Re: Developer Registration Request

>
> On 2 Jun 2009, at 21:12, dmelliott wrote:
>
>>
>> Dear Søren,
>>
>>   More thoughts:
>>
>>
>>   The idea of this routine is that it integrates existing data in  
>> its most
>> ususal form.  That is, it does not need a function to call, and,  
>> since, for
>> better or for worse, most data comes on a domain of equally sized  
>> steps,
>> this is what it uses.  This is an attempt to make it compatable  
>> whith most
>> experimantal data, and the outputs of other routines, e.g. sales by  
>> week,
>> Fourie analysed amplitudes by frequency, actuarial data by age,  
>> etc.  The
>> patrial interval integrals, including cumulative, can be achieved  
>> by looping
>> or otherwise manipulating the endpoint values.
>>
>>   The best way to see its worth is to compare it to the available  
>> routine
>> for existing data, by editing the "integrator_verifier" to use the  
>> "trapz"
>> routine.  The latter is only of first order, and even for as many  
>> as 500
>> points on [0,1] yields a 2.01e-6 relative error for the next higher  
>> order,
>> the second.  For higher orders higher than second it is even more  
>> useless.
>> There probably should a caution appended to the instructions about  
>> this.
>
> Douglas,
>
> It seems to me there is one detail you are missing here:
> experimental data are not always polynomials, actually most often  
> you don't even have a way to tell wether they are smooth, if all you  
> have are equispaced samples.
>
> Using a high order quadrature rule only makes sense if you know a-
> priori your data can be
> accurately approximated by a polynomial, e.g. by Taylor expansion.
> If you have no a-priori knowledge about the smoothness of your data,  
> using a low order quadrature rule is a safer bet. For example  
> consider this well known example:
>
> >> x = linspace (-1, 1, 10);
> >> f = @(x) 1./(1+25.*x.^2);
> >> F = f(x);
> >> integrator (F, -1, 1)
> ans =  0.47972
> >> sum ((F(2:end)+F(1:end-1)).*diff(x)/2)
> ans =  0.54437
>
> while the exact value of the integral is
>
> >> .4 * atan (5)
> ans =  0.54936
> >> quad(f, -1, 1)
> ans =  0.54936
>
> so in this case even the simple trapezoidal rule beats your 10th  
> order method because it's based on approximating f by means of a  
> highly-order and thus highly oscillating polynomial.
>
> So I believe your function would actually be more useful if the  
> order of the quadrature rule to use was to be given as an input  
> option rather than be set automatically depending on the size of the  
> sample.
>
> HTH,
> c.