PHZ 3113 Fall 2010Exam 1Name:This is a closed-book exam. Some possible useful information:(1 + x)p=1+px+12p(p − 1) x2+16p(p − 1)(p − 2) x3+ ···=∞Xn=01n!p!(p − n)!xnex=1+x +12x2+16x3+ ···=∞Xn=01n!xnZx0sec3x0dx0=12[sec x tan x +log(secx +tanx)]1. Let an=nnn!.Letbn=n!nn.(a) Which (if either) of the series∞Xn=0an,∞Xn=0bnconverges? (Explain why or why not.)Look at anspread out in factors,an=nnn!=nnn(n − 1)n(n − 2)···n1> 1:Pandoes not even pass the preliminary test.Or, look at the ratiolimn→∞an+1an=(n +1)n+1/(n +1)!nn/n!=(n +1)n(n +1)n!nn(n +1)!=n +1nn→ e.The last step was a homework problem (with α = 1), or look at the log of the limit andapply l’Hˆopital’s rule,limn→∞n ln(1 + 1/n)=ln(1 + 1/n)1/n=(−1/n2) / (1 + 1/n)(−1/n2)=11+1/n→ 1.Since an+1/an> 1, by the ratio test,Pandiverges; andPbncorrespondingly converges(Pbn=2.87985).(b) For the series in part (a) that has the worst divergence, or the slowest convergence (orif they behave the same, pick one), does the power seriesPcnxn(cn= anor bn)convergefor some values of x? If not, explain why; if yes, find the interval of convergence.The radius of convergence comes fromlimn→∞an+1xn+1anxn= e |x| < 1,and soPanxnconverges for x<e−1.2. The logarithmic probability distribution has probabilities pn= Cpnn(where p is aconstant, 0 <p<1) that a discrete random variable takes the value n.(a) What is the value of C so that∞Xn=1pn=1?The sum of the pnis∞Xn=1Cpnn= C [− ln(1 − p)],C= −1ln(1 − p).This is a series that you “should” remember, but even if you don’t remember anything butthe geometric series, you can get∞Xn=11npn=Zp0∞Xn=1xn−1dx =Zp0dx1 − x= − ln(1 − p).(b) Compute the average or expected value of n, h n i =Pnpn. (This should be one of theeasier calculations on this exam). Compute the second factorial moment h n(n − 1) i.h n i =∞Xn=1npn=∞Xn=1Cpn= C (p + p2+ p3+ ···)=Cp1 − p= −p(1 − p)ln(1− p).h n(n − 1) i =∞Xn=1n(n − 1) pn=∞Xn=1C (n − 1)pn= C (p2+2p3+3p4···)=Cp2ddp11 − p= −p2(1 − p)2ln(1 − p).(c) The generating function G(z) is defined as G(z)=∞Xk=1pkzk.FindG(z) for the logarith-mic distribution.This is the same sum as in (a), with argument pz instead of p,G(z)=∞Xk=1pkzk=∞Xk=1C1k(pz)k=ln(1 − pz)ln(1 − p).(d) Show in general that pnis related todnG(z)dznz=0. Show that in general the momentsh n i, h n(n − 1) i are related to derivatives of G(z) evaluated at z =1.UseyourG(z)to findthe moments computed in part (b).Evaluated at z = 0, the only nonzero term in n derivatives applied to the sum that definesG(z)hask = n: for those terms with k<n, dn/dznvanishes (six derivatives of x3vanishes);and terms with k>nhave factors zn−k→ 0. Thus,dndznG(z)z=0= n! pn.From the definition,ddzG(z)z=1=∞Xn=1pnnzn−1z=1=∞Xn=1npn= h n i .d2dz2G(z)z=1=∞Xn=1pnn(n − 1) zn−2z=1=∞Xn=1n(n − 1) pn= h n(n − 1) i .(This clearly generalizes to “factorial moments” of any order.)Applied for G(z)=ln(1 − pz)ln(1 − p),h n i =dGdzz=1=1ln(1 − p)−p(1 − p),h n(n − 1) i =d2Gdz2z=1=1ln(1 − p)−p2(1 − p)2.3. The function i(x)=(3 + x2)sinhx − 3x cosh xx3is a modified spherical Bessel function.How does i(x) behave for small x? (Find at least the first nontrivial term of an expansionof i(x) for x near 0). How does i(x) behave for large x? Sketch a plot of i(x).For small x, expansions of sinh x and cosh x givei(x) ≈(3 + x2)(x +16x3+1120x5+ ···) − 3x(1 +12x2+124x4+ ···)x3=(3x +12x3+140x5+ ···)+(x3+1120x5+ ···) − (3x +32x3+18x5+ ···)x3=(140+1120−18) x2+ ···=115x2+ ···.For large x,thefactorx2dominates the numerator, andi(x) ≈sinh xx≈ex2x1 −3x.4. The (volume) thermal expansion coefficient α isα =1v∂v∂Tp.(a) An ideal gas has equation of state pv = RT . Compute α for an ideal gas.α =1v∂v∂T=1v∂∂vRTp=Rpv=1T.(b) The virial equation of state ispvRT=1+Bv,whereB is constant (the virial constant).Compute the thermal expansion coefficient α for a virial gas. For positive B,isα larger orsmaller than for an ideal gas?Writev =RTp1+Bv;then∂v∂T=Rp1+Bv−RTpBv2∂v∂T,∂v∂T1+RT Bpv2=Rp1+Bv=vT,andα =1T1(1 + RT B/pv2)=1T(1 + B/v )(1 + 2B/v).For B>0, this is smaller than for the ideal gas.Or, writeT =pv/R1+B/v,∂T∂v=p/R1+B/v+(pv/R)(B/v2)(1 + B/v)2=(p/R)(1 + 2B/v)(1 + B/v)2=∂v∂T−1.This leads to the same result.5. Use the method of Lagrange multipliers to find the point on the circle x2+ y2=1thatis closest to the point (1,34).The squared distance is d2=(x − 1)2+(y −34)2, the constraint is φ = x2+ y2− 1 = 0, andthe method of Lagrange multipliers say find extrema ofF =(x − 1)2+(y −34)2+ λ(x2+ y2− 1).Extrema are where the derivatives vanish:∂F∂x=2(x − 1) + 2λx =0,∂F∂y=2(y −34)+2λy =0,∂F∂λ= x2+ y2− 1=0The first two of these sayx =11+λ,y=3/41+λ,andinthethirdthesegive1(1 + λ)21+916=1(1 + λ)22516=1,1(1 + λ)= ±45.This then gives the positions of the extremax = ±45,y= ±35.Clearly, + is the closest, − is the furthest. A plot of the geometry appears at the end.Bonus: use the method of Lagrange multipliers to find the point on the circle x2+ y2=1that is closest to the line 3x +4y =6.This time we need to find a point (x, y) on the circle and also another point, call it (x0,y0),on the line,F =(x − x0)2+(y − y0)2+ λ1(x2+ y2− 1) + λ2(3x0+4y0− 6).The extrema occur at∂F∂x=2(x − x0)+2λ1x =0,∂F∂y=2(y − y0)+2λ1y =0,∂F∂x0=2(x0− x)+3λ2=0,∂F∂y0=2(y0− y)+4λ2=0,∂F∂λ1= x2+ y2− 1=0,∂F∂λ2=3x0+4y0− 6=0.This is six equations in six unknowns. The middle pair can be solved forx − x0=3λ22,y− y0=2λ2;with this information the first pair givex = −3λ22λ1,y= −2λ2λ1;and in turn the fifth equation givesx2+ y2=9λ224λ21+4λ22λ21=25λ224λ21=1,λ2λ1= ±25.Thus,x = ±35,y= ±45.Once again, + is the nearest, − is the furthest. This works as long as λ16=0;ifλ1=0,then x = x0, y = y0, λ2=0,andx, y must solve both ofx2+ y2=1, 3x +4y =6,but this pair has only complex solutions.The blue points are at extremal distance (nearest and furthest) from the point (1,34), whichhappens to be on the line in the Bonus. The red points extremize distance from the line tothe