From: tltran@zeus.calpoly.edu (ta^m)
Subject: Re: (none)
Followup-To: just a try 
Organization: California Polytechnic State University, San Luis Obispo
Date: Mon, 09 Nov 1992 03:57:09 GMT

In article <8v-qsnl@lynx.unm.edu> ngoclu@triton.unm.edu (ngoc-dung t  luntanondana) writes:
>>From lynx!news.cs.indiana.edu!usenet.ucs.indiana.edu!ux1.cso.uiuc.edu!sdd.hp.com!decwrl!netsys!agate!dog.ee.lbl.gov!news!manta!dam Fri Nov  6 13:02:03 MST 1992
>>Article: 21275 of soc.culture.vietnamese
>>Newsgroups: soc.culture.vietnamese
>>Path: lynx!news.cs.indiana.edu!usenet.ucs.indiana.edu!ux1.cso.uiuc.edu!sdd.hp.com!decwrl!netsys!agate!dog.ee.lbl.gov!news!manta!dam
>>From: dam@nosc.mil (Khanh Dam)
>>Subject: Re: (none)
>>Message-ID: <1992Nov4.183520.3088@nosc.mil>
>>Summary: just a try
>>Organization: Naval Ocean Systems Center, San Diego, CA
>>References: <9211040511.AA193825@nike.calpoly.edu>
>>Date: Wed, 4 Nov 1992 18:35:20 GMT
>>Lines: 51
>>
>>In article <9211040511.AA193825@nike.calpoly.edu>, tltran@nike.calpoly.edu (ta^m) writes:
>>> 
>>>       To^i co' ba`i toa'n na`y la`m hoa`i kho^ng ddu*o*.c. Ne^'u co' anh
>>> chi. na`o bie^'t la`m o*n help me.(especially part b)
>>>   -----
>>> 
>>>       Given T(1) = 2,
>>>               T(n+1) = [T(n)]^2 - T(n) + 1,    n > 0.
>>>       Prove:
>>> 
>>>     a,  If m != n, T(m) and T(n) have no common factor greater than 1
>>>     b,  calculate
>>>              1/T(1) + 1/T(2) + 1/T(3) + ...
>>> 
>>>     note:
>>>         T(n) mean T subscript n
>>>         m != n mean m not equal to n
>>> 
>>>                                                        Thankx
>>>                                                         Ta^m
>>> 
>>
>>Ho^m nay ra?nh ro^?i, -do.c ba`i na`y tha^'y hay hay ne^n Kh. la`m thu+? nhe':
>
>>T(n+1) = T(n)[ T(n) - 1 ] +1,   n > 0
> 
>>b).
>>	 T(1) = 2
>>n = 1 => T(2) = T(1)[ T(1)-1 ] + 1
>>	      = 2[1] + 1 = 3
>>
>>n = 2 => T(3) = 3[2] + 1 = 7
>>n = 3 => T(4) = 7[6] + 1 = 43
>>n = 4 => T(5) = 43[42] + 1 = 1807
>>n = 5 => T(6) = 1807[1806] + 1 = 326,3443 
>>.
>>.
>>.
>>  => 1/T(1) + 1/T(2) + 1/T(3) + ... = 1/2 + 1/3 + 1/7 + 1/43 + 1/1807 + ....
>>				    = 1.0000
>
>>a).
>>     for any m != n => T(m) != T(n) (or T(m+1) != T(n+1)) and the terms
>>		       {T(m)[ T(m)-1 ] +1}  and  {T(n)[ T(n)-1 ] +1}
>>		       will have no common factor > 1 (because of the +1 term).
>>
>>     this can be shown by induction of the above terms for n = 1, 2, 3, ...

       ba.n Kh.
          Ta^m kho^ng nghi~ la\ ba.n co' the^? du\ng Induction cu\ng
        mo^.t lu'c vo*'i ca? m va\ n.

>>
>>
>>C. Kh.
>    I have a comment: Ba.n Khanh  tuy ra(`ng ti`m ddu'ng dda'p so^' 
>cu?a to^?ng song ca'ch tri`nh ba`y va` ti'nh nhu' va^.y co' ve? e'p qua'.
>To^i co' ca'ch na`y, kho^ng bie^'t Ta^m va` Khanh xem co' ddu'o'.c kho^ng
>( To+' mo+'i vu+`a ddo.c ba`i toa'n na`y tho^i)
>   b-
>    let's call S(n)=1/T(1)+...+1/T(n)
>    We have
>     T(n+1)=T(n)**2-T(n)+1=T(n)[T(n)-1]+1
>  => T(n+1)-1=T(n)[T(n)-1]
>  Hence,
>     T(2)=T(1)[T(1)-1]+1 => T(2)-1=T(1)[T(1)-1]
>     T(3)=T(2)[T(2)-1]+1
>         =T(2).T(1).[T(1)-1]+1
>     .
>     .
>     .
>     T(n-1)=T(n-2)...T(1)[T(1)-1]+1
>     T(n)  =T(n-1).T(n-2)...T(1)[T(1)-1]+1
>  From this we have:
>
>     T(n)-T(n-1)...T(1)(T(1)-1)= 1                   (*)
>
>     Let's call A(i) an element of the series S(n); then A(k)=1/T(k).
>
>     We set up another series P(n) = sum (from 1 to n) of B(k)
>with B(k)=1/[T(k-1).T(k-2)...T(1).(T(1)-1)] for k>1, and
>B(1)=1/[T(1)-1] ( for obviously T(1)#1).
>( notice that there is no extra 1 in the denominator of B(k)).
>
>     Then, for any k > 1 we have:
>     B(k)-A(k)=1/[T(k-1)...(T(1)-1)] - 1/T(k)   
>              =[T(k)-T(k-1)...(T(1)-1)] / [T(k)...T(1).(T(1)-1)]
>              = 1/[T(k)...T(1).(T(1)-1)]   ( because of (*))
>              = B(k+1) ( because of the way we set up B(k))
>     Therefore:
>     B(1) - A(1) = 1/[T(1)-1]-1/T(1)=B(2)
>     B(2) - A(2) = B(3)
>     .
>     .
>     .
>     B(n) - A(n) = B(n+1)
>  _______________________________________
>=>   B(1) - S(n) = B(n+1)

  Ba.n ngoc dung: (cho phe'p go.i "ba.n" nhe')
         Ta^m tha^.t kho^ng hie^?u cho^? na\y vi\ theo ca'ch set up cu?a
  ba.n thi\:
           S(n) = 1/T(1) + ... + 1/T(n)  
           A(n) = 1/T(n)
         tha\nh ra:  
           A(n) != S(n) 
         nhu*ng theo hai phe'p ti'nh tre^n thi\
           A(n) = S(n)
         va^.y cha(?ng ho'a ra la\ contradiction sao?

>=>   S(n) = B(1)-B(n+1)
>     It is easy to see that as  n goes to infinity, B(n+1) goes to 0.
>     Therefore,
>     S(n) = 1/T(1)+...+ 1/T(n) goes to 1 as n goes to infinity
>      or
>1/T(1)+ 1/T(2)+........=1
>     
>    
>
>
>


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