### The silver ratio and a continued fraction for log(2)

Define the three sequences,

$C_n = \sum_{k=0}^n {\binom n k} {\binom {n+k}k} = 1, 3, 13, 63, 321,\dots$

$B_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k} = 1, 3, 19, 147, 1251,\dots$

$A_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k}^2 = 1, 5, 73, 1445, 33001, \dots$

The last two are Apery numbers and have been discussed previously. The first are the central Delannoy numbers which obeys the limit,

$\lim_{n \to \infty} \frac{C_{n+1}}{C_n} = \left(1+\sqrt{2}\right)^2 = 5.8284\dots$

which is the square of the silver ratio.  (The ratios for the others have already been mentioned.)  These have the recurrence relations,

$n C_n = 3(2n-1)C_{n-1}-(n-1)C_{n-2}$

$n^2 B_n = (11n^2-11n+3)B_{n-1}+(n-1)^2B_{n-2}$

$n^3 A_n = (34n^3-51n^2+27n-5)A_{n-1}-(n-1)^3A_{n-2}$

To recall, the polynomials $P(n) =11n^2-11n+3$ and $P(n) = 34n^3-51n^2+27n-5$  generated numbers for the continued fractions of $\zeta(2), \zeta(3)$, so I was curious if $P(n) = 3(2n-1)$ could be used in an analogous manner.  It turns out, depending on what sign to use, it gives either $\log(2)$ or $\arctan(\frac{1}{3})$,

$\frac{1}{2}\log(2) = \cfrac{1}{v_1 - \cfrac{1^2}{v_2 - \cfrac{2^2}{v_3 - \cfrac{3^2}{v_4-\ddots}}}}$

or,

$\arctan(\frac{1}{3}) = \cfrac{1}{v_1 + \cfrac{1^2}{v_2 + \cfrac{2^2}{v_3 + \cfrac{3^2}{v_4 + \ddots}}}}$

where, starting with n = 1,

$v_n = 3(2n-1)$

This can be partly demystified since one continued fraction for the natural logarithm, and arctan, is,

$\frac{1}{2}\log\left(1+\frac{2x}{y}\right)= \cfrac{x}{1(x+y) - \cfrac{(1x)^2}{3(x+y) - \cfrac{(2x)^2}{5(x+y) - \cfrac{(3x)^2}{7(x+y) - \ddots}}}}$

and,

$\arctan(\frac{x}{z}) = \cfrac{x}{1z + \cfrac{(1x)^2}{3z + \cfrac{(2x)^2}{5z + \cfrac{(3x)^2}{7z + \ddots}}}}$

Hence, if $x+y = z$, as with the case {x, y, z} = {1, 2, 3}, then there will be identical-looking continued fractions that differ only in the signs.  But it remains interesting how the recurrence relations of these three binomial sums are involved in the continued fractions of $\log(2), \zeta(2), \zeta(3)$.  Later, we shall see there is a recurrence relation for the cfrac of $\zeta(4)$ as well.

### The equation ap^2+bq^2+cr^2+ds^2 = 0

In the previous post, it was discussed that an initial non-trivial integer solution to the diagonal quartic surface,

$ax_1^4+bx_2^4+cx_3^4+dx_4^4 = 0$

apparently does not help in determining if the equation has an infinite number of primitive integer solutions.  However, it is different for its quadratic and cubic counterparts.

Theorem 1:  “In general, given one non-trivial solution to the quadratic,

$a_1y_1^2+a_2y_2^2+\dots+a_ny_n^2 = 0$

then an infinite more can be found.”

Theorem 2:  “Likewise, given one non-trivial solution to the cubic,

$a_1y_1^3+a_2y_2^3+\dots+a_ny_n^3 = 0$

then an infinite more can be found.”

Proof of Theorem 1  (Theorem 2 will be discussed in the next post):

We’ll start with n = 3,4 and prove the rest by induction. There is the identity,

$ax_1^2+bx_2^2+cx_3^2 = (ay_1^2+by_2^2+cy_3^2)\,(az_1^2+bz_2^2+cz_3^2)^2$

where,

$(x_1, x_2, x_3) = (uy_1-vz_1, uy_2-vz_2, uy_3-vz_3)$

and,

$(u,v) = (az_1^2+bz_2^2+cz_3^2,\, 2(ay_1z_1+by_2z_2+cy_3z_3)\,)$

Thus, if one has an initial solution $(y_1, y_2, y_3)$, then the RHS of the identity becomes zero, and one gets a parametrization in the $(x_1, x_2, x_3)$ for three free variables $(z_1, z_2, z_3)$. An example is given in this page, form 2b.  For n = 4, it is just a generalization,

$ax_1^2+bx_2^2+cx_3^2+dx_4^2 = (ay_1^2+by_2^2+cy_3^2+dy_4^2)(az_1^2+bz_2^2+cz_3^2+dz_4^2)^2$

where,

$(x_1, x_2, x_3, x_4) = (uy_1-vz_1, uy_2-vz_2, uy_3-vz_3, uy_4-vz_4)$

and,

$(u,v) = (az_1^2+bz_2^2+cz_3^2+dz_4^2,\, 2(ay_1z_1+by_2z_2+cy_3z_3+dy_4z_4)\,)$

The pattern is easily seen for n = 5,6, ad infinitum.  If only it was that easy for the quartic case.

### Quartic equations

Composite degrees n are better solved by polynomial decomposition.  However, for $n = 2^m$, there is an alternative method.  Thus, we’ll give two methods to solve the quartic.  Given the depressed quartic,

$x^4+px^2+qx+r = 0$

Its resolvent cubic is simply,

$y^3+2py^2+(p^2-4r)y-q^2=0$

Method 1:

Given any root y of the cubic, then,

\begin{aligned} x_{1,2} &=\tfrac{1}{2} \left(\sqrt{y} \pm \sqrt{-(2p+y)-\tfrac{2q}{\sqrt{y}}}\,\right)\\ x_{3,4} &= \tfrac{1}{2} \left(-\sqrt{y} \pm \sqrt{-(2p+y)+\tfrac{2q}{\sqrt{y}}}\, \right)\end{aligned}

For convenience, one can choose the real root y.  Notice that {$x_3, x_4$} are the negative case of $\pm \sqrt{y}$.  The formula can easily be proven by squaring the variable y, forming the quadratic in {$x_1,\,x_2$}, and eliminating y between them using resultants,

$\text{Factor}[\text{Resultant}[y^6+2py^4+(p^2-4r)y^2-q^2,\,x^2+xy+\frac{y^3+py-q}{2y},\,y]]\\ = (x^4+px^2+qx+r)^3$

proving any root of the sextic (a cubic in $y^2$) will suffice.

Method 2:

Given the three roots {$y_1, y_2, y_3$} of the resolvent cubic, then,

\begin{aligned} x_{1,2} &= \frac{\sqrt{y_1} \pm (\sqrt{y_2}+\sqrt{y_3}\,)}{2}\\ x_{3,4} &= \frac{-\sqrt{y_1} \pm (\sqrt{y_2}-\sqrt{y_3}\,)}{2}\end{aligned}

where signs are chosen such that,

[1]  $\sqrt{y_1}\sqrt{y_2}\sqrt{y_3} = -q$

Proof:  Expanding,

$(x-x_1)(x-x_2)(x-x_3)(x-x_4) = 0$

using the formulas, we get,

$x^4-\frac{1}{2}(y_1+y_2+y_3)x^2-\sqrt{y_1}\sqrt{y_2}\sqrt{y_3}\,x\\+\frac{1}{16}(y_1^2+y_2^2+y_3^2-2(y_1y_2+y_1y_3+y_2y_3)) = 0$

From [1] and the properties of elementary symmetric polynomials, we find that it is in fact equal to,

$x^4+px^2+qx+r = 0$

A similar method was used to solve the octic $x^8-x^7+29x^2+29 = 0$ in this post, though now one had to take square roots of its resolvent septic.

### Cubic equations

Recall that solving the quadratic equation,

$x^2+ax+b = 0$

gives,

$x = \frac{1}{2}\,(-a \pm \sqrt{y}\,)$

where y is the linear,

$y = a^2-4b$

The solution to the cubic equation is simply a generalization. Given,

$x^3+ax^2+bx+c=0$

then,

$x =\frac{1}{3}\,(-a+y_1^{1/3}+y_2^{1/3})$

where the $y_i$ are now the two roots of the quadratic,

$y^2+(2a^3-9ab+27c)y+(a^2-3b)^3 = 0$

Notice that the constant term of its resolvent is a perfect cube.  For prime degrees, if the irreducible equation is solvable, then it can be solved analogously. For depressed quintics (an easy linear transformation to make a = 0), then,

$x = \frac{1}{5}\,(y_1^{1/5}+y_2^{1/5}+y_3^{1/5}+y_4^{1/5})$

where the $y_i$ are now the four roots of its resolvent quartic. For non-binomial quintics, this quartic will have a constant term that is a perfect fifth power. And so on for other prime degrees.

### Solvable sextics

In a previous post, we discussed the Bring-Jerrard quintic,

$x^5+ax+b=0$

and rational parametrizations for {a, b}.  It turns out this has a sextic version. To illustrate, these irreducible equations are solvable,

$\text{(1)}\;\; x^6+3x+3 = 0$

$\text{(2)}\;\; x^6-7^2x-7^2 = 0$

However, each needs a different approach. The first factors into two cubics over $\sqrt{-3}$, while the second factors into three quadratics whose coefficients are determined by a cubic.  These two examples belong to two infinite (but not complete) families.

(1)  1st family:

$x^6+4mn(m^2-3n)x+n(m^4+18m^2n+n^2) = 0$

This factors over $\sqrt{-n}$.  Let {m, n} = {-1/2, 3/4}, and it will yield (1).

(2)  2nd family:

The second is trickier.  Given,

$x^2+vx+(pv^2-qv-nr) = 0$

$v^3-rv-r = 0$

Eliminating v between them using resultants (easily done in Mathematica or Maple), results in a sextic in the variable x of form,

$x^6+ax+b = 0$

if,

\begin{aligned}p &=\frac{3n+1}{2}\\q &=\frac{9n+1}{4}\\r &= \frac{(9n+1)(9n-7)}{4(3n+1)(n-1)} \end{aligned}

Let n = 3, we get {p, q, r} = {5, 7, 7}, and it yields (2).  Thus, to solve,

$x^6-7^2x-7^2 = 0$

entails solving the cubic in v,

$v^3-7v-7 = 0$

though other n do not show this nice symmetry.  Another good value is at n = 1/3 which gives,

$x^6+3x+5 = 0$