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Re: approximation by rationals
*To*: mathgroup at smc.vnet.net
*Subject*: [mg23493] Re: [mg23487] approximation by rationals
*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>
*Date*: Sun, 14 May 2000 16:59:57 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
If you read your own statement of the theorem the answer should become
obvious. The theorem quite correctly states that if (a/b) satisfies the
inequality then (a/b) is a convergent of z, but it does not state that every
convergent of z satisfies the inequality. This is the difference between a
necessary condition and a sufficient condition. That's all. It is in fact
untrue that the necessary condition (to be a convergent) holds for Pi but
not for E. It does not hold for either. Here is a proof that it also does
not hold for Pi using Mathematica::
Load in the continued fractions package:
In[1]:=
<< NumberTheory`ContinuedFractions`
define the test:
In[2]:=
test[f_, Rational[c_, d_]] := Abs[f - c/d] < 1/(2d^2) && d >= 1
Let's take the first 30 convergents of Pi:
In[3]:=
conv = Convergents[Pi, 30];
Let's check how many satisfy the test:
In[4]:=
Length[conv1 = Select[conv, test[Pi, #] &]]
Out[4]=
21
We can see the ones that do not:
In[5]:=
Complement[conv, conv1] // InputForm
Out[5]//InputForm=
{3, 333/106, 103993/33102, 208341/66317, 833719/265381,
4272943/1360120, 5706674932067741/1816491048114374,
5371151992734/1709690779483, 14885392687/4738167652}
on 5/13/00 11:54 AM, Matt Herman at Henayni at hotmail.com wrote:
>
> hi,
>
> there is a number theory theorem which states that if |z-(a/b)|< 1/(2
> b^2), then a/b is a convergent of z. If this is true, then why when I go
> into mathematica and set z=E and a/b = 106/39, which is a convergent
> of E, the inequality is false? Is this a problem in mathematica's
> numerical evaluation, or a problem in the theorem? Is the theorem only
> valid when z is a quadratic irrational? Because the theorem works for
> z=Pi. Also the stronger form of the theorem is |zb-a|<1/2b^2. This is
> false when z=E. Again the same question as above.
>
>
> Matt
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