∫dx/x

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 a_{n}. 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 c_{m} as a
factor and the n-th column contains all terms that have a_{n} as a
factor. Now imagine rotating the matrix forty-five degrees clockwise, putting
the a_{0}c_{0} 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 a_{0}c_{0}=1, because if we had a_{0}c_{0}=0,
then all of the a_{n} and all of the c_{m} would equal zero.
Thus we have

(Eq’n 10)

Because all of the a_{n} take a plus sign in their
series, that equation can only stand true to mathematics if at least some of the
c_{m} 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 c_{m} 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 a_{k}c_{j}=-a_{j}c_{k}.
In order for the series representing B to converge successive terms must become
progressively smaller, so a_{k}≠a_{j}
if k≠j. Thus we have a_{k}=±c_{k}, the minus sign occurring when
k is an odd number. Because all of the a_{k} 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 a_{k} 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,

e=2.7182818284590...,

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 T_{0} to T_{1}, 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 T_{0} at which that occurs
cannot equal zero: that fact necessitates that there be a temperature below T_{0}
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=(T_{1}/T_{0})^{N}
and thus, by substituting that expression into Equation 18 we have Boltzmann's
Equation

S=klnW.

(Eq'n 19)

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