hello everyone.
i’m dealing with some struggles when programming in Julia since I come from matlab.
I need to extrapolate this plot so I can get rho(0).
any help?
version 1.6.3
http://juliamath.github.io/Interpolations.jl/dev/interpolations/
btw, I think you may want a “fit”, do you have a formula you want to fit?
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yeah, a fit would really do.
actually this is a plot of two vectors rho x a
a = [1 2 4 6 8 16 32]
rho = [ 680 610 415 295 235 190 180]
i was wondering if there is a command to do this instead of finding the line equation.
In order to extrapolate, you really need to have some model in mind, even if you don’t explicitly fit.
For example, if you want to extrapolate using a straight line from the first two points, you can use extrapolate(..., Line()) from Interpolations.jl.
If you think that your data models a peak, so that 1/rho is approximately a low-degree polynomial, you can use Richardson extrapolation from Richardson.jl
For example, with your data I get an extrapolated value of 651 ± 7:
julia> val, err = extrapolate(zip(reverse(inv.(rho)), reverse(a)))
(0.0015350496476734183, 1.664001386944276e-5)
julia> 1/val
651.4447278728993
julia> err / val^2 # convert error estimate in val to error in 1/val via the derivative
7.061692970914665
If you additionally know that your data is symmetric around a=0, \rho(-a) = \rho(a), then you can do even-order Richardson extrapolation (i.e. force it to use symmetric polynomials), which predicts \rho(0) = 742 \pm 6:
julia> val, err = extrapolate(zip(reverse(inv.(rho)), reverse(a)), power=2)
(0.0013481200793223414, 1.1069973063407247e-5)
julia> 1/val
741.7736856962099
julia> err / val^2
6.091011361487163
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Expanding on this: if for instance you think \rho = \frac{A}{a+B} then you can do
using LsqFit
model(x, p) = p[1] ./ (x .+ p[2])
fit = curve_fit(model, a, rho, [1.0, 1.0])
model(0, fit.param)
The same can be done for any (sensible) model.
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Without knowing the underlying model, will draw one more number in the lottery:
using Dierckx
a = [1, 2, 4, 6, 8, 16, 32]
rho = [ 680, 610, 415, 295, 235, 190, 180]
spl = Spline1D(a, rho; k=2, bc="extrapolate")
ρ₀ = evaluate(spl, 0) # ρ₀ = 723
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Not sure if I understand the scope of the problem. Is the underlying function supposed to be analytic, are the points exact or can they contain noise, …
Do you mind sharing the Matlab code?
Honestly if there’s a function that lets you do this without specifying a model by which to do it, that function is lying to you. And if such a function is built into MATLAB, that’s a flaw of MATLAB, not a strength.
2 Likes