[Date Index]
[Thread Index]
[Author Index]
Re: Proving inequalities with Mathematica
*To*: mathgroup at smc.vnet.net
*Subject*: [mg52499] Re: [mg52491] Proving inequalities with Mathematica
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Tue, 30 Nov 2004 05:24:01 -0500 (EST)
*References*: <200411290622.BAA27997@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
On 29 Nov 2004, at 15:22, Toshiyuki (Toshi) Meshii wrote:
> Hi,
>
> I was wondering whether Mathematica is useful for proving a problem of
> inequality.
> My problem is as follows:
>
> Let An and Bn (n=1,2,3...) be real sequences.
> Some characteristics of these sequences are known.
>
> i) Abs[An+1/An] < 1
> ii) Abs[Bn+1/Bn] < 1
> iii) Sum[An, {1,Infinity}] = 0
> iv) Sum[An*Bn, {1, Infinity}] = alpha (note: a real number)
>
> Then I want to prove with Mathematica that
> 0 < Abs[alpha] < Abs[A1*B1]
>
> Does anyone have an idea?
>
> -Toshi
>
First of all, you have written your condition as
> Abs[An+1/An] < 1
Presumably you meant n+1 to be a subscript (if not the condition is
always false).
Assuming that, Mathematica can certainly be helpful in proving your
claim to be false. Here is an example.
We define the sequence A[n] as follows:
A[1] = -1; A[2] = Log[2];
A[n_] = (-1)^(n - 1)/(n - 1);
You can check that the condition Abs[A[n+1]/A[n]]<1 is always
satisfied. We also define the sequence B[n] by
B[i_] := A[i]
so that the condition
Abs[B[n+1]/B[n]]<1
is satisifed.
We check that (iii) is satisfied:
A[1]+A[2]+Sum[A[n],{n,3,Infinity}]//Simplify
0
Let's now compute alpha:
alpha=A[1]*B[1]+A[2]*B[2]+Sum[A[i]*B[i],{i,3,Infinity}]//N
2.12539
Let's see if your inequality holds:
Abs[alpha]<Abs[A[1]*B[1]]
False
Oops ...
Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/~andrzej/
http://www.mimuw.edu.pl/~akoz/
Prev by Date:
**pair functions applied to Zeta function in two ways**
Next by Date:
**Re: sort procedure**
Previous by thread:
**Proving inequalities with Mathematica**
Next by thread:
**Re: Proving inequalities with Mathematica**
| |