Enmeshments

with some comments on proof

We already know how to condense an harmonic series into a formula that lets us calculate the value of the series without going through the tedium of calculating the value of each term in the series and then summing up those values. That condensation comes to us as a true blessing, especially when we have an especially long series to evaluate. Now I want to devise analogous condensations for enmeshments of harmonic sequences and for enmeshments of sequences of regular numerical coefficients with harmonic sequences.

Two harmonic sequences of equal length, one
ascending and one descending, give us our first enmeshment. We can combine (1,
x, x^{2},..., x^{n-1}, x^{n}) with (y^{n}, y^{n-1},...,
y^{2}, y, 1) by multiplying together the terms with the same order
number and adding the products to get the series

(Eq'n 1)

To condense that series we multiply each term by x/y to obtain

(Eq'n 2)

Solving that equation for σ_{1} gives
us the condensation that we want; that is,

(Eq'n 3)

Of course, our enmeshments won't always
originate in sequences that begin or end in one. We might have an harmonic
sequence that begins with x^{m} and ends with x^{m+n}. In such a
case we need only factor out x^{m}, treat the enmeshment as above, and
end up with a formula equal to the right side of Equation 3 multiplied by x^{m}.

For our next example I want to condense the
enmeshment of the harmonic sequence (x, x^{2}, ..., x^{n}) with
the sequence of the first n natural numbers. We have

(Eq'n 4)

Multiply that equation by x and subtract the product from it:

(Eq'n 5)

But the positive terms on the right side of that equation comprise the basic harmonic series minus one and the condensation of the basic harmonic series simply equals Equation 3 with y=1, so we can solve Equation 5 and obtain

(Eq'n 6)

We can test the validity of that formula easily enough. Just let x=2 and n=3. By direct calculation from Equation 4 we get

_{σ2}
= 2+8+24 = 34.

(Calc 1)

From Equation 6 we get

_{σ}_{2}
= (48+1) - (16-1) = 34.

(Calc 2)

This may seem a trivial point, but the
necessary introduction of ellipses into our descriptions of these series obliges
us to take special care in devising the algebraic description of the last term
in each series. This issue came up when I carried out the above calculations to
validate the condensations that I had devised for enmeshments more complicated
than the ones above and found that I had gotten them wrong. So now I take the
extra step of examining the first terms comprising my series, discerning the
relationship among the order indices and the exponents of the variables, and
then devising the last term in the series in accordance with that relationship:
I have thus made explicit what I had originally done intuitively. In Equation 4,
for example, I found that the order index of each term equaled the coefficient
and the exponent of x, so the n-th term, the last term, came out as nx^{n}.

Now let's enmesh the sequence of the first n natural numbers with the descending harmonic series minus one. We have

(Eq'n 7)

In this case the sum of the order index and the exponent of wye equals a constant, n+1. Divide that equation by y and subtract the quotient from it:

(Eq'n 8)

The positive terms on the right side of that
equation comprise the basic descending harmonic series minus one, so we can
replace those terms by Equation 3 (with x=1) minus one and then solve the result
for _{3} to obtain

(Eq'n 9)

Again, we can test the validity of that formula via calculation. In this case let y=3 and n=4. By direct calculation from Equation 7 we get

_{σ3}
= 81+54+27+12 = 174.

(Calc 3)

From Equation 9 we get

_{σ3}
= (729 - 3)/4 - 15/2 = 174.

(Calc 4)

Next let's enmesh the first n positive odd integers with a one-less, ascending harmonic sequence. We thus get the series

(Eq'n 10)

In this case the order index of each term equals the exponent of eks and the coefficient equals twice the order index minus one. Now multiply that series by x and subtract the result from it to get

(Eq'n 11)

On the right side of the equality sign the positive terms comprise two times the basic ascending harmonic series minus x+2, which doubled series we can replace with twice Equation 3 (with y=1), so we have

(Eq'n 12)

For our validity test of that equation let x=2 and n=4. By direct calculation we have

_{σ4}
= 2+12+40+112 = 166.

(Calc 5)

And by formula calculation we have

_{σ4}
= 224+4-62 = 166.

(Calc 6)

I'll leave condensing the enmeshment of the positive odd integers with the descending harmonic series to others because now I want to take the next step up in complexity and condense the enmeshment of the first n square numbers with the one-less ascending harmonic series. I thus get the series

(Eq'n 13)

As before, multiply that equation by x and subtract the product from it to get

(Eq'n 14)

That equation contains Equation 10, so I can solve it readily:

(Eq'n 15)

For the validity test let x=5 and n=3. By direct calculation we have

_{σ5}
= 5+100+1125 = 1230.

(Calc 7)

By formula calculation we have

(Calc 8)

With those examples I have made the procedure of condensing enmeshed series clear enough that the interested reader may condense yet other examples that I need not enumerate. I have also omitted something that we readily associate with the discipline of mathematics - I have offered no proof of my derivations, only pairs of calculations meant to verify that one formula yields the same results as does another. But we only need proof when we explore theoretical mathematics.

No proper mathematician would accept this essay as an exercise in theoretical (or pure) mathematics nor do I offer it as such. As in many of the essays in this series, I offer it as an example of applied mathematics. We tend to think of number theory, the subject of these essays, as belonging to the aetherial realm of theoretical mathematics, but every now and then we must visit the gritty realm of applied mathematics to forge new logical tools in the foundries of our imaginations. On that basis I can say that this essay serves two purposes:

a. that of devising condensed formulae equal to certain series made by enmeshing sequences of numbers, preparing them and the methods used to produce them for use in subsequent essays; and

b. that of exploring and refining our knowledge of the boundary between theoretical mathematics and applied mathematics.

We often think of applied mathematics as comprising only those mathematical disciplines that we apply to science, engineering, and other practical uses. But we can, indeed must, also apply mathematical knowledge within mathematics itself, so I need to make the following distinctions:

1. Theoretical mathematics comprises a search for significant new information about numbers and/or their inter-relationships. The search proceeds via predicate logic to find assertions of mathematical fact and necessarily requires logical proof, as by reductio ad absurdum, to ensure that those assertions can do nought else but come true to mathematics.

2. Applied mathematics consists of using the methods of calculation, however abstract, to transform one mathematical formula into another. We have no need for proof, but rather employ accounting methods to ensure that the new formula has properly absorbed all of the terms and factors in the original formula.

But surely those accounting methods constitute a kind of proof; after all, proof means testing and we have used simple accounting methods to test the validity of our condensations. So on what basis can I claim that those accounting methods do not constitute a proper proof?

Put most simply, predicate logic tells us to take two statements, P and Q, unite them in a compound statement, P implies Q, and then put that compound statement to the proof, either verifying it or falsifying it with complete certainty. Classically, proof shows three aspects:

1. Direct proof; that is, given P, we deduce Q
along a path that admits no alternatives that would negate our deduction. For
example, let's make P = "the n-th square number, n^{2}, equals the
number of markers in a uniform square array that has n markers on each of its
sides". Now we can disassemble that square array by moving the markers
diagonally to produce an array of chevrons: the numbers of markers in the
chevrons form a sequence; 1, 3, 5, 7, 9, and so on. Thus we infer the statement
Q = "the n-th square number, n^{2}, equals the sum of the first n odd
integers in the sequence of the natural numbers".

2. Indirect Proof; that is, given P, we assume
not-Q and deduce a necessary contradiction that leaves "P implies Q" as the only
possible statement true to mathematics. For this example, let P = "the set of
counting numbers contains an infinite number of prime numbers". That statement
implies Q = "the counting numbers contain no last (largest) prime number". Now
assume that not-Q is true to mathematics; "the set of counting numbers contains
a largest prime number, N_{p}, and no prime numbers larger". Take the
set of all prime numbers n_{p} up to and including N_{p},
multiply those numbers together, and add one to create the number N_{p}'.
If we divide N_{p}' by any of the prime numbers n_{p} that we
multiplied together to create it, we always get a fractional remainder, 1/n_{p};
therefore, N_{p}' is a prime number (note that any composite number can
always be evenly divided by some prime number smaller than itself). Thus, from
not-Q we have deduced a contradiction of not-Q as a necessary consequence, which
makes not-Q false to mathematics; therefore, Q must be true to mathematics and
that truth entails P also being true to mathematics.

3. Proof by Contraposition; that is, given P and Q, we assume not-Q, deduce not-P, and invoke the statement "if not-Q implies not-P, then P implies Q" (if not getting to work on time implies that I did not catch the 6:30 bus, then my catching the 6:30 bus implies that I get to work on time). For a mathematical example of a contrapositive of an implication we have:

A number n does not evenly divide a number m
[not-Q] implies that the square of that number, n^{2}, does not evenly
divide m [not-P]; therefore, the statement that n^{2} evenly divides m
[P] implies the statement that n evenly divides m [Q].

In contrast to the deductive method of proof
we have in our accounting methods more induction than deduction and our level of
certainty is different.

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