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Herman,
my method is a bit different.
I begin with the 1st digit accuracy [10 steps maximum, 1.0 to 1.9]
and localize
1.4<sqrt(2)<1.5
Then, for the 2nd digit, 10 steps [maximum] again to obtain
1.41<sqrt(2)<1.42
and so on.
In the average, I need 5 steps per digit and, for a 3-decimal
accuracy it will take about
5*5*5=125 steps instead of the normal 1000.
Archimedes, the copyright of the method, was not working with
decimals.
For some reason [not explained anywhere], his first choice was the
sevenths of the unity 1/7, 2/7, 3/7 etc .
So, in his famous "Measurement of a Circle", he proves that
3+1/7>pi>3+10/71
As you see, he slightly increases the denominator [he does not
increase the numerator].
There are some obscure statements, he never explained, for example,
how did he came to the [useful approximation]
265/153<sqrt(3)<1351/780 [!!!].
but, we can not always have what we want...
Dimitris Tsokakis
--- In amibroker@xxxxxxxxxxxxxxx, "Herman vandenBergen" <psytek@xxxx>
wrote:
> thanks DT, I have to study your code but I think you have the
general idea.
> I have an impossible formula to transform (it contains HHV and LLVs,
> stochastic mutation) and want to find the x that would give me the
given y.
> Right now I linearly increment x untill I hit my y-target, this is
awfully
> slow.
>
> like y = function(x); // y ranges 0-100 and I want 2 decimal places
for x
> that gives me a given y
>
> I thought cutting the range in half, see whether it is greater or
less, cut
> the result in half again, etc. I am not a math guy but I have used
> AD-converters that worked like that and what we can do in hardware
we can do
> in software :-)
>
> thanks for the starter DT,
> herman
>
>
>
>
> -----Original Message-----
> From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@x...]
> Sent: January 28, 2004 9:18 PM
> To: amibroker@xxxxxxxxxxxxxxx
> Subject: [amibroker] Re: Successive Approximation in afl
>
>
> Herman,
> You mean a procedure like this
>
> a=1;b=2;z=0;
> st0=0.1;st1=0.01;st2=0.01;
> //1st decimal
> for(i=a;i<b;i=i+st0)
> {
> if(i^2<2 AND (i+st0)^2>2)
> z=i;
> }
> //2nd decimal
> for(i=z;i<z+st0;i=i+st1)
> {
> if(i^2<2 AND (i+st1)^2>2)
> z=i;
> }
> //...etc
> Plot(z,"sqrt(2)",1,1);
>
> to find sqrt(2) without using all the values from 1 to 2 ?
> Dimitris Tsokakis
> --- In amibroker@xxxxxxxxxxxxxxx, "Herman vandenBergen"
<psytek@xxxx>
> wrote:
> > Hello,
> >
> > has anybody come accross a successive approximation routine in
afl?
> Or
> > perhaps js?
> >
> > thanks,
> > herman
>
>
>
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