Posts Tagged ‘quadratic’

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.

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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