrote is true without me telling you so; if you still have questions in a day, I will be glad to answer them.

Out of curiosity, is that how you constructed the problem?  Because an integral test doesn't work very well - I'm not sure but I suspect that we need to sum a large number of terms by hand because the integral gives us an estimate that is not very sharp.

That indeed is how it was derived. It was actually a problem in an IrMO, a few years ago... How does the integral solution go? Im not so sharp on that kind of thing... :(

We have

$\sum_{k=n}^{\infty} \frac {1}{k^3}<\int^{\infty}_{n-1} \frac {dt}{t^3}=\frac{1}{2(n-1)^2}$

This method actually works if we take $n=6$ and sum the first three terms manually.

How'd you get the first bit? Or is it something general like

$\displaystyle \sum_{k=f}^\infty f(k) < \int_{f-1}^\infty f(k) dx$

See if you can prove that it holds for any monotonically decreasing $f$

Woah you did some serious digging! Dspite massive attempts, no one can calculate this value! Can anyone think of a function that has $\frac{1}{k^{3}}$ as roots for all $k \in \mathbb{N}$?