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Identities: ex=n=0n!xn=1+x+2!x2+3!x3+4!x4ex=n=0n!xn=1x+2!x23!x3+4!x4 ln(AB)=ln(A)+ln(B)ln(ab)=bln(a)xln(f(x))=f(x)1xf(x)xln(1eax)=(eax1)a0ceax2x=21aerf(ca)0eax2x=21a0xeax2x=2a10xeaxx=a10x2eaxx=a320x3eaxx=a460x4eaxx=a524x2eax2x=21a3eax2x=a0x2eax2x=41a3 5. There is a subject called statistical mechanics that studies the normalizer of the Boltzmann equation: Norm=ekBTE We call =kBT1 for convenience, which makes Norm =eE. Statistical mechanics shows that the expression: ln(Norm) is equal to the average energy (E). For this question we will demonstrate that ln(Norm)=E. (On a side note, E) is the same thing as internal energy U that you learned about in PChem I). a. First please simplify the log derivative xln(f(x)) to remove the In function. b. Now apply the formula from pt. a to ln(Norm()). Not only do you have to apply the identity from pt. a you also need to evaluate Norm(), given that: Norm=eE. Hint: a derivative isn't affected by the sum- you can pretend it's not there. Here is an example: catnsin(ncat)=ncatsin(ncat)=nncos(ncat) c. Can you show that your expression from pt. b is equal to (E) given that E)=EP(E), where P(E) is the probability defined by the Boltzmann equation P(E)=NormeAE