Cardano Tangent

This is the eigth part of a twelve-part series for comparing real estate to other investments like stocks. This part improves the accuracy of the optimum point calculation by revisiting assumptions 3-5 of the last article.

For $ln (G + 1)$ , we assume it can be approximated by a first order Taylor Series about a performance of zero ( $G \approx 0$ ). Thus the optimization function becomes:

n \frac{d G}{d n} = (G + 1) G

3. The performance $G = g + 1$ was defined such that it is cumulative over time, not a geometric average. Thus the further the optimum point from when transaction fees are made up, the further G will deviate from 0 and the less accurate substituting $ln g \approx G$ becomes. We improve the accuracy by using a second order Taylor series about $g = 1$ , thus the optimization function becomes:

n \frac{d G}{d n} = (G + 1) (G - \frac{1}{2} G^{2})

The power functions that make up G can be approximated by first order Maclaurin series, and assumption (2) for the resulting logarithm can be reapplied:

x^{n} \approx 1 + n ln x \approx 1 + X n

4. We replace first-order Maclaurin series with a first-order Taylor series about the quadratic optimum point (τ) for all power functions, but retain assumption (2) to replace instances of $ln x$ with $X = x - 1$ for inflation, nominal and real opportunity cost (π, ω and ρ).

x^{n} \approx x^{τ} + x^{τ} ln x (n - τ) \approx x^{τ} (X n + 1 - X τ)

We assume real interest rates of zero, thus

\frac{Σ_{n}}{p} = n

5. Instead of assuming the real interest rate ( $R = i - π$ ) is zero, we assume the nominal interest rate (i) is not one but close to one such that geometric series edge case can be ignored but interest rate power series can be approximated like inflation and opportunity costs above. Thus the rate of equity buildup becomes:

\frac{Σ_{n}}{p} = n + \frac{i R}{I} [\frac{i^{1 - N}}{I} (i^{n} - 1) - n] where I = i - 1

The performance of a mortgage over n months was found previously.

G_{n} = \frac{L}{R_{N}} (c ω (\frac{ω^{n} - 1}{ω - 1}) - ρ (\frac{ρ^{n} - 1}{ρ - 1}) + \frac{\sum_{n}}{p} - T) for ω > π > 1

First we substitute the rate of equity buildup.

G_{n} = \frac{L}{R_{N}} (c ω (\frac{ω^{n} - 1}{ω - 1}) - ρ (\frac{ρ^{n} - 1}{ρ - 1}) + n (1 - \frac{i R}{I}) + \frac{R i^{2 - N}}{I} (\frac{i^{n} - 1}{I}) - T)

\frac{d G_{n}}{d n} \approx \frac{L}{R_{N}} (c ω^{n + 1} - ρ^{n + 1} + 1 - \frac{i R}{I} (1 - \frac{i^{n + 1}}{i^{N}}))

Next we make the first-order Taylor series approximations for each power series. To help keep things managable, for ω, ρ and i in place of x, we define alternate time-shifts, $τ_{x} = \frac{x^{τ} - 1}{x - 1}$ .

G_{n} = \frac{L}{R_{N}} [c ω (ω^{τ} n + τ_{ω} - ω^{τ} τ) - ρ (ρ^{τ} n + τ_{ρ} - ρ^{τ} τ) + n (1 - \frac{i R}{I}) + \frac{R i^{2 - N}}{I} (i^{τ} n + τ_{i} - i^{τ} τ) - T]

\frac{d G_{n}}{d n} \approx \frac{L}{R_{N}} [c ω^{τ + 1} (Ω n + 1 - Ω τ) - ρ^{τ + 1} (Ρ n + 1 - Ρ τ) + 1 - \frac{i R}{I} (1 - \frac{i^{τ + 1}}{i^{N}} (I n + 1 - I τ))]

We now have linear equations $G_{n} = A (a n - b)$ and $\frac{d G_{n}}{d n} = A (a + d n - d τ)$ , where

A = \frac{L}{R_{N}}

a = c ω^{τ + 1} - ρ^{τ + 1} + 1 - \frac{i R}{I} (1 - i^{τ + 1 - N})

b = T - [c ω (τ_{ω} - ω^{τ} τ) - ρ (τ_{ρ} - ρ^{τ} τ) + \frac{R i^{2 - N}}{I} (τ_{i} - i^{τ} τ)]

d = c ω^{τ + 1} Ω - ρ^{τ + 1} Ρ + R i^{τ + 2 - N}

and

τ_{x} = \frac{x^{τ} - 1}{x - 1}

for x = ω, ρ and i.

Next is substituting $G_{n} = A (a n - b)$ and $\frac{d G_{n}}{d n} = A (a + d n - d τ)$ into the cubic optimization equation.

0 = G^{3} - G^{2} - 2 G + 2 \frac{d G}{d n}

0 = A^{3} a^{3} n^{3} - 3 A^{3} a^{2} b n^{2} + 3 A^{3} a b^{2} n - A^{3} b^{3}

- (A^{2} a^{2} n^{2} - 2 A^{2} a b n + A^{2} b^{2})

- (2 A a n - 2 A b) + 2 A a n + 2 A d n^{2} - 2 A d τ n

Removing the common term, collecting equal power terms and dividing through by (Aa)³ leaves a normalized cubic equation with three new coefficients.

0 = n^{3} + B n^{2} + C n + D

where

B = \frac{- 3 A^{3} a^{2} b - A^{2} a^{2} + 2 A d}{A^{3} a^{3}}

C = \frac{3 A^{3} a b^{2} + 2 A^{2} a b - 2 A d τ}{A^{3} a^{3}}

D = \frac{- A^{3} b^{3} - A^{2} b^{2} + 2 A b}{A^{3} a^{3}}

The Cardano method is based on two observed factorizations of a cubic root.

{(x - y)}^{3} = x^{3} - 3 y x^{2} + 3 y^{2} x - y^{3}

{(x - y)}^{3} = x^{3} - 3 x y (x - y) - y^{3}

From the first observed factorization, the quadratic term is depressed by shifting the cubic equation right by $n = m - \frac{b}{3}$ .

0 = m^{3} - 3 \frac{B}{3} m^{2} + 3 \frac{B^{2}}{9} m - \frac{B^{3}}{27} + B (m^{2} - 2 \frac{B}{3} m + \frac{B^{2}}{9}) + C (m - \frac{B}{3}) + D

and this depressed cubic is rewritten into a standard form, $x^{3} + 3 p x + 2 q = 0$ , where:

p = \frac{1}{3} (C - \frac{B^{2}}{3}) = \frac{C}{3} - {(\frac{B}{3})}^{2}

q = \frac{1}{2} (\frac{2 B^{3}}{27} - \frac{B C}{3} + D) = {(\frac{B}{3})}^{3} - \frac{C}{2} (\frac{B}{3}) + \frac{D}{2}

The next step of the Cardano method utilizes the second observed factorization to separate the depressed cubic equation into two simultaneous equations with two new variables, by first letting $x = u - v$ .

u^{3} - 3 u v (u - v) - v^{3} + 3 p (u - v) + 2 q = 0

Since we have introduced an independent variable (either u or v), we can assert $u v = p$ as a second independent equation.

u^{3} + 2 q = v^{3}

u^{3} v^{3} = p^{3}

This is not a linear system of equations, but is quadratic and can be solved by substitution and completing the square.

u^{3} (u^{3} + 2 q) = p^{3}

u^{6} + 2 u^{3} q + q^{2} = p^{3} + q^{2}

{(u^{3} + q)}^{2} = p^{3} + q^{2}

u^{3} = - q \pm \sqrt{p^{3} + q^{2}}

v^{3} = + q \pm \sqrt{p^{3} + q^{2}}

If $p^{3} + q^{2} \geq 0$ then the result is real and because both solutions shift from the same absolute values about zero, it can be traced through both cases that either choice of square root plus or minus lead to the same value x.

x = \sqrt[3]{- q - \sqrt{p^{3} + q^{2}}} - \sqrt[3]{q - \sqrt{p^{3} + q^{2}}}

Finally we undo the shift that depressed the equation.

n = - (\frac{B}{3} + \sqrt[3]{q + \sqrt{p^{3} + q^{2}}} + \sqrt[3]{q - \sqrt{p^{3} + q^{2}}})