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Let's start with the infinite series

(Eq'n 1)

in which we have used the usual technique of evaluating an infinite series to get the last term. Next integrate the series with respect to x and get

(Eq'n 2)

We now want to explore the properties of F(1+x).

We know right away that for x = 0, F(1) = 0 and that for x = -1, F(0) = -∞.

If we make the substitution 1+x = (1+y)(1+z), we have

(Eq'n 3)

which we divide by (1+y)(1+z) to get

(Eq'n 4)

so now we know that

(Eq'n 5)

That statement combined with the statement that F(1) = 0 tell us that

(Eq'n 6)

Thus, we infer that the function F has the properties of a logarithm.

To what base must we refer that logarithm? That is, for what number B does F(B) = 1? We know that we can always represent a number as a convergent infinite series, so let's define our base as

(Eq'n 7)

We know that we have an infinite range of possible values for the numbers an. Most of those values will equal very small fractions and some may be negative numbers. In order to determine what those values are we define

(Eq'n 8)

and multiply it by Equation 7 to get a partial product matrix

(Eq'n 9)

    If we carry out the indicated multiplication in that equation, we can array the partial products in a matrix in which the m-th row contains all terms that have cm as a factor and the n-th column contains all terms that have an as a factor. Now imagine rotating the matrix forty-five degrees clockwise, putting the a0c0 term at the apex of an infinite triangle. Each row of that triangle consists of the terms for which m+n=p.

    Now we have an infinite set of equations. All of the triangle’s rows except one must add up to zero. That exception must be the zeroth row, which gives us a0c0=1, because if we had a0c0=0, then all of the an and all of the cm would equal zero. Thus we have

(Eq’n 10)

    Because all of the an take a plus sign in their series, that equation can only stand true to mathematics if at least some of the cm take a minus sign. We can rewrite the series of Equation 7 as

(Eq’n 11)

so that we have Equation 8 as

(Eq’n 12)

Because f contains only plus signs we can say that alternate terms in the series take a minus sign; specifically, the odd-indexed cm take the minus sign, so we can simply multiply each term in the series by (-1)m.

    For all j+k=p yielding an odd number (the p-th row having an even number of elements) we have akcj=-ajck. In order for the series representing B to converge successive terms must become progressively smaller, so akaj if k≠j. Thus we have ak=±ck, the minus sign occurring when k is an odd number. Because all of the ak are positive numbers, we rewrite that result as

(Eq’n 13)

    Alternating plus and minus signs in rows that add up to zero is the pattern we see in Pascal’s triangle for calculating the binomial coefficients of (1-1)p. Further, the values of the numbers in one row determine the values of the numbers in the next row, as in Pascal’s triangle, so we assume that our triangle is some version of Pascal’s triangle. In Pascal’s triangle the q-th element of the p-th row has the form p!/(p-q)!q!. In our triangle we divide p! out of the elements of each row because the fact that the row adds up to zero allows us to do so (as we would do in solving any algebraic equation) and because we must do so in order that ak have the same values for all values of p. Thus we have for any given element

(Eq’n 14)

That gives us the obvious result


(Eq'n 15)

which means that we have at last

(Eq'n 16)

We recognize that equation as expressing Euler's number,


the base of the natural (or Naperian) logarithms.

Now we can make the substitution t=1+x and dt=dx and write the logarithm as mathematicians do to make the definition

(Eq'n 17)

Example Application: Entropy

If we have a thermodynamic system at some absolute temperature T and we change the system's energy by the amount dE and the system's temperature changes by dT in consequence, then we can ascribe to the system an entropy defined as S = dE/dT. If we change the energy of the system without changing the temperature, we have dS = dE/T. Thus, we have taken E=ST, differentiated it (dE=SdT+TdS), and taken the cases dS=0 and then dT=0.

We know that the temperature of a thermodynamic system reflects the average energy of the particles comprising the system; that is, E=kT, in which k, Boltzmann's constant, through the indicated multiplication converts units of temperature into units of energy. For an N-particle system, then, we can describe the energy contained by the system as E=NkT. If we change the energy of the system, we thus have dE=NkdT. If we change the energy of the system by an amount that changes the system's temperature from T0 to T1, then we change the system's entropy by

(Eq'n 18)

In that system the number of ways in which a particle can participate in sharing the energy of the system, from having all of the energy to having none at all, however large that number may be, must be proportional to the average energy of the system, proportional to the system's temperature. If the system were to contain no energy, then each particle would have only one way to participate in the energy share-out. We know by Nernst's theorem that the temperature T0 at which that occurs cannot equal zero: that fact necessitates that there be a temperature below T0 separated from it by some quantity of energy that absolutely cannot be removed from the system (the zero-point energy). So the number of ways that the system can distribute its energy among N particles is W=(T1/T0)N and thus, by substituting that expression into Equation 18 we have Boltzmann's Equation


(Eq'n 19)


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