The general solution of the linear difference equation of degree-2 and the continued fraction made from this equation

Nikos Bagis Department of Informatics Aristotle University of Thessaloniki Greece
Email: bagkis@hotmail.com

Abstract In this article we give, for the fist time the solution of the general difference equation of 2-degree. We also give as application the expansion of a continued fraction into series, which was first proved, found in the past by the author.

Introduction It is well known that the linearly difference sequence

2 1 n n n n n x b x a x + +

: (1)

produces a continued fraction K .i.e.

0 1 1 0 2 1 2 …. n n n a a K a b b a b b ∞

  Κ =

    + + + :(2)

It is also well known that the tail sequence for this fraction is:

1 1 ,
,
n n n n n n n a x t n t b t x + +

∈ = − + ℕ :(3)

and if (xn) is a “minimal” solution of (1) see [L,W] , then t0=K.

Theorems Theorem 1. The general solution of (1) is : 1, 1, 2, 2, 2 1 2 1 ( , ) ( , , ) 1 2 0 0 2 ( , ) ( , , ) 0 1 …

1 … n n n n n n i i j i j k n i S i j S i j k i i n i i j i j k i S i j S i j k x x g g g g g g b a x g g g g g g b − −

= −

 

⋅ + + + + +         + ⋅ ⋅ + + + +       ∑ ∑ ∑ ∏ ∑ ∑ ∑

:(4)

2

Where:

{

}

, ( , , ,…, ,

)

, ,…,

… 2 ,
2 ,
2 ,…,
2 q n S i j k l m i j m  q i j k l m n j i k j m l

= ∈ ≤< < < < < ≤ − −≥ − ≥ −≥

q =1, 2, and 1 ,
1,2,3,… i i i i a g i b b −

=

Knowing that xn is a zero limit sequence we take the limits in (4) and calculate the ratio -x1/x0 = t0 = K. After some calculations we can see that:

Theorem 2.

1 1 2 2 2 1 2 1 1 0 0 1 0 0 1 1 2 2 1 1 1 1 1 1 1 2 1 … 1 … …. i i i j i i j i j k i i j i j k i i i j i i j i j k i i j i j k g g g g g g a a a b b g g g g g g a b b ∞ ∞ ∞ ∞ ∞ ∞

=

=

=

∞ ∞ ∞ ∞ ∞ ∞

=

=

=

= ⋅ + + + + + + + ∑ ∑∑ ∑∑∑ ∑ ∑∑ ∑∑∑ :(5)

Without the care of convergence.

(For another proof see [Ba]).

At this point we can see some applications of this continued fraction expansion.

Applications Application 1 For am= z and bm= m+c , z∈C we find:

( ) ( ) 0 1 0 1 1; ; 1 2 3 4 …. F c z c z F c z c z c z c z c c +

Where 0F1 the usual Hypergeometric function.

Application 2 For am= qm and bm= 1 we find the Rogers-Ramanujan continued fraction expansion:

3 2 ( 1) 0 2 0 3 ( ) 1 1 ( ) 1 1 1 … k k k k k k k k k q z q qz q z q z q q z + ∞

∞

= + +

∑ ∑

where: ( ) ( ) ( ) ( ) 2 0 ( ) 1 1 … 1 ,
1 k k q q q q q

− ⋅ − ⋅ ⋅ −

Application 3 ( ) 0 1 ( 1)

,
, m m k k c k m m For a z and b q c

    − ⋅ + −     ∑

= ∈ℝ we find (after some elementary calculations) ( 1) 2 0 ( 1) 2 0 2 2 3 ( 1) ( ) 1 1 ( 1) ( ) … k k k k k k k k k k k k z q q z z q z q q z q z q z q q − + ∞

− − ∞

− ⋅ ⋅

− ⋅ ⋅ + + + + + ∑ ∑

where 1 q > . This is the Rogers Ramanujan continued fraction in another form.

References Books. [B2]: B. C. Berndt. “Ramanujan`s Notebooks Part IΙ”. Springer-Verlag, New York Inc.1989.

[B3]: B. C. Berndt. “Ramanujan`s Notebooks Part IΙI”. Springer-Verlag, New York Inc.1991

[L,W]: Lisa Lorentzen, Haakon Waadeland. ”Continued Fractions with Applications”. North Holland. Amsterdam, London, New York, Tokio.

[W]: H. S. Wall “Analytic Theory of Continued fractions”. CHELSEA PUBLISHING COMPANY BRONX, N.Y.

Articles [Ba]: N. Bagis. “On Series Integrals and Continued Fractions”. Hardy Ramanujan Journal, Vol. 26, pg. 23-29. 2003