Some simple continued fraction expansions for an in nite product Part 1. Peter Bala, January ax 4n+3 1 ax 4n+1. (a; x) =

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1 Soe sile continued fraction exansions for an in nite roduct Part. Introduction The in nite roduct Peter Bala, January 3 (a; x) = Y ax 4n+3 ax 4n+ converges for arbitrary colex a rovided jxj <. Let and be ositive integers with > 4. Let x denote the real algebraic nuber x = 4 : We shall nd the sile continued fraction exansion of the in nite roducts ( Y ; x) = x 4n+3 x 4n+ and ( ; x) = Y + x 4n+3 + x 4n+ Our results were otivated by conjectures ade by Paul Hanna in sequences A754 through A7543, who considers the articular cases of the above corresonding to = and = 4; 6; 8; or =. Hanna works with the real nuber! X ex n (x n + x n ) n= but it is not di cult to show that this is equal to (; x):. Preliinaries on continued fractions We adot the standard coact notation a + a a a 3 b + b + b 3 + to denote the general continued fraction a + b + a a b + a 3 b 3 +. () We refer to the ters a n ; n ; in () as the artial nuerators and the ters b n as the artial denoinators of the continued fraction. A sile continued fraction is a continued fraction in which a is an integer, all the artial nuerators are equal to and each artial denoinator is a ositive integer. We recall (see for exale [, Theore 4]) that every ositive irrational real nuber has

2 a unique exansion as a sile continued fraction (with an in nite nuber of ters). Rational nubers have nite sile continued fraction exansions. Given a sequence n of non-zero colex nubers, the continued fraction a a + b + a b + 3 a 3 3 b 3 + ( n 6= ) () is said to be obtained fro () by eans of an equivalence transforation. The continued fractions () and () are equivalent in the sense that the n-th convergents of both fractions have the sae value for all n [3,.9]. If in () the artial nuerators a n are all nonzero then we can choose colex nubers ; ; 3,::: so that = a = a = 3 a 3 = : : :. By this eans we can arrange that the artial nuerators in the equivalent continued fraction () are all equal to. In the next section we introduce another continued fraction transforation (Lea ) that converts a continued fraction with artial nuerators equal to into a continued fraction with all artial nuerators equal to Soe continued fraction transforations In order to rove Lea we will need the following reliinary result. Proosition. If a ; a ; : : : ; a n is a sequence of colex nubers then + = a + a + + a n : (3) a + a + + a n Proof. By induction on n. The result is easily veri ed for n =. Assue that (3) is true for a xed integer n > : Then by induction + = + a + a + + a n+ a + a + + a n + = = and the induction goes through. a + a + + a n + a + a + + a n+ Lea. If a ; a ; :::; a n is a sequence of colex nubers then a a a 3 = + a n a + a n + a n+ a n + a n+ + a a n + : (4)

3 Proof. By induction on n. The result (4) is easily veri ed for n =. Assue that (4) is true for a xed integer n > : Let F (n) denote the rhs of (4). Then F (n + ) = + a + + a a n + + a n+ + where = + a + ; A A = + a a n + + a n+ + = a a 3 by the induction hyothesis. Thus F (n + ) = + = + a a + a a 3 a a 3 a n+ a n+ a n+ = + a + ( a ) + a (( ) n a n+ ) = = a + ( a ) + a (( ) n a n+ ) a a a n+ ; [equivalence transforation] [by Proosition ] where we have used another equivalence transforation to obtain the nal exression. This coletes the roof by induction. 4. Sile continued fraction exansions The following continued fraction exansion is a articular case of a ore general result due to Raanujan. For a roof consult [, Entry with b = and a relaced with a]. (a; x) = Y ax 4n+3 ax 4n+ = valid for arbitrary colex a rovided jxj < : An equivalence transforation yields Y ax 4n+3 ax 4n+ = a ax ax 3 ax 5 + x + x 4 + x 6 (5) x + x x + x a x + x 3 (6) 3 x + x 4 4 3

4 valid for < jxj < : There are several ways in which we can choose values for a and x so that the artial denoinators of this continued fraction becoe integers. We deal with two cases here and leave consideration of a further two cases to Part of these notes. Case. Let and be ositive integers with > 4 and set a = : Let x denote the real algebraic nuber so that < x < and x = 4 x + x = : (7) A well-known roerty of T n (x), the n-th Chebyshev olynoial of the rst kind, is the identity x + x T n = xn + x n x 6= : Thus fro (7) x n + x n = T n and the continued fraction (6) becoes Y x 4n+3 x 4n+ = T = T 3 n = ; ; ; 3; ::: T T : Alying Lea to this continued fraction we obtain the continued fraction exansion Y x 4n+3 x 4n+ = T T T T 4 + : (8) 4

5 ext we will show that the artial denoinators of this continued fraction are ositive integers, with three ossible excetions, so that, aart fro these cases, (8) is the sile continued fraction reresentation of the real nuber given by the in nite roduct. In the case of the three excetions we can still obtain a sile continued fraction exansion fro (8) after a little extra work. Proosition. Let ; be ositive integers such that > 4: Then for k both T k and T k+3 are ositive integers. Proof. The Chebyshev olynoials T n (x) satisfy the recurrence equation T n+ (x) = xt n (x) T n (x) [T (x) = ; T (x) = x]: Calculation gives T = and T = ; an integer. A straightforward induction roof, aking use of the recurrence equation, shows that the nubers T k are integers, whilst the nubers T k+ equal an integer ultilied by : Therefore the quantities T k and T k+3 are integers. We now show that they are ositive integers greater than or equal to 3. An exlicit forula for the Chebyshev olynoials of the rst kind is b n X c n T n (x) = (x ) k x n k : k k= An easy consequence of this result is the inequality T n (x) > T 3 (x) = 4x 3 3x [x > and n > 3]: (9) Thus for k T k > T 3 = ( 3) if we recall that > 4: In addition T = > : 5

6 Hence, for k ; T k is a ositive integer. Siilarly, for k we have fro (9) T k+3 > T 3 = ( 3) : and hence the artial denoinator T k+3 is a ositive integer. The only artial denoinators of the continued fraction (8) that are not ositive integers are T = when = or = and T 3 = ( 3) when = and = 5 (recall the assution > 4): In these cases we have to do a further sili cation to get the continued fraction (8) into the for of a sile continued fraction. We state the nal result in the for of a theore. Theore. Let ; be ositive integers such that > 4: (a) When 3 we have the sile continued fraction exansion Y 4n+3 4 4n+ = T T T T 4 = ( 4) + + : + 6

7 (b) When = and > we have the sile continued fraction exansion Y 4n+3 4n+ = + T ( ) + + T 3 ( ) + + T 4 ( ) + + T 5 ( ) + + = : (c) When = and > 4 we have the sile continued fraction exansion 6 Y 4n+3 4 4n+ = ( 3) T 4 + T T = ( 3) ( 4) ( 4)( ) + : Hanna has recorded four articular cases of art(a) of Theore in A745 ( = and = 4), A745 ( = and = 6), A745 ( = and = 8) and A7453 ( = and = ). Case We return to Raanujan s continued fraction exansion (5). Let and be ositive integers with > 4; as before, but now we set a = : As before, let x denote the real algebraic nuber x = 4 : 7

8 Then with these choices Raanujan s continued fraction (5) becoes ( ; x ) = Y + x 4n+3 + x 4n+ x x 3 x 5 = + + x + + x x 6 + = + x + x + x + x + x + x x + x by an equivalence transforation. Exressing the artial denoinators of this continued fraction in ters of the Chebyshev olynoials of the rst kind we have the following result. Theore. Let ; be ositive integers such that > 4: There holds the sile continued fraction exansion Y + + 4n n+ = + T + T 3 + T T = : In the articular case = ; the sequence of artial denoinators of this sile continued fraction becoes the sequence [; T ; T ; T3 ; : : :]. There are any sequences of this tye currently in the OEIS database (but with initial ter rather than ): see A548 ( = 3), A35 ( = 4), A35 ( = 5), A3499 ( = 6), A56854 ( = 7), A8693 ( = 8), A5698 ( = 9), A87799 ( = ), A5776 ( = ), A878 ( = ), A78363 ( = 3), A679 ( = 4), A78365 ( = 5), A977 ( = 6), A78367 ( = 7), A875 ( = 8), A78369 ( = 9), A978 ( = ), A979 ( = ), A973 ( = ), A973 ( = 3), A973 ( = 4), A9733 ( = 5), A947 ( = 6), A948 ( = 7), A949 ( = 8) and A95 ( = 9). In Part of theses notes we nd the sile continued fraction exansions of the 8

9 in nite roducts Y x 4n+3 + and x 4n+ Y + x 4n+3 ; x 4n+ where now x denotes an algebraic nuber of the for + 4 x = : REFERECES. B. C. BERDT, Raanujan s otebooks, Part III, Sringer-Verlag, ew York, 99.. A. Ya. KHICHI, Continued Fractions, Dover Publications Inc. 3. H. S. WALL, Analytic Theory of Continued Fractions, AMS Chelsea Publishing,