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#1 2018-02-11 20:48:31
- George,Y
- Member
- Registered: 2006-03-12
- Posts: 1,379
the Mythical formula to calculate Numerical Derivatives!
I come across this method when trying to calculate the derivative of an unknown curve which I have only sample points at equal grids.
The simplest way to calculate numerical derivative is three-point formula:
dy/dx = (y[1]-y[-1]) /2/dx
ddy/dx/dx = (y[1]+y[-1]-2*y[0]) /dx/dx
But I found this one on five points :
dy/dx = ( y[-2] -8*y[-1] +8*y[1] -y[2]) / (12*dx)
ddy/dx/dx = (-y[-2] +16*y[-1] -30*y[0]+ 16*y[1] -y[2]) / (12*dx^2)
It is surprisingly accurate when I test the derivative on exp(3) using exp(2) exp(2.5) exp(3) exp(3.5) exp(4)
the result of 1st numerical derivative and 2nd are
20.04243 & 20.07127
remember the correct answer are both exp(3) = 20.08554
Can anyone prove how this works?
4th polynomial?
X'(y-Xβ)=0
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#2 2018-02-12 08:14:24
Re: the Mythical formula to calculate Numerical Derivatives!
There are analogous formulae for higher order derivatives too, and several published papers about the error term.
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#3 2018-02-12 22:42:05
- George,Y
- Member
- Registered: 2006-03-12
- Posts: 1,379
Re: the Mythical formula to calculate Numerical Derivatives!
These coefficients are part of a more general phenomenon called the Savitzky-Golay filter in numerical analysis. In fact the traditional five-point 'stencil' reads:where . You can derive this result simply by playing around with the Taylor series for with . It's a little tedious, but you can see the derivation here: https://en.wikipedia.org/wiki/Five-poin … he_formulaThere are analogous formulae for higher order derivatives too, and several published papers about the error term.
That is very informative, zeta. Thanks for the reply.
I wonder if there is a way to calculate the midpoint differentials by 4 points around it:
dy/dx ]x=0
ddy/dx/dx ]x=0
given y[-0.5], y[0.5], y[-1.5] & y[1.5]
X'(y-Xβ)=0
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