Some sufficiet coditios of a give arxiv:0807.376v2 [math.nt] 8 Jul 2008 series with ratioal terms covergig to a irratioal umber or a trascdetal umber Yu Gao,Jiig Gao Shaghai Putuo college, Shaghai Jiaotog Uiversity Abstract I this paper, we propose various sufficiet coditios to determie if a give real umber is a irratioal umber or a trascedetal umber ad also apply these coditios to some iterestig examples,particularly,oe of them comes from complex aalytic dyamics.
Itroductio I series theory, there is well kow Cauchy covergece test which is used to determie covergece of a give series,but Cauchy covergece test usually is ot practical i most applicatios,so there come out various covergece test such as D Alembert covergece test,itegral covergece test ad so o. I Diophatie approximatio theory, we are i totally differet situatio that we already have ecessary ad sufficiet coditio to determie if a give real umber is a irratioal umber or a trascedetal umber such as well kow Roth theorem but seems to be lack of practical test just as various coveiet test i series theory. The purpose of this paper is to propose some sufficiet coditios for coveiet use i determiig if a give real umber is a irratioal umber or a trascedetal umber ad also give out various iterestig examples to illustrate how to apply these coditios,particularly, we will explai a example comig from complex aalytic dyamics i detail. At the ed of this paper, we propose a cojecture about ratioal approximatio of ay irratioal umber. Theorem Assume that series =, = a ratioal umbers ad satisfy followig coditios: 0, ( =, 2, ) are () +, ( =, 2, )
(2) lim a + = 0 (3)for ay atural umber N, =N 0 the the series =N coverges to a irratioal umber. c Proof. First of all, sice coditio (2) implies lim + = 0, the series = is coverget ad set the coverget result to be θ. We will use idirect method to show θ is a irratioal umber. Suppose that θ = s r is a ratioal umber. By (), whe k,b k, we have A = r ( s r a b a a ) = r ( a + + + a +2 +2 + ) = r (+ + +2 + ) 0 ad A is a iteger umber, we otice that A = r + ( + +2 + + +3 + + ) ad lim + = 0, thus there exists N, whe N we have + < 2, +2 + < 2,, so +3 + = +3 +2 +2 + < ( 2 )2 +4 + = +4 +3 +2 +3 +2 + < ( 2 )3 2
Therefore, A r a c + ( + c +2 + c +3 + ) c c + c + < r a c + c ( + 2 + 2 2 + ) = 2 r a c + c By (2),for 2 r > 0,there exists N 2,whe N 2, a c + c < 2 r. Set N = max(n, N 2 ),whe N,we get A < 2 r a c + c < 2r 2r = which cotradicts the fact that A is a iteger ad A 0. That follows the theorem. Whe the series just cotais positive terms, coditio (3) is aturally satisfied, we have Theorem 2 Assume that series =, = a ratioal umbers ad satisfy followig coditios: > 0, ( =, 2, ) are () +, ( =, 2, ) (2) lim a + = 0 the the series = coverges to a irratioal umber. Remark. I the above theorem, the coditio lim a + = 0 is ot sufficiet because =0 +! = 2e is a irratioal umber ad a + = +2 ( ) + 3
Example. e = =0! = +! + 2! + 3! + +! + is a irratioal umber. Because by the theorem 2, Where a = +! lim a = lim ( + )! = lim + = 0 Example 2. θ = = 4 (!) 5 = + 24 (2!) 5 + + 4 (!) 5 + is a irratioal umber, because by the theorem 2, a + = a + = 0( ) + + where a = 4, = (!) 5. Let s look at a more complicated example as follows: Example 3. Suppose that r is a iteger, the si r is a irratioal umber. Sice si r = ( ) (2 )!r = c 2 = = where a = ( ), = (2 )!r 2, the = = a lim a + = lim a + = lim ( ) (2 )!r 2 (2 + )!r 2+ = 0. The remaiig is to verify =N 0 = =N ( ) N (2N )!r 2N + ( ) N (2N + )!r 2N+ + ( ) N+ (2N + 3)!r 2N+3 + 4
Whe N is odd = ( =N (2N )!r 2N (2N + )!r 2N+)+( (2N + 3)!r 2N+3 (2N + 5)!r 2N+5)+ = (2N + )!r2 (2N )! (2N )!(2N + )!r 2N+ + (2N + 5)!r2 (2N + 3)! (2N + 3)!(2N + 5)!r 2N+5 + > 0 Similarly,whe N is eve = ( =N (2N )!r 2N (2N + )!r 2N+) ( (2N + 3)!r 2N+3 (2N + 5)!r 2N+5) < 0 Thus for ay atural umber N, =N 0,by the theorem,si r is a irratioal umber. Sice the sum of two irratioal umbers is ot ecessarily a irratioal umber, the followig theorem is iterestig. Theorem 3 Assume that α = =,where = a > 0 ad β = = c, wherec = a b > 0 ad above two umber are irratioal umbers determied by the theorem 2 ad satisfy the followig coditios: lim a + b + = 0 ad lim a + b b b + = 0 the α + β is also a irratioal umber. Proof. Let γ = α+β = = d ad d = +c = a + a b where ã = a b + a, = b the = ab +ba b = añ lim ã d + d = lim = lim (a + b + ++a + ) b + b + bñ a + + a + b a + = lim +lim b + b + I additioal,we otice that +, so by the theorem 2 we get α + β is a irratioal umber Example 4.Let α = = 2!, β = = 3!,we ca use above theorem to 5 = 0
verify that α + β is a irratioal umber as follows: First of all, α, β are irratioal umbers because of theorem 2,secodly, let = 2!, b = 3!,the b b + b b + so α + β is a irratioal umber. = 3! 2! 0, ( ) = 2! 3! 0, ( ) Essetially, we ca replace coditio of theorem 2 by a more geeral coditio as follows: Theorem 4 Let series = a b where a > 0 are ratioal umbers ad lim a + + [b, b 2, ] = 0 where [b, b 2, ] deotes least commo multiple of b, the = a b Proof. Let = a ad coverges a irratioal umber = a [b,b 2, ] [b, b 2, ] = ã = where ã = [b,b 2, ], = [b, b 2, ] ad + a + a + lim ã = lim b = lim + [b, b 2, ] = 0 + Obviously, +, =, 2 ad the series = satisfies coditios of theorem 2,ad the we fiish the proof. = a = = = = Example 5. θ = p 2 2! + p 2 2 2! + + p 2 2! + 6
is a irratioal umber, where p is 2 2 -th prime umber. Let s show it as! 22! follows: We eed the famous result[]: let p is -th prime umber, there exists two positive umbers such that c l < p < c 2 l,the a + [b, b 2, ] = p 2 2!p 2 p 22! 2 2! + p 2 2 (+)! Where k = c, c = c 2 Sice < c 22 2! l 2 2! c 2 2 22! l 2 22! c 2 2 2! l 2 2! c 2 2(+)! l2 2(+)! = 22! +2 2! + +2! + log 2 C 2 2(+)! 2!+2!+ +! 2 (+)! k(l 2)! + 2! + +!! < ( + )!, 2!+2!+ +! 2 (+)! <. Also, there exists N such that whe N log 2C < 2!. Therefore 2 2! +2 2! + +2! +log 2 C 2! + 2! 22! 2 (+)! we get 2 2! +2 2! + +2! +log 2 C 2 2(+)! Thus whe N lim a + + [b, b 2, ] k(l 2) 0( ),by theorem 4,θ is a irratioal umber The followig theorem shows that coditio 2 of theorem 2 is also ecessary i some special case. 7
Theorem 5 Assume that the sequece = a Pm(),where a 2 is a iteger ad P m (x) = b 0 x m +b x m + +b m x+b m is a polyomial with positive iteger coefficiets,the series = coverges to a irratioal umber if ad oly if lim + = 0 Proof. Obviously, P m () < P m ( + ),so a Pm() a Pm(+), =, 2, which satisfies coditio of theorem 2 ad P m ( + ) P m () = b 0 ( + ) m + + b m (b m 0 + + b m + b m ) = b 0 ( m + m m + ) + + b m (b m 0 + + b m + b m ) = mb 0 m + l m 2 + l 2 m 3 + () the Thus + = apm() a Pm(+) = a Pm(+) Pm() = a mb 0 m +l m 2 + c + lim = 0 m 2 a b 0 m = (2) By theorem 2, whe m 2 series m= coverges to a irratioal umber. ad lim + is a ratioal umber. 0 meas m = the = = = a b 0+b = Theorem 6 Let θ = = a b,where a a b (a b 0 ) > 0( =, 2 ) are ratioal umber, ad assume that f( ) is a fuctio of ad f( ) > 0,furthermore if 8
the followig coditios are satisfied: ()b < b 2 < ad +, =, 2, (2)f( ) > 0 ad f(+) f() (3) f(b) + a + < 2 (4) b f() the 0( ) < 2 ( is big eough) ()θ is a irratioal umber ( is big eough) (2) Whe is big eough, there exists ifiite fractios c such that θ c < f() Proof. Accordig coditio 3,whe is big eough, we have f(b) + a a + < 2 or equivaletly a + + < b 2f(),so lim a + = lim a + + = 0 I the last step we use coditio 4.By theorem 2, θ is a irratioal umber. Let s prove the secod part. θ a b a 2 b 2 a = a + + + a +2 +2 + a +3 +3 + = f( ) (f()a + + + f()a +2 +2 + f()a +3 +3 + ) = f( ) (f()a + + + f() f(+ ) f(+ )a +2 + f() f(+2 ) f(+ ) f(+ ) f(+2 ) f(+2 )a +3 + ) f(+3 ) < f( ) ( 2 + 2 2 + 2 3 + ) = f( ) 9
We use coditio 2 ad 3 i the last two steps. Let = ( a b + a 2 b 2 + + a ),sice a Thus there are ifiite umber of c satisfy θ = θ a b a 2 b 2 a < f( ) (whe is big eough )That proves the theorem > 0, ( =, 2, ), c < + + Remark. I above theorem, coditio 2 ad 3 ca be replaced by lim f() f(+ ) = l < 2 ad lim f(b) + a + = k < 2 Usig theorem 6 ad two followig kow results, we ca get two useful theorems, oe is about how to determie a give umber is trascedetal umber, the other is about complex aalytic dyamics. Theorem(K.Roth). Let θ be a 2 degree algebraiumber, the for ay give ǫ > 0, there exists oly fiite positive iteger pairs x, y such that θ x y < y 2+ǫ Theorem (H.Cremer)[2] If irratioal umber θ satisfies the coditio that there exists ifiite positive itegers such that θ m m dm, idifferet fixed poit z = 0 of polyomial f(z) = z d + +e 2πiα belogs to Julia set. First of all, we use theorem 6 ad Roth theorem to derive followig theorem: Theorem 7 Let θ = = a b,where a > 0( =, 2 ) are ratioal umbers which satisfy ()b < b 2 < ad +, =, 2, (2) for some ǫ > 0, a +b 2+ǫ + the θ = = a b < 2 ( is big eough) is a trascedetal umber. 0
Proof. I the theorem 6,let s take f( ) = b 2+ǫ,the it s easy to verify f( ), =, 2, satisfy coditio ad 3 of theorem 6, we oly eed to check coditio 2 ad 4. Sice ǫ > 0 ad a, are positive itegers, we have f( ) f(+ ) = b2+ǫ b 2+ǫ + < b2+ǫ < a +b 2+ǫ < + + 2,thus we pass coditio 2 of theorem 6. Also because b < b 2 < ad ǫ > 0, f( ) = b 2+ǫ = b +ǫ 0( ), the we fiish checkig all coditios of theorem 6 get satisfied. By theorem 6, θ is a irratioal umber, or equivaletly, it s ot a first order algebraic umber,by the coclusio 2, whe is big eough, there exists ifiite fractios c satisfy θ c < 2+ǫ,by Roth theorem we get θ is a trascedetal umber. Example 6. m= 0 m! is a trascedetal umber. Because by takig a =, =, 2,, = 0! ad ǫ =, (0! ) 3 0 +! = umber. a + b 2+ǫ + = 0( ),by theorem 7, 0!( 2) m= is a trascedetal 0 m! Example 7. 3 = 23 is a trascedetal umber. Because by takig a = 3, =, 2,, = 2 3 ad ǫ = 2, a + b 2+ǫ 3 + = 9 3 0( ),by theorem 7, 3 2 3 = 23 is a trascedetal umber. Theorem 8 Let θ = = a b,where a > 0( =, 2 ) are ratioal umbers which satisfy ()b < b 2 < ad +, =, 2,
(2) a +b db + < 2 (d 2 is a iteger ad is big eough) The idifferet fixed poit z = 0 of polyomial f(z) = z d + +e 2πiα belogs to Julia set Proof. Set f( ) = b db,it s easy to verify f( ) satisfy coditio ad 3 of theorem 6. We oly eed to verify coditio 2 ad 4. Sice +, ( =, 2, ) ad + 2,we have f( ) f(+ ) = bdb b d+ + (b 2 +) d 2 b+ b d+ + = ( 2 )d 2 b+ b d 2 b+ + b d+ + < ( 2 )d 2 + < 2 thus coditio (2) is satisfied. Let s verify coditio (4), Sice whe 2, 2 ad d 2,d b 3, thus whe 2 f( ) = b db b 2 0( ) By theorem 6,θ is a irratioal umber ad there exists ifiite fractios c such that θ c < get our result db b ( is big eough), the by Cremer theorem, we Example 8. I order to illustrate this example, we eed some otatio to describe a special series so called th expoetial floor as follows: Set [a, a,, a ] = f where f is defied iductively by f k+ = (a k+ ) f k, k =, 2, ad f = a For ay positive iteger d 2, let = [d,, d, d] 2 ad θ = =, we will show that idifferet fixed poit z = 0 of polyomial g(z) = Z d + 2
+ e 2πiθ z belogs to Julia set. Let s check satisfy coditios of theorem 8, coditio is obvious ad by oticig [d,, d, d] 2 = d [d,,d,d] 2,we have a + b db + = ([d,, d, d] 2) [d,,d,d]2+ [d,, d, ( + )d] 2(+) < ([d,, d, d] 2) [d,,d,d] 2(+) [d,, d, ( + )d] 2(+) = d([d,,d,d] 2 )([d,,d,d] 2+ ) [d,, d, ( + )d] 2(+) = dd [d,,d,d] 2 2 +[d,,d,d] 2 [d,, d, ( + )d] 2(+) = dd [d,,d,d] 2 2 +[d,,d,d] 2 d d[d,,d,(+)d] 2 We otice that [d,,d,d] 2 2+[d,,d,d] 2 [d,,d,(+)d] 2() 0( ) That meas whe is big eough, a +b db + < 2 ad we fiish checkig this example satisfies all coditios of theorem 8 ad thus idifferet fixed poit of g(z) belogs to Julia set. At the ed of paper, we are goig to propose a cojecture which relates to theorem 2. To do this, we eed followig defiitio firstly. Defiitio 9 Let α be a irratioal umber,if α satisfies followig coditios: () α = =, where = a, ( =, 2, ) ad a, are positive itegers. (2) +, ( =, 2, ) 3
(3) Cojecture.Every positive irratioal umber has E ratioal approximatio. lim a + = 0 We call the irratioal umber α has E ratioal approximatio. Remark. The positive aswer of above cojecture will give a explicit character of ay positive irratioal umber. Refereces [] L. K. Hua Itroductio to umber theory, Scietific Press, 979. [2] H. Cremer Zum Zetrumproblem, Math. A. 98 5-63 4