## Archive for the ‘algebra’ Category

### Odd powers of Fibonacci numbers

The Fibonacci numbers $F_n$,

$F_n = 0, 1, 1, 2, 3, 5, 8, 13, \dots$

obey the following recurrence relations,

\begin{aligned}&F_n-F_{n-1}-F_{n-2} = 0\\[1.5mm]&F_n^2-2F_{n-1}^2-2F_{n-2}^2+F_{n-3}^2 = 0\\[1.5mm]&F_n^3-3F_{n-1}^3-6F_{n-2}^3+3F_{n-3}^3+F_{n-4}^3 = 0\\[1.5mm]&F_n^4-5F_{n-1}^4-15F_{n-2}^4+15F_{n-3}^4+5F_{n-4}^4-F_{n-5}^4 = 0\\[1.5mm]&F_n^5-8F_{n-1}^5-40F_{n-2}^5+60F_{n-3}^5+40F_{n-4}^5-8F_{n-5}^5-F_{n-6}^5 = 0\end{aligned}

and so on.  As a number triangle, the coefficients are,

\begin{aligned} &1;\; \bold{1, -1,-1}=0\\ &2;\; 1, -2, -2, \;1=0\\ &3;\; \bold{1, -3, -6, \;3, \;1}=0\\ &4;\; 1, -5, -15, 15, \;5, \;-1=0\\ &5;\; \bold{1, -8, -40, 60, 40, -8, -1}=0 \end{aligned}

See Ron Knott’s article on the fibonomials, so-called since the above is reminiscent of the binomial triangle.  However, I found another set of recurrence relations can be given as,

\begin{aligned} &F_{n-1}^2-F_{n+1}^2 = -F_{2n}\\[1.5mm] &F_{n-1}^3-F_{n}^3-F_{n+1}^3 = -F_{3n}\\[1.5mm] &F_{n-2}^4+3F_{n-1}^4-3F_{n+1}^4-F_{n+2}^4 = -6F_{4n}\\[1.5mm] &F_{n-2}^5-3F_{n-1}^5-6F_{n}^5+3F_{n+1}^5+F_{n+2}^5 = 6F_{5n}\\[1.5mm] &F_{n-3}^6+4F_{n-2}^6-20F_{n-1}^6+20F_{n+1}^6-4F_{n+2}^6-F_{n+3}^6 = -120F_{6n}\\[1.5mm] &F_{n-3}^7-8F_{n-2}^7-40F_{n-1}^7+60F_{n}^7+40F_{n+1}^7-8F_{n+2}^7-F_{n+3}^7 = -240F_{7n}\\[1.5mm]\end{aligned}

etc.  As a number triangle,

\begin{aligned} &2;\; 1, -1 =-1\\ &3;\; \bold{1, -1,-1}=-1\\ &4;\; 1, \;\;\;3, -3, \;-1=-6\\ &5;\; \bold{1, -3, -6, \;3, \;\;1}\;=\;6\\ &6;\; 1, \;\;\;4, -20, 20, -4, -1=-120\\ &7;\; \bold{1, -8, -40, 60, 40, -8, -1}=-240 \end{aligned}

Compare the two triangles.  Notice how, for odd powers, the same coefficients appear, though moved up by one odd power.  I have no explanation for the phenomenon, other than the fact that I’ve seen several instances already of a “recycled” polynomial appearing in many contexts.

### Ramanujan’s continued fraction for Catalan’s constant

Ramanujan was a goldmine when it came to continued fractions (and many others).  In this post, two families will be given: they involve pi and Catalan’s constant as special cases.  However, versions will be given that roughly double the rate of convergence.

Recall that Catalan’s constant C is given by,

$C = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = 0.915965\dots$

Ramanujan gave the beautiful pair of continued fractions.  Let $|x| > 1$, then,

$f(x) = 2\sum_{n=0}^\infty \frac{(-1)^n}{(x+2n+1)^2} = \cfrac{1}{x^2-1 + \cfrac{2^2}{1 + \cfrac{2^2}{x^2-1 + \cfrac{4^2}{1 + \cfrac{4^2}{x^2-1 +\ddots}}}}}$

$g(x) \; = \;\frac{\Gamma^4 (\frac{x+1}{4})}{\Gamma^4 (\frac{x+3}{4})}\; =\; \cfrac{16}{x^2-1 + \cfrac{1^2}{1 + \cfrac{1^2}{x^2-1 + \cfrac{3^2}{1 + \cfrac{3^2}{x^2-1 +\ddots}}}}}$

One can just admire Ramanujan’s artistry — one continued fraction uses even numerators, the other, odd numerators.  For even x > 0, it is easily seen the first one involves Catalan’s constant.  For example,

$f(2) = 2(1-C)$

On the other hand, for odd x > 1, both involve pi,

$f(3) = \frac{1}{24}(12-\pi^2)$

$g(3) = \pi^2$

Notice though that numerators are repeated. Thanks to the insight of J.M. from a mathstackexchange post about Apery’s constant, we can speed up the rate of convergence of this particular form by getting rid of every other level and extracting even convergents.  After some slightly tedious algebraic manipulation, given,

$y = \cfrac{a_1}{b_1 + \cfrac{a_2}{1 + \cfrac{a_3}{b_2 + \cfrac{a_4}{1 +\ddots}}}}$

then,

$y_{even} = \cfrac{a_1}{b_1+a_2 - \cfrac{a_2\, a_3}{b_2+a_3+a_4 - \cfrac{a_4\, a_5}{b_3+a_5+a_6 - \cfrac{a_6\, a_7}{b_4+a_7+a_8 -\ddots}}}}$

So Ramanujan’s continued fractions are now the more compact,

$f(x) = 2\sum_{n=0}^\infty \frac{(-1)^n}{(x+2n+1)^2} = \cfrac{1}{u_1 - \cfrac{2^4}{u_2 - \cfrac{4^4}{u_3 - \cfrac{6^4}{u_4 -\ddots}}}}$

where, starting with n = 1,

$u_n = (2n-2)^2 + (2n)^2 + x^2-1$

and,

$g(x) \; = \; \frac{\Gamma^4 (\frac{x+1}{4})}{\Gamma^4(\frac{x+3}{4})}\; =\; \cfrac{16}{-1+v_1 - \cfrac{1^4}{v_2 - \cfrac{3^4}{v_3 - \cfrac{5^4}{v_4 -\ddots}}}}$

$v_n = (2n-3)^2 + (2n-1)^2 + x^2-1$

Thus, we have the slightly faster continued fraction for Catalan’s constant C,

$f(2) = 2(1-C) = \cfrac{1}{u_1 - \cfrac{2^4}{u_2 - \cfrac{4^4}{u_3 - \cfrac{6^4}{u_4 -\ddots}}}}$

where,

$u_n = 8n^2-8n+7$

(There is an even faster one by Zudilin given in An Apery-like difference equation for Catalan’s constant though he states this still does not prove C is irrational.)

### Continued fractions for Zeta(2) and Zeta(3)

It seems there is a nice “pattern” between the continued fractions for the Riemann zeta function,

$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\dots$

at s = 2, and s = 3.  First though, a little introduction.

The origins of this function go back to 1644 when, at the tender age of 18, the Italian mathematician Pietro Mengoli (1626-1686) first proposed what would be later known as the Basel Problem, namely to determine the exact value of the sum of the reciprocals of the squares.  Euler would later find that, for n a positive integer, then $\zeta(2n)$ is a rational multiple of $\pi^{2n}$, with the first case s = 2 as,

$\zeta(2) = \frac{\pi^2}{6}$

which obviously is irrational.  However, the status of odd s was not as easy to resolve.  It was only in 1979 that Apery proved that $\zeta(3)$ is irrational, which henceforth was called Apery’s constant.

I. Zeta(2)

One of its infinite number of continued fractions can be given as,

$m\,\zeta(2) = \cfrac{1}{u_1+\cfrac{1^4}{u_2+\cfrac{2^4}{u_3+\cfrac{3^4}{u_4+\ddots}}}}$

where $m = \frac{1}{2}$ and the $u_n$, starting with n = 1, are generated by,

$u_n = 2n-1 = 1, 3, 5, 7,\dots$

or simply the odd numbers.  The convergence is slow, but Apery found it can accelerated by using a quadratic function,

$u_n = 11n^2-11n+3 = 3, 25, 69, 135,\dots$

but now $m = \frac{1}{5}$.

II. Zeta(3)

Likewise, this also has an infinite number of continued fractions (see Ramanujan’s versions here), but one important form is,

$m\,\zeta(3) = \cfrac{1}{v_1-\cfrac{1^6}{v_2-\cfrac{2^6}{v_3-\cfrac{3^6}{v_4-\ddots}}}}$

where $m = 1$ and the $v_n$, again starting with n = 1, are,

$v_n = (n-1)^3+n^3 = (2n-1)(n^2-n+1) = 1, 9, 35, 91, \dots$

Apery again found an accelerated version,

$v_n = 34n^3-51n^2+27n-5 = (2n-1)(17n^2-17n+5) = 5, 117, 535, 1463, \dots$

where now $m = \frac{1}{6}$, and established that its rate of convergence was such that $\zeta(3)$ could not be a ratio of two integers.

III. Connection between Zeta(2) and Zeta(3)

Define the two sequences,

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

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

then it was established that these Apery numbers have the recurrence relations,

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

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

Interesting that the same polynomials pop up, isnt’ it?  Furthermore, their limiting ratios have the common form,

$\left(\frac{b+\sqrt{b^2+4}}{2}\right)^{6-b}$

thus for b = 1, 2,

$\lim_{n \to \infty} \frac{B_{n+1}}{B_n} = \left(\frac{1+\sqrt{5}}{2}\right)^5 = 11.0901\dots$

$\lim_{n \to \infty} \frac{A_{n+1}}{A_n} = \left(1+\sqrt{2}\right)^4 = 33.9705\dots$

hence the golden ratio and the silver ratio surprisingly turn up in the continued fractions for $\zeta(2)$ and $\zeta(3)$, respectively.  It is easy to check other sequences,

$C_n = \sum_{k=0}^n {\binom n k}^p {\binom {n+k}k}^q$

for some small {p, q}, but there are no other recurrence relations similar in form to the two above.

IV. Zeta(5)

There is an orderly non-simple continued fraction for all $\zeta(s)$, with the next odd s as,

$m \zeta(5) = \cfrac{1}{w_1-\cfrac{1^{10}}{w_2-\cfrac{2^{10}}{w_3-\cfrac{3^{10}}{w_4-\ddots}}}}$

where $m = 1$ and the $w_n$ are,

$w_n = (n-1)^5+n^5 = (2n-1)(n^4-2n^3+4n^2-3n+1) = 1, 33, 275, 1267, \dots$

Unfortunately, no one has yet found an accelerated version where the $w_n$ are generated by a 5th degree (or higher) polynomial, and m is rational.

Can you find one?

For further reading, refer to Alfred van der Poorten’s excellent, A Proof That Euler Missed.

### Some of Ramanujan’s continued fractions for pi

The digits of pi go on forever apparently with no discernible pattern. However, there are beautifully simple patterns in its (ironically) non-simple continued fraction expansions. Examples are,

$\pi = \cfrac{4}{1 + \cfrac{1^2}{2 + \cfrac{3^2}{2 + \cfrac{5^2}{2 + \ddots} }}}$

known by Lord Brouncker (1620-1684), and,

$\pi = 3 + \cfrac{1^2}{6 + \cfrac{3^2}{6 + \cfrac{5^2}{6 + \ddots} }}$

One can see the affinity between the two.  They in fact belong to the same family.  Given complex numbers {n, x} with $Re[x] > 0$, then,

$F(n,x) = \Large{\frac{\Gamma\left(\frac{x+n+1}{4} \right)\Gamma\left(\frac{x-n+1}{4} \right) }{\Gamma\left(\frac{x+n+3}{4} \right)\Gamma\left(\frac{x-n+3}{4} \right)}} = \cfrac{4}{x + \cfrac{1^2-n^2}{2x + \cfrac{3^2-n^2}{2x + \cfrac{5^2-n^2}{2x + \ddots} }}}$

where $\Gamma(\tau)$ is the gamma function.  This is Entry 25 in Ramanujan’s Second Notebook (Chapter 12) though this result was also known by Euler. For the case = 0, the continued fraction assumes the form of the examples and the function simplifies as,

$F(0,x) = \frac{\Gamma^2\left(\frac{x+1}{4}\right)}{\Gamma^2\left(\frac{x+3}{4}\right)}$

For x an odd integer, then (0, x) is a rational multiple of $\pi$ or $1/\pi$.  Specifically, for x = {1, 3, 5, 7, 9}, we have,

$F(0,x) = \pi,\; \frac{4}{\pi}, \frac{\pi}{4}, \frac{16}{9\pi}, \frac{9\pi}{64}$

More generally,

$F(0,4m+1) = \big(\frac{(2m)!}{m!^2}\big)^2\, \frac{\pi}{2^{4m}}$

$F(0,4m+3) = \big(\frac{m!^2}{(2m+1)!}\big)^2\, \frac{2^{4m+2}}{\pi}$

See p. 178 of Annie Cuyt’s HandBook of Continued Fractions for Special Functions.  For more examples, see also this article Ramanujan’s Continued Fractions, Apery’s Constant, and more.

### 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.)