Thank you Luc, this is very useful. Definitely much easier than what I was thinking of doing
:).
sujit
On Mar 31, 2012, at 10:13 AM, Luc Maisonobe wrote:
> Le 31/03/2012 01:31, SUJIT PAL a écrit :
>> Thanks Luc, thanks for getting back so quickly, and will wait for your information
tomorrow.
>
> Hi Sujit,
>
> Sorry for the delay.
>
> If you know your sample points (xi, yi) are all separated by the same
> interval h (i.e x(i+1) = x(i) + h for all i), then you can compute any
> derivatives using the following method:
>
> First write down the series expansion of the function y = f(x):
>
> y(x + k * h) = y(x)
> + k * h * y'(x)
> + (k * h)^2 / 2 y''(x)
> + ...
> + (k * h)^n / n! yn(x)
> + ...
>
> Then you truncate this expression to at least include the derivative you
> want, and apply it to a number of points equal to your truncation order.
> Here, as an example, we will truncate just to order 2 in order to have
> y'', so we have three components (y, y' and y''), so we need three
> points. Let's take k = 1, k = 0 and k = 2. We have the following
> expressions:
>
> y(x  h) = y(x)  h y'(x) + h^2/2 y''(x)
> y(x) = y(x)
> y(x+h) = y(x) + h y'(x) + h^2/2 y''(x)
>
> Note that the first and last expression differ only in the sign of the
> odd terms h y'(x), the even terms (h^0 and h^2) have the same sign.
>
> You can rewrite these expressions as a simple square matrix vector
> multiplication:
>
> [ y(xh) ] [ 1 h h^2/2 ] [ y(x) ]
> [ y(x) ] = [ 1 0 0 ] * [ y'(x) ]
> [ y(x+h) ] [ 1 h h^2/2 ] [ y''(x) ]
>
> Then you invert this system and get:
>
> [ y(x) ] [ 0, 1, 0 ] [ y(xh) ]
> [ y'(x) ] = [ 1 / 2h, 0, 1 / 2h ] * [ y(x) ]
> [ y''(x) ] [ 1 / h^2, 2 / h^2, 1 / h^2] [ y(x+h) ]
>
> The third row is the one you want:
>
> y''(x) = (y(xh)  2 * y(x) + y(x+h)) / h^2
>
> You can check this formula using again the initial expansion, and you
> will see this combination basically canels out the y(x) and y'(x) part
> and only y''(x) remains.
>
> You could also have used more points, which could improve accuracy as it
> fits your function to an higher degree. Beware that you should *not*
> increase the number of points too much, otherwise you would get Gibbs
> oscillation problems (i.e a polynomial that has very large changes
> between the sampling points).
>
> Hope this helps
> Luc
>
>
>>
>> sujit
>>
>> On Mar 30, 2012, at 2:36 PM, Luc Maisonobe wrote:
>>
>>>
>>>
>>>
>>> SUJIT PAL <sujit.pal@comcast.net> a écrit :
>>>
>>>> Hi,
>>>>
>>>> I have a (newbie) question about how to go about solving the problem
>>>> below with commonsmath.
>>>>
>>>> 1) I have histogram data (equal x intervals) based off some
>>>> distribution.
>>>> 2) I need a way to calculate the second differential between two given
>>>> x points in the distribution.
>>>>
>>>> Here is what I think I should do:
>>>> 1) Feed the (x,y) values represented by the top of each histogram entry
>>>> within my xrange into a UnivariateRealInterpolator between ranges
>>>> (x1,y1) and (x2,y2).
>>>> 2) The interpolate method gives me back a PolynomialSplineFunction
>>>> (call if psf).
>>>> 3) I get the second differential UnivariateRealFunction (call it urf)
>>>> by calling psf.getPolynomialSplineDerivative().derivative()
>>>> 4) I compute the second derivative value as the difference of
>>>> urf.value(x2)  urf.value(x1)
>>>>
>>>> Does this sound like a reasonable/accurate approach? Any
>>>> suggestions/gotchas or pointers to better approaches?
>>>
>>> No, this is not an approach I would recommend. A much more straightforward
>>> way to compute this derivative is to apply a dedicated finite differences formula
>>> on three consecutive points. I'll give ou such a formula tomorrow (I am on a
mobile
>>> device right now and it is difficult to write a complète mail...)
>>>
>>> Luc
>>>
>>>>
>>>> Thanks very much,
>>>>
>>>> Sujit
>>>>
>>>>
>>>> 
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>>>
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>>
>>
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