## Posts Tagged ‘identities’

### Algebraic values of the Hypergeometric function

\begin{aligned} &\,_2F_1(a,b,c,z) = \sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n} \frac{z^n}{n!}\end{aligned}

where $(a)_n$ is a Pochhammer symbol is, for rational {a, b, c, z}, generally not a rational value.  In the link given above, equations (18) and (19) are,

$\,_2F_1\big(\frac{1}{3},\frac{2}{3},\frac{5}{6},\frac{27}{32}\big) =\frac{8}{5}$

$\,_2F_1\big(\frac{1}{4},\frac{1}{2},\frac{3}{4},\frac{80}{81}\big) =\frac{9}{5}$

There are in fact an infinite number of such equalities.  One given by M. Glasser is, let $0 < x < \frac{1}{\sqrt{3}}$, then,

$\,_2F_1\big(\frac{1}{3},\frac{2}{3},\frac{3}{2},\frac{27x^2(1-x^2)^2}{4}\big) =\frac{1}{1-x^2}$

Another, based on eq. (42) of Vidunas’ “Transformations of algebraic Gauss hypergeometric functions” is, let $0 < y < 1$, then,

$\,_2F_1\big(\frac{1}{4},\frac{-1}{12},\frac{2}{3},\frac{(9-y^4)^3(-1+y^4)}{64y^{12}}\big) =\frac{1}{y}$

A third which yields not a rational but an algebraic number is remarkable for its connection to the Rogers-Ramanujan continued fraction. Let $0 < z < v_1$ where,

$v_1 = \text{Root}[z^4+228z^3+494z^2-228z+1=0] = 0.004428\dots$

then,

$\,_2F_1\big(\frac{19}{60},\frac{-1}{60},\frac{4}{5},\frac{1728z(z^2-11z-1)^5}{(z^4+228z^3+494z^2-228z+1)^3}\big) =\frac{1}{(z^4+228z^3+494z^2-228z+1)^{1/20}}$

If the polynomials are familiar, it is because they are invariants of the icosahedron.  They also appear in the j-function formula,

$j(\tau) = \frac{-(r^{20}-228r^{15}+494r^{10}+228r^5+1)^3}{r^5(r^{10}+11r^5-1)^5}$

where,

$r = r(\tau) = \cfrac{q^{1/5}}{1+ \cfrac{q}{1 + \cfrac{q^2}{1+ \cfrac{q^3}{1 + \ddots}}}}$

and

$q = e^{2\pi i \tau}$

### On Bailey and Crandall’s unusual sum

In page 20 of Bailey and Crandall’s On the Random Character of Constant Expansions, they give the wonderfully unusual sum,

\begin{aligned} x_1 = \sum_{n=0}^\infty \frac{1}{5^{5n}}\left(\frac{5}{5n+2}+\frac{1}{5n+3}\right)&=\frac{5^2}{2}\ln\left(\frac{781}{256}\left(\frac{57-5\sqrt{5}}{57+5\sqrt{5}}\right)^{\sqrt{5}}\right)\\ &= 2.833601\dots\end{aligned}

I didn’t think this was an isolated result so set about to find a generalization.  I found its counterpart,

\begin{aligned} x_2 = \sum_{n=0}^\infty \frac{1}{5^{5n}}\left(\frac{5^3}{5n+1}+\frac{1}{5n+4}\right)&=\frac{5^3}{2}\ln\left(\frac{781}{256}\left(\frac{57+5\sqrt{5}}{57-5\sqrt{5}}\right)^{\sqrt{5}}\right)\\ &= 125.256703\dots\end{aligned}

Note that,

$(57+5\sqrt{5})(57-5\sqrt{5}) = 4(781) = 5^5-1$

We can demystify the sum a bit by splitting the log function into parts. After some algebraic manipulation, we find that the first one becomes,

\begin{aligned}&x_1 = 25\Big(-5\ln(2)+\tfrac{1-\sqrt{5}}{2}\,\ln(57+5\sqrt{5})+\tfrac{1+\sqrt{5}}{2}\,\ln(57-5\sqrt{5})\Big)\end{aligned}

Thus it can be expressed in the form,

$x_1 = r\ln\big({p_1}^{q_1}\,{p_2}^{q_2}\big)$

where {$p_1, p_2$} are roots of the same equation, {$q_1, q_2$} are roots of another, and r is a rational.  The fact that,

$2\cos(\pi/5) = \frac{1+\sqrt{5}}{2}$

was my clue that trigonometric functions may be involved.  Define,

$c_k = -2\cos(2k\pi/p)$

then for p = 5,

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{5^{5n}}\left(\tfrac{5}{5n+2}+\tfrac{1}{5n+3}\right) &= 5^2\ln\big((5c_1+26)^{c_2}(5c_2+26)^{c_1}/4^2\big)\\&=2.833601\dots\end{aligned}

p = 7

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{7^{7n}}\left(\tfrac{7}{7n+3}+\tfrac{1}{7n+4}\right) &= 7^3\ln\big((7c_1+50)^{c_3}(7c_2+50)^{c_1}(7c_3+50)^{c_2}/6^2\big)\\&=2.583334\dots\end{aligned}

p = 9

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{9^{9n}}\left(\tfrac{9}{9n+4}+\tfrac{1}{9n+5}\right) &= 9^4\ln\big((9c_1+82)^{c_4}(9c_2+82)^{c_1}(9c_3+82)^{c_3}(9c_4+82)^{c_2}/8^2\big)\\&=2.450000\dots\end{aligned}

with the constants {$26, 50, 82$} easily ascertained as {$5^2+1, 7^2+1, 9^2+1$}, and so on.  On the other hand, their counterparts are easier as the exponent $c_k$ has the same subscript as the base.  Still defining,

$c_k = -2\cos(2k\pi/p)$

then for p = 5,

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{5^{5n}}\left(\tfrac{5^3}{5n+1}+\tfrac{1}{5n+4}\right) &= 5^3\ln\big((5c_1+26)^{c_1}(5c_2+26)^{c_2}/4^2\big)\\&=125.256703\dots\end{aligned}

p = 7

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{7^{7n}}\left(\tfrac{7^5}{7n+1}+\tfrac{1}{7n+6}\right) &= 7^5\ln\big((7c_1+50)^{c_1}(7c_2+50)^{c_2}(7c_3+50)^{c_3}/6^2\big)\\&=16807.169\dots\end{aligned}

p = 9

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{9^{9n}}\left(\tfrac{9^7}{9n+1}+\tfrac{1}{9n+8}\right) &= 9^7\ln\big((9c_1+82)^{c_1}(9c_2+82)^{c_2}(9c_3+82)^{c_3}(9c_4+82)^{c_4}/8^2\big)\\&\approx 4.7829\, {\rm x}\, 10^6\end{aligned}

etc.

### The equation ap^3+bq^3+cr^3+ds^3 = 0

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

$a_1y_1^3+a_2y_2^3+a_3y_3^3+\dots+a_ny_n^3 = 0$

then an infinite more can be found.”

Proof:  As before, we will start with a particular example and derive the rest by induction. The following is identically true,

$ax_1^3+bx_2^3+cx_3^3+dx_4^3 = (ay_1^3+by_2^3+cy_3^3+dy_4^3)\,(ay_1^3-by_2^3)^3$

where,

\begin{aligned} x_1 &=y_1(ay_1^3+2by_2^3)\\ x_2 &= -y_2(2ay_1^3+by_2^3)\\ x_3 &= y_3(ay_1^3-by_2^3)\\ x_4 &= y_4(ay_1^3-by_2^3) \end{aligned}

Thus, if one has initial {$y_1, y_2, y_3, y_4$} such that the RHS is zero, this leads to a second, the {$x_1, x_2, x_3, x_4$}. By iteration, these can be used to generate a third, and so on.  The reason why {$x_3, x_4$}  have a common factor will be clear in a moment.

This is by A. Desboves, but it is not hard to generalize it to n cubes. The basis is the identity,

$\text{(1)}\;\; a(ax^4+2bxy^3)^3 + b(-2ax^3y-by^4)^3 = (ax^3+by^3)(ax^3-by^3)^3$

If the first factor of the RHS, $ax^3+by^3$, can be expressed as a sum of n cubes,

$ax^3+by^3 = c_1z_1^3+c_2z_2^3+\dots +c_nz_n^3$

it is a simple matter of distributing the second factor of the RHS among the n cubes so that (1) assumes the form,

$au_1^3+bu_2^3 = c_1v_1^3+c_2v_2^3+\dots +c_nv_n^3$

which explains the common factor of {$x_3, x_4$} in the 4-cube identity.  Hence from {$x, y, z_i$}, we get new solutions {$u_i, v_i$} leading to further solutions, ad infinitum.

There doesn’t seem to be any example of a diagonal quartic surface,

$a_1y_1^4+a_2y_2^4+a_3y_3^4+\dots+a_ny_n^4 = 0$

which has been proven to have only one non-trivial and primitive integer solution.  Based upon the quadratic and cubic cases, it is tempting to speculate that if it has one, then there may be in fact an infinity.

### 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.

### A Fermat’s Last Theorem near-miss

By Fermat’s Last Theorem, the quartic equation,

$x^4+y^4 = z^4$

has no non-trivial rational solutions.  In fact, the same can be said for the less strict,

$x^4+y^4 = z^2$

So how do we explain the near-equalities,

$24576^4+48767^4 \approx (49535.000000000006\dots)^4$

$419904^4 + 1257767^4 \approx (126155.000000000000001\dots)^4$

A search for others with z < 8,000,000 will not yield better approximations.  Noam Elkies showed that an identity is behind it, namely,

$(192v^8-24v^4-1)^4+ (192v^7)^4= (192v^8+24v^4-1)^4+12(2v)^4$

Since the second term on the RHS is small compared to the others, this gives an excellent near-miss to Fermat’s Last Theorem.  Inspired by Elkies’ quartic identity, Seiji Tomita found similar ones as,

$(48v^8-12v^4-1)^4 + 2(48v^7)^4 = (48v^8+12v^4-1)^4 + 6(2v)^4$

$(12v^8-6v^4-1)^4 + 4(12v^7)^4 = (12v^8+6v^4-1)^4 + 3(2v)^4$

It can be shown all three belong to the same family.  For arbitrary m let,

$b = \frac{8}{m},\;\; d = \frac{3m}{2}$

then,

$(3m^2 v^8-3m v^4-1)^4+b(3m^2 v^7)^4=(3m^2v^8+3mv^4-1)^4+d(2v)^4$

Note that,

$bd = 12$

In general, these are diagonal quartic surfaces of form,

$ax^4+by^4 = cz^4+dt^4$

The case {a, b, c, d} = {1, 2, 1, 4}, or the Swinnerton-Dyer quartic surface, was discussed in the previous post and, in the link above, Elsenhans gives first-known but large solutions to other positive {a, b, c, d}.  From Elkies’ and Tomita’s results, it is tempting to speculate that if,

$x^4+2y^4 = z^4+4t^4$

has a non-trivial solution, then does

$x^4+y^4 = z^4+8t^4$

have one as well?  Unfortunately, this particular form does not seem to be in Elsenhans’ list.  (For positive {a, b, c, d}, the only other case I know of that has an infinite number of solutions is a = b = c = d = 1.)

### The Swinnerton-Dyer quartic surface

Does the Diophantine equation,

$\text{(1)}\,\,\, a^4+2b^4=c^4+4d^4$

first suggested by Swinnerton-Dyer, only have a finite number of primitive solutions? Note that, by Sophie Germain’s identity, the RHS factors as,

$c^4+4d^4 = \big((c+d)^2+d^2\big)\big((c-d)^2+d^2\big) = (c^2+2cd+d^2)(c^2-2cd+d^2)$

Before discussing (1), we can compare it to other quartic Diophantine equations.  Any primitive integral solution to,

$\text{(2)}\,\,\, a^4+b^4+c^4+d^4 = e^4$

must obey the congruential constraint that there is an addend such that either {$d^2+e^2,\, d^2-e^2$} is divisible by $5^4$.  The smallest was found by Norrie in 1911 as,

$(a,b,c,d,e) = (30, 120, 315, 272, 352)$

and indeed,

\begin{aligned}d+e &= 272+353 = 5^4\\ -d+e &= -272+353 = 3^4 \end{aligned}

On the other hand, for,

$\text{(3)}\,\,\, a^4+b^4+c^4 = d^4$

Morgan Ward proved the added constraint that either {c+d, -c+d} is divisible by $2^{10}$. The smallest was found by Roger Frye in 1988,

$(a,b,c,d) = (95800, 414560, 217519, 422481)$

and we can see that,

$c+d = 217519+422481 = 2^{10}\, 5^4$

Going back to the Swinnerton-Dyer equation, the only solution less than a hundred million ($10^7$) is,

$(a,b,c,d) = (1484801, 1203120, 1169407, 1157520)$

found by A. Elsenhans and J. Jahnel in 2004.  On a hunch, I decided to test the values and found that,

$a+c = 1484801+1169407 = 2^{15}\,3^4$

I don’t know if this is a general congruence valid to all primitive solutions of (1), and an email to the authors revealed neither did they.  In fact, to quote their paper, “... it is unknown whether this equation admits ﬁnitely or inﬁnitely many primitive solutions. If their number were actually ﬁnite then this would settle a famous open problem in the arithmetic of K3 surfaces…

So many questions, so little time.