4. Suppose that T is a normal operator on V and that 3 and 4 are eige.pdf

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4. Suppose that T is a normal operator on V and that 3 and 4 are eigenvalues of T. Prove that there exists a vector v epsilon V such that ||v|| = root 2 and ||T(v)|| = 5. HINT: Think about what we know about the eigenspaces of normal opera tors, then create a vector from the two eigenspaces and use the Pythagorean theorem to verify the vector has the desired properties. Solution Given that 3 and 4 are eigen values of T. Then there exists a vector v such that v = xi+yj such that |x|=1 and |y| =1 as T is normal Hence ||V|| = rt of x^2+y^2 = rt 2 ||T(v)|| = rt of 3^2+4^2 = 5.

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3. Transportation technology (15 points) a. Explain both generally and with specific examples how changes in transportation technology have affected the development of urban areas. b. Indicate two ways in which governmental policy has influenced the shape of urbanization in the U.S. c. Do you believe that new telecommunication and other forms of technology will make distance an unimportant factor in economic development and, thus, make cities unimportant economic factors? Support your answer. Solution Transportation enbles activity of connecting people, businesses and resources.Transportation improvement is necessary for the economicdevelopment and transportation policies help to support the economic objectives.it helps to improve economic efficieny. Economic efficiency refers to the ratio total benifit to cost. Increase economic efficieny, increase productivity which leads to economic development. The ultimate goal of transportation is accessabilty, access the people and the company abilities including raw material, labor, worksites, proffessional services, distrbutors etc. Increased accessibilty means reduction in time,money, risk require recouces reaches in less time so the productivity of a company increases. cost saving in an organisation---Economic efficiency more output per unit of input---productivity more goods and services produced in less cost---Ecoonomis development. Example- if in a company a manager has to submit an assignment cost company say 10 crores and he has to submit it with in 2 days then the manager priority is to travel by Air not by train etc. because the loss of the company is veyr high then the cost of transportation. In transportation priortization is imortant point to improve productivity hence economic development. Urbanisation means the population shifted rural to urban areas and how the economy adopt the change.Recent development of commiucation technology in US, such as broadband services , package express delivery services, cheap telecommunication, courier services , easy ways to transprotation, highways development promotes people to work in the different ares of the country. But now days the USA government initiating suburbanisation..

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3. Suppose for f(z) holomorphic C there are constants a, b, e > 0: Prove that f is a polynomial of degree n less than equal to c . Solution A holomorphic function is a complex-valued function of one or more complex variables that is complex differentiable in a neighborhood of every point in its domain. Hence f(z) is holomorphic means f\'(z) exists for all z 2x= z+z bar and 2iy = z-z bar f(z,zbar) = =?m?0,n?0amnzmz.

3. Suppose for f(z) holomorphic C there are constants a, b, e 0 Pr.pdf

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3. Joe is considering pursuing an MBA degree. He has applied to two different universities. The acceptance rate for applicants with similar qualifications is 25% for University A and 40% for University B. What is the probability that Joe will not be accepted at either university? (Points : 4) 0.75 0.45 0.90 0.65 0.60 Solution Given P(A)=0.25, P(B)=0.4 So the probability is P(A\' and B\') = P(A\')*P(B\') =[1-P(A)]*[1-P(B)] =(1-0.25)*(1-0.4) =0.45 Answer:0.45.

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3. Let On be the set of odd permutations in Sn. Is On a subgroup? Solution No, On is not a subgroup.. to see that the odd permutations don\'t form a subgroup of Sn is a bit easier to resolve. Recall that any subgroup of Sn has the same identity element as the identity element in Sn. But since the identity element in Sn is not an odd permutation (a result we saw in class), it follows that the collection of odd permutations cannot be a subgroup of Sn. (One could also argue that the odd elements aren\'t closed under the group operation.).

3. Let On be the set of odd permutations in Sn. Is On a subgroup .pdf

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3. (4 points) Consider the function (in the domain {|z ? i| Solution Consider f(z) It got a denominator of 1+(z-i)2 Equate to 0 the denominator to get singular points (z-i)2 =-1 Take square root z-i =i, -i z=2i or 0 As |z-i|<1 we can expand f(z) by binomial series f(z) = 1-(z-i)2+(z-i)4+...+(-1)n(z-i)2n+... As odd powers are 0 we have derivative of f^m(z) =0 for m = odd For n even f\"m(z) = (-1)n(2n)(2n-1)....(1)(x2n-2n) = (-1)n(2n)! Proved..

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3. Find currents (values and directions) in all resistors of the circuit given below. Find total electric power developing in this circuit Neglect internal resistances of the batteries. Solution As all the Voltage sources are DC (Batteries) there by the capacitor acts as short circuit . I1 = Current through R1 =E1-E2)/R1 = (6-3)/5 = 3/5 Amps (direction from E1 to E2) I2= Current Throuh R3 = E2/R3 = 3/4 Amps (direction from Top to bottom) I3 = Current Through R2 = E3-E2)/R2 = (5-3)/5 = 2/5 Amps (direction from E3 to E2) Power delivered = (I1*E1) + (E3*I3) =(3/5)*6 + (2/5)*5 = 28/5 Watts.

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3. Assume that the DS register contains 0100 and register SI contains 0020. What is the result of following instruction LES DI, [SI] Where the content of memory Location is as follows: Solution the command LES DI , [SI] does the following: it takes the pointer of SI , i.e the address of the memory location SI and breaks it into lower bits and upper bits. then the lower bits are stored in DI and the upper bits are stored in ES. so address of SI is 1027 . lower bit is 27 and upper bit is 10. 27 will be stored in DI . and 10 will be saved in ES. now DI is the lower bit of the register DS. as it is said initially DS 0100 so, initially DI is 00. which is changed to 27. now DS is 0127..

3. Assume that the DS register contains 0100 and register SI contains.pdf

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2a Briefly describe the 1st law of thermodynamics. You can use either.pdf

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3 (a) calculate thy potential at point P located a distance z above the center of a charged disk of radius a and surface charge density p, Coulombs per square meter. Please show all work An integral that you may need is The elemental area in cylindrical coordinates is dA = (b) Calculate the electric field at P. Use back of page if necessary Solution function [Ef] = Loop_Ef() %(Main M-File function) clc, clear all, close all u.N1 = 1e21; %This is the number of states u.NA = 1e15; %This is the number of acceptors u.ND = 0; %This is the number of donors u.Z1 = 281.73e19; %This term in front of the square root of the energy (this contains mass of electron) u.Z2 = 757.989e19; %This is the mass of hole u.Eg = 1.12/2; %This is the energy gap u.a = 0.0259; %u.rat_a = 2 %scaling factor that modifies the steepness of the oxide exponential density of states u.es = 11.9 * 8.854187817e-14; %permittivity measured C / cm * V u.qe = 1.602e-19; %This is the value of charge u.psi_0_range = linspace(-0.4, 1, 160)\'; %This defines the range of Psi_0 u.display = true; %psi is in Volts Not ev? %% %ratN_range = 1e-2; ratS_range = [ 0.5: 0.2: 2]; ratN_range = 1; ratS_range = 1; %This doesn’t matter for Parabolic DOS for iN = 1 : length(ratN_range) for iS = 1 : length(ratS_range) u.ratN = ratN_range(iN); u.ratS = ratS_range(iS); u.N2 = u.ratN * u.N1; try % Ef_(iN,iS) = Calculate_Ef(u); %Here its calculating the f_function Ef_(iN,iS) = -0.356324 %REMOVE-TESTING u.Ef_ = Ef_(iN,iS); %Here Ef equals u.Q = Calculate_El(u);%Here is calculating the charge DiffQ(u); catch Ef_(iN,iS) = NaN; %If there is nothing then NaN end end end clf, plot(u.psi_0_range, u.Q) %Its plotting the graph u.Ef_all = Ef_; save \'u.mat\' %It saves all the data in u.file save \'Ef_.dat\' Ef_ -ascii %Exports the data of Ef as ascii keyboard %This command gives the control to the user end %================================================ function [Ef_] = Calculate_Ef(u) clc time = cputime; %rescale volumes for numerical stability u.Nscale = u.N1 ^ 2; Nscale = u.Nscale; u.N1 = u.N1 / Nscale; u.N2 = u.N2 / Nscale; u.Z1 = u.Z1 / Nscale; u.Z2 = u.Z2 / Nscale; %get rid of the u.\'\' N1 = u.N1; N2 = u.N2; Z1 = u.Z1; Z2 = u.Z2; Eg = u.Eg; a = u.a; NA = u.NA/Nscale; ND = u.ND/Nscale; P_1 = @(E) Z1.*sqrt(-Eg-E); %This is the parabolic function(Valance band) P_2 = @(E) Z2.*sqrt(E-Eg); %This is the parabolic function (Conduction band) Fn = @(E,Ef) 1 ./ ( exp((E-Ef)/a) ); %This is the Fermi-dirac for electrons Fp = @(E,Ef) 1 ./ ( exp((Ef-E)/a) ); %This is the Fermi-dirac for holes dp = @(E,Ef) P_1(E) .* Fp(E,Ef); %the integrands dn = @(E,Ef) P_2(E) .* Fn(E,Ef); p = @(Ef) Nscale * (IntegrateInf_p( @(E) dp(E,Ef), u ) + ND); %define our functions p & n (Ef) n = @(Ef) Nscale * (IntegrateInf_n( @(E) dn(E,Ef), u ) + NA); %%we need to make sure that multiplication/division with Nscale is consistent all the way through to DiffQ Ef_ = Solve_Ef(-0.5, p, n); %Ef* is obtained here fprintf(\'The value Ef*= %f, gives equal val.

3 (a) calculate thy potential at point P located a distance z above .pdf

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