## 10.12.10

### On Equality and Inequality

Recall from the previous post that:

Equation (A)
$$L(x) + [(x - 2)/(x(x - 1))] = U(x) + [(x - 1)/(x(x + 1))] = 3$$

Furthermore, recall that:

$$L(q) < I(q^k) + I(n^2) <= U(q)$$

where $N = {q^k}{n^2}$ is a (hypothetical) Odd Perfect Number (OPN).

Lastly, recall that:

$$I(q) + I(n^2) = U(q)$$

I have some "pressing" questions at this point:

[[ #1 ]]

What happens to Equation (A) if I change $x$ to $x + 1$? (i.e. What happens if I do a "translation"?)

$$L(x + 1) + [(x - 1)/(x(x + 1))] = U(x + 1) + [x/(x + 1)(x + 2))] = 3$$

[[ #2 ]]

What happens to Equation (A) and the last equation if I let

$$x = P$$
and
$$x + 1 = Q$$
(i.e., $P + 1 = Q$)?

(i.e. What happens if I do a "restriction" after doing the "translation" in [[ #1 ]]?)

$$L(P) + [(P - 2)/(P(P - 1))] = U(P) + [(P - 1)/(P(P + 1))] = 3$$

$$L(Q) + [(Q - 2)/(Q(Q - 1))] = U(Q) + [(Q + 1)/(Q)(Q + 3))] = 3$$

Consequently:

$$L(P) + [(P - 2)/(P(P - 1))] = U(P) + [(P - 1)/(P(P + 1))]$$

equals

$$L(Q) + [(Q - 2)/(Q(Q - 1))] = U(Q) + [(Q + 1)/(Q)(Q + 3))]$$

[[ #3 ]]

What will happen to the last two equations if you interchange the "roles" of $P$ and $Q$? (In this sense, we are trying to check if $L(x)$ and $U(x)$ are "invertible" functions when restricted to integers $P$ and $Q$ satisfying $Q = P + 1$.)

So we go like:

$$L(Q) + [(Q - 2)/(Q(Q - 1))] = U(Q) + [(Q - 1)/(Q(Q + 1))]$$

$$L(P) + [(P - 2)/(P(P - 1))] = U(P) + [(P + 1)/(P)(P + 3))]$$

So it does seem to be the case (at least, to myself) that $L(x)$ and $U(x)$ are indeed "invertible".

For the record, the inverses of

$$L(x) = (3x^2 - 4x + 2)/[x(x - 1)]$$
and
$$U(x) = (3x^2 + 2x + 1)/[x(x + 1)]$$

are:

$${L^{-1}}(x) = [\sqrt{x^2 - 8} + (x - 4)]/[2(x - 3)]$$

$${U^{-1}}(x) = [\sqrt{x^2 - 8} - (x - 2)]/[2(x - 3)]$$

Note that the inverses satisfy the equation:

$${L^{-1}}(x) = {U^{-1}}(x) + [(x - 1)/(x - 3)]$$

or equivalently:

$${L^{-1}}(x) - {U^{-1}}(x) = 1 + [2/(x - 3)]$$
for all $x$ in $[5, \infty)$.

In particular, compare the resulting inequalities for the inverses:

$$1 < {U^{-1}}(x) < {L^{-1}}(x) < 2$$

with the inequalities for the functions:

$$2\sqrt{2} < L(x) < U(x) < 3$$

IN GENERAL, we have:

$$1 < [U^{-1}](x) < [L^{-1}](x) < L(x) < U(x) < 3$$
for all $$x \in [5, \infty).$$

LASTLY, note that:

$$L(x + 1) = U(x)$$
for all $x$ not in the set $\{-1, 0, 1\}$.

Graphically, we have:

where

=    OR   ===

both denote the usual equality symbol, and

|
|

denotes the usual inequality between the one below and the one above.

(i.e.

$b$
|
|
$a$

means that $a < b$).