Hello all. I've been trying to use `LsqFit` to fit a hyperbolic tangent function t

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Hello all. I’ve been trying to use LsqFit to fit a hyperbolic tangent function to some data about terminal velocity in free fall. Using SciPy’s curve_fit function, it seems to work fine. However, in Julia, it’s nowhere near the correct fit.
Here’s the function I’ve written to fit the curve -

function fit_terminal_velocity(data::CSV.File)
	# Create base model with random initial parameters
	p0 = rand(2)
	model(t, p) = p[1] .* tanh.(p[2]/p[1] .* t)
	
	# We are fitting v(t), so lets get those values
	# We also want to go from t = 0 to t = t_max
	t = data.t .- minimum(data.t)
	y = data.y
	
	v  = diff(y) ./ diff(t)
	tv = t[1:end-1]
	
	
	fitted_model = lsqfit.curve_fit(model, v, tv, p0)
	
	return fitted_model, model
end

And it gives

p[1] = -1.04293e-10
p[2] = 0.00917912

Whereas Python gives p = [-623.23056677, -319.5100746 ] , which fit the model correctly. For reference, the code in python is

f = lambda t, g, vt : vt * np.tanh(g/vt * t)
popts, pcov = curve_fit(f, t[1:], v)

Where curve_fit is scipy.optimize.curve_fit.

I think it may be something to do with the fact that LsqFit requires p0 but SciPy doesn’t? I would really appreciate if anybody can help me figure out why curve fitting isn’t working in Julia properly. Thank you so much!

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Hello, I am the poster on slack. If anybody has a solution to this it would be greatly appreciated
Here is the data I’m trying to fit - https://transfer.sh/seiAb/0.0grams.csv

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Hello everybody!

I figured out the answer! Python’s scipy.optimize.curve_fit sets the array of initial parameters to ones if not supplied. LsqFit on the other hand does not try to determine the parameters automagically and thus needs them passed in. SciPy sets the initial parameters all to ones if not passed, so if I set the initial parameters to ones in Julia as well, I get the exact same fit. So the new code looks like -

function fit_terminal_velocity(data::CSV.File)
	# Create base model with random initial parameters
	p0 = ones(2)
	model(t, p) = p[1] .* tanh.(p[2]/p[1] .* t)
	
	# We are fitting v(t), so lets get those values
	# We also want to go from t = 0 to t = t_max
	t = data.t .- minimum(data.t)
	y = data.y
	
	v  = diff(y) ./ diff(t)
	tv = t[1:end-1]
	
	
	fitted_model = lsqfit.curve_fit(model, v, tv, p0)
	
	return fitted_model, model
end

(changed rand(2) to ones(2))
Solved!

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