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I have two sets of experimental data X and Y. The data in both sets are related by $y = f(x),\hspace{0.2cm} x \in X,\hspace{0.2cm}y \in Y$ the function $f(x)$ is unknown. The data is in the interval $x,y\in[0,1]$ or better put $X\in[0,1]$ and $Y\in[0,1]$. The relation between $X$ and $Y$ is highly non linear, and by highly nonlinear (I know that is not a mathematical term) I mean that the function can't be properly fitted by polynomial series like Taylor series, Legendre polynomials, Laguerre polynomial, Hermite polynomials, I even try Bernoulli polynomials. The function apparently can't be fitted by Fourier series either. I'm currently trying wavelets techniques but as far as I get the results apparently say that it can't be fitted by wavelets either. The function $f(x)$ although closed in the interval $[0,1]$ shows a extremely non linear behavior given that the derivatives approximations are high valued and change in sign in some points. The function don't oscillate much, in some sub intervals looks like a line. I have to say that the quality of data is very good. There is no doubt on the values of the data.

The values of the underlying function can't just be approximated because the values have a precisely physical meaning and a significance, that means that small error in the approximations incurred in a big error in the physical model. That's the reason why I need not a good, but a very good approximation to the function values. Also I need that the resulting fitted model can be reasonable computed in a normal desktop machine. After doing the fitting the resulting function is going to be massively evaluated in another algorithm so I need the most efficient evaluation of $f(x)$.

The question finally is: there is another advanced or new technique to perform a curve fitting to data with these characteristics?

Data and approximation

Detail of data near zero

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  • $\begingroup$ You are showing a curve that looks like exponential or power so have you tried $e^{ax}$ and $x^\gamma$ type of regression, because you do not talk about it in your post ? Maybe you could also share a .csv if the data is not confidential. $\endgroup$
    – zwim
    Jun 10, 2021 at 16:56
  • $\begingroup$ Not confidential at all.....it's just values of specific volume of water in saturated liquid transformed by de difference between critical point and triple point. $y=\frac{v-v_t}{v_c-v_t}$ and $x=\frac{P-P_t}{P_c-P_t}$ $\endgroup$
    – Carlos
    Jun 10, 2021 at 23:18
  • $\begingroup$ I used both, $e^{ax}$ don't fit much, and $x^\gamma$ work partially for $\gamma=0.0008$. The best result I did get was with a linear combination of $x^\gamma$, $\tanh{(x)}$ and odd powers of $x$ $\endgroup$
    – Carlos
    Jun 10, 2021 at 23:25

1 Answer 1

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I do not think that the fitting model would be simple. We have $$y=\frac{V_s-V_t}{V_c-V_t} \qquad \text{and} \qquad x=\frac{P_s-P_t}{P_c-P_t}$$

Assume that we use the simplest possible models (using $T_r=\frac T{T_c}$).

  • for the saturated liquid volume, Rackett equation (the exponent has been simplified) $$V_s=\frac {R\,T_c}{P_c} Z_c ^{1+(1-T_r)^{1/3}}$$
  • for the saturation pressure, Van der Waals equation $$P_s=P_c \, e^{k \left(1-\frac{1}{T_r}\right)}$$

and we know the values of $(V_c,V_t,P_c,P_t)$. Introducing reduced volume and pressure at saturation $$V_r=\frac {V_s}{V_c} \qquad \text{and} \qquad P_r=\frac {P_s}{P_c}$$

$$y=\frac{V_r-a}{1-a} \qquad \text{and} \qquad x=\frac{P_r-b}{1-b}$$ where $a=\frac{V_t}{V_c}$ and $b=\frac{P_t}{P_c}$ are known constants.

So, from $V_r$, we have $$T_r=1-\frac{\log ^3(V_r)}{\log ^3(Z_c)}$$ and, from $P_r$ $$T_r=\frac{k}{k-\log (P_r)}$$

Making them equal would lead to $$\log(V_r)=-\sqrt[3]{\frac{\log (P)}{k-\log (P)}} \log(Z_c)$$

In practice, I think that is would be much safer to fit independently $V_s$ and $P_s$ as function of $T$.

Edit

Reworking what we did at the time of Nist/Asme project about water properties, I reworked the problem as stated in the question. What I did is the following $$y=\frac{V_s-V_t}{V_c-V_t}=\frac{1}{1+\sum _{n=1}^p a_n\, \tau ^{n/3}}\qquad \text{where} \qquad\tau=1-T_r$$ Using $p=6$ gives an $R^2=0.999879$ and a maximum absolute error equal to $0.004$. Notice that this could be improved a lot (using a stepwise regression for finding the best exponents).

As a function of $\tau$ the function $y$ is quite nice. So, what I suggest is to use as a model $$y=\frac{1}{1+\sum _{n=1}^p a_n\, \Big[\tau(x)\Big] ^{n/3}}$$ and it is quit simple to make a model (inversion of the vapor pressure) $$\tau(x)= F[x]$$

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  • $\begingroup$ Surprisingly the final equation has the characteristics of the curve, although not enough to fit to desired relative error. I had to say that I'm looking more for something that be like the reference equations of state. If you read about them they are as random terms combined linearly, but at a close look they can be some sort of series expansion of a generalized function. Something like a Taylor series. $\endgroup$
    – Carlos
    Jun 15, 2021 at 5:42
  • $\begingroup$ For that reason I'm looking for a series expansion that fit the data. To be a reference equation of estate the data has to be fitted with an error no more than 0.01% (or even less in some intervals depending of the quality and confidence of the data) for normalized values in the interval [0, 1], which means that the absolute error must be incredible small. This kind of normalization is needed to extend the technique to fitting other similar curves of the PvT surface. The reasoning behind it is, if I can fit a curve like that in [0, 1], then I can do the same with other curves just normalizing $\endgroup$
    – Carlos
    Jun 15, 2021 at 5:49
  • $\begingroup$ @Carlos. I am a thermodynamicist; so I do not consider that it is surprising ! I plan to add more within a couple of hours using the Nist/Asme EoS. Comme back ! Cheers :-) $\endgroup$ Jun 15, 2021 at 5:49
  • $\begingroup$ ....and fitting to the new curve values. That's because the other curves although different they are more or less of the same behavior $\endgroup$
    – Carlos
    Jun 15, 2021 at 5:52
  • $\begingroup$ Mr. Claude Leibovici I'm waiting more like you for those results...but I don't think that Rackett and Van der Walls equations accomplish what I need, I already doing some studies by myself and those equation are far far from the accuracy of prediction I'm looking for. I don't need for now a theoretical correct equation from the physical point of view, I need a mathematical way of reproduce the data with the precision needed. $\endgroup$
    – Carlos
    Jun 15, 2021 at 6:00

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