golden mean, golden section, or extreme and mean ratio, n. the proportion of the division of a line so the the smaller is to the larger as the larger is to the whole, or of the sides of a rectangle so the the ratio of their difference to the smaller equals that of the smaller to the larger, supposed in classical aesthetic theory to be uniquely pleasing to the eye. This yieldsG = (√5 − 1) / 2 = 0.618033988... of which the inverse is 1.618033988... = G+1, which is also sometimes referred to as the golden ratio. It is a consequence of the definition that if one draws a rectangle with sides in the golden ratio (a golden rectangle), and then removes from it a square, the rectangle that remains has the same proportions as the original. If this process is repeated as shown in Fig. 172, then the successive points of division lie on a logarithmic spiral. The golden mean is also the limit both of the continued fraction1/(1 + 1/(1 + 1/(1 + ...))), and of the ratio of successive terms of the Fibonacci numbers.
—Dictionary of Mathematics
Define F(x) and its complement G(x) from the equations F(x)−1 = F(x) − x, and G(x)−1 = G(x) + x, to give the formulae:
F(x) = ( √(x²+4) + x ) / 2
G(x) = ( √(x²+4) − x ) / 2
Where G(1) = G, the golden mean, and F(1) is the golden ratio. The rest of this document goes on to explore the properties for F(x) and G(x). For the most part, details of the derivation of terms have been left out if they consist only of simple algebra (I assumed that readers could replicate these elementary results themselves).
F(−x) = F(x) − x = G(x)
G(−x) = G(x) + x = F(x)
F(x) + G(x) = √(x²+4)
F(x) − G(x) = x
F(x)−1 = F(x) − x = G(x)
G(x)−1 = G(x) + x = F(x)
F(x)² = 1 + xF(x)
G(x)² = 1 − xG(x)
F(x) × G(x) = 1
F(x) ÷ G(x) = F(x)²
G(x) ÷ F(x) = G(x)²
Let Pβ(α) be a polynomial of α to the order of max(β−1,0), such that Pβ+1(α) = αPβ(α) + Pβ−1(α).
If F(x)n = Pn−1(x) + Pn(x)F(x) (which holds when n is 1 and 2 from above),
then F(x)n+1 = Pn(x) + [Pn−1(x) + xPn(x)]F(x) = Pn(x) + Pn+1(x)F(x) (which therefore holds for all positive integer n's).
It turns out the Pβ(α) is constructed such that P0(α)=0 and the coefficient of α of the order p∈{0,1,2,...} in Pβ(α) (β>0) is the (p+1)th entry in the ½(p+β+1)th row of Pascal's triangle if (p+β) is odd, and is 0 if (p+β) is even.
Rearranging from above, we can also say Pβ−1(α) = Pβ+1(α) − αPβ(α).
Now, if F(x)n = Pn−2(x) + Pn−1(x)F(x) (which holds when n is a positive integer),
then F(x)n−1 = [Pn−1(x) − xPn−2(x)] + Pn−2(x)F(x) = Pn−3(x) + Pn−2(x)F(x) (which therefore holds for all negative integer n's).
For β≤0, Pβ(α) is constructed such that Pβ(α) = P−β(α) if β is odd, and Pβ(α) = −P−β(α) if β is even.
To obtain the similar formulae for G(x), we can simply use the substitution F(x) = G(x)−1.
G(x)−n = Pn−1(x) + Pn(x)G(x)−1 = Pn−1(x) + Pn(x)[G(x) + x] = [xPn(x) + Pn−1(x)] + Pn(x)G(x) = Pn+1(x) + Pn(x)G(x)
G(x)n = P−n+1(x) + P−n(x)G(x) = (−1)nPn−1(x) + (−1)n+1Pn(x)G(x)
In general, for integer n,
F(x)n = Pn−1(x) + Pn(x)F(x)
F(x)−n = (−1)n+2Pn+1(x) + (−1)n+1Pn(x)F(x)
G(x)n = (−1)nPn−1(x) + (−1)n+1Pn(x)F(x)
G(x)−n = Pn+1(x) + Pn(x)G(x)
F(x)m × G(x)n = (F(x)n × G(x)n) × F(x)m−n = F(x)m−n = G(x)n−m