Proof Of Functional Smarandache Iterations Of Kind1

PROOF OF FUNCTIONAL SMARANDACHE ITERATIONS OF KINDS

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ZHENG JIANFENG Shaanxi Financi & Economics Professional College Xianyang,Shaanxi, P.R.China ABSTRACT: The paper makes use of method of Mathematics Analytic to prove Functional Smarandache Iterations of three kinds. 1.Proving Functional Smarandache Iterations of First Kind. Kind 1. Let f : A → A be a function, such that f ( x) ≤ x for all x, and min { f ( x), x ∈ A } ≥ m0 , different from negative infinity. Let f have p ≥ 1 fix points: m0 ≤ x1 < x 2 <

< xp.

[The point x is called fix, if

f ( x) = x .]. Then: SI1( x) = the smallest number of iterations k such that f

f ( x) )) = constant.

f(f(

iterted k times

Proof: I.When A ⊆ Q or A ⊆ R, conclusion is false. Counterexample: Let A=[0,1] with f ( x) = Denote:

An (x) =

x

2

, then f ( x) ≤ x , and

f ( f (… f ( x) …))

,

A1 ( x) =

x1 = 0 , x2 = 1

are fix points.

f ( x) , ( n=1,2,…).

n −times

then

An ( x) = x2

n

( n=1,2,… ).

For any fixed x ≠ 0, x ≠ 1 , assumed that the smallest positive integer k exist, such that An ( x) = a (constant), hence, Ak +1 ( x) = f ( Ak ( x)) = f (a) = a , that is to say a be fix point. So

x2

k +1

=0 or 1, ⇒

x=0 or 1, this appear contradiction. If

A ⊆ Z, let A be set of all

rational number on [0,1] with f ( x) = x2 , using the same methods we can also deduce contradictory result. This shows the conclusion is false where A ⊆ Q or A ⊆ R. II. when A ⊆ Z, the conclusion is true. (1). If x = xi ( xi is fix point, i=1,…p ). Then f ( x) = f ( xi ) = xi = A1 ( x) . So for any positive integer n,

An (x) = xi

( i=1,…p ), ⇒

SI1( x) =1. f

Keywords and phrases. Functional iterations; fix point; limit. 1

(2). Let x ≠

xi

( x is fixed ,i=1,… p ), if f ( x) =

f ( x) ≠ xi but f ( f ( x)) =

A2 ( x) = xi

positive integer k, if

( i=1,…p ), then

xi

( i=1,…p ), then

SI1( x) =1,

if

f

SI1( x) =2.

In general, for fixed

f

A1 ( x) ≠ xi , A2 ( x) ≠ xi

…

Ak −1 ( x) ≠ xi ,

but

Ak ( x) = xi

then

SI1( x) =k. f

(3). Let x ≠

xi

( x is fixed ), and for ∀ n ∈ N

An ( x) ≠ xi

( i=1,…p ), this case is

no exist. Because x is fix point, m0 < < An ( x) < < A2 ( x) < A1 ( x) < x . So sequence { An (x) } is descending and exist boundary, this makes know that { An (x) } is convergent. But, each item of { An (x) } is integer, it is not convergent, this appear contradiction. This shows that the case is no exist. (4). Let x ≠ xi ( x is fixed ,i=1,… p ), if exist the smallest positive integer k such ( a ≠ xi Ak ( x) = a Ak +1 ( x) = f ( Ak ( x)) = f (a) = a ,

that

),

it

is

yet

unable.

Ak +1 ( x) = Ak ( x) = a , namely, a = xi , this also

Because

this shows that a is fix point ,

appear contradiction. Combining (1),(2),(3) and (4) we have SI1( x) = the smallest number of iterations k such that f

f(f(

f ( x) )) =

xi

( xi is fix point, i=1,…p ).

iterted k times

This proves Kind 1. We easily give a simple deduction. Let f : A → A be a function, such that f ( x) ≤ x for all x, and

min{ f ( x), x ∈ A } ≥ m0 ,

different from negative infinity. Let f (m0) = m0 , namely, m0 is fix point, and only one. Then:

SI1( x)

= the smallest number of iterations k such that

f

f(f(

f ( x) )) = m0 .

iterted k times

2.Proving Functional Smarandache Iterations of Second Kind. Kind 2. Let g : A → A be a function, such that g ( x) > x for all x, and let b > x. Then: SI 2( x,b) = the smallest number of iterations k such that g

g(g(

g ( x) )) ≥ b .

iterted k times

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Proof: Firstly, denote:

Bn (x) =

g(g(

g ( x) )) , ( n=1,2,…). n times

I. Let A ⊆ Z, for ∀ x < b , x ∈ Z , assumed that there are not the smallest positive integer k such that B k ( x) ≥ b ,then for ∀ n ∈ N have B n ( x) < b , so x < B1 ( x) < B 2 ( x) < < B n ( x) < < b . This makes know that { B n (x) } is convergent, but it is not convergent. This appear contradiction, then, there are the smallest k such that II. Let A ⊆ Q or A ⊆ R. (1). For fixed x b , then if g ( x) < g (b) but general, if

Bn ( x) ≥ g ( x) > b

B2 ( x) ≥ g (b) > b ,then Bn ( x) ≥ g (b) > b

B1 ( x) < g (b)

,

B2 ( x) < g (b)

,…

B n ( x) ≥ b .

Bk −1 ( x) < g (b)

(

n ∈ N ),

SI 2( x,b) =1, g

( n ≥ 2 ),

SI 2( x,b) =2.

In

g

, but

Bk ( x) ≥ g (b) > b

, then

*

≤ g (b) .

SI 2( x,b) =k. g

(2).For fixed x < b , x < B1 ( x) < B 2 ( x) <

Bn ( x) < g (b) , < B n ( x) <

( n∈ N )

< g (b) ,

so { B n (x) } is convergent. Let lim B n ( x) = b∗ ∵ →∞

1).

*

b

= g (b) . ∵

lim B n ( x) = b* ∴ n →∞

then

Bn ( x) < g (b)

( n ∈ N ), ∴

b

for ε = g (b) − b > 0 , ∃ positive integer k, when n>k such

Bn ( x) > g (b) − ε = g (b) − ( g (b) − b) = b . That is to say there are the smallest k such that B n ( x) > b . 2). b* < g (b) . ∵ g (b*) > b* , ∴ { B n (x) } does not converge at g (b*) . So ∃ ε 0 > 0 , for ∀ N, ∃ n1 , when n1 > N , such that B ( x) − g (b*) ≥ ε 0 , then, n

that

Bn ( x) − g (b)

< ε . So

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* ( n ∈ N ), ∴ B n1 ( x) ≥ g (b ) + ε 0 ∴ B n1 ( x) > b + ε 0 . On the other hand, B n ( x) ≤ b * * * B n1 ( x) ≤ b then b + ε 0 < Bn1 ( x) ≤ b , but this is unable. This makes know that there is not *

*

the case. By (1) and (2) we can deduce the conclusion is true in the case of A belong to Q or R. Combining I. and II., we have: for any fixed x >b there is SI 2( x,b) = the smallest number of iterations k such that g

g(g(

g ( x) )) ≥ b .

iterted k times

This proves Kind 2. 3.Proving Functional Smarandache Iterations of Second Kind. Kind 3. Let h: A → A be a function, such that h(x) < x for all x, and let b < x. Then:

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SI 3( x, b) = the smallest number of iterations k such that h

h(h( h( x) )) ≤ b . iterted k times

Using similar methods of proving Kind 2 we also can prove Kind 3, we well not prove again in the place. We complete the proofs of Functional Smarandache Iterations of all kinds in the place. REFERNECES 1.“Functional Iterations” at http://www.gallup.unm.edu/~smarndache/bases.txt 2. East China Normal University Department of Mathematics Writing, Mathematics Analytic, People`s Education Press, Shanghai,1982-4.

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