1
$\begingroup$

In an actual Project, I have to find a function that best-fits given data. The Parameter-estimation of such a function should be hopefully not the Problem, but now I am struggling with which type of function I should choose for the Regression.

I added a Picture, where I plotted my gained data against the time, see this

enter image description here

It seems that there might be some function behind. Unfortunately I don’t know which type it could be.

Does anyone of you can give a hint or has an idea what type of function this could be?

Thank you in advance, questionmarkengineer


EDIT

Unfortunately, the path I tried to follow this weekend did not help me. So I add a few more details what I am trying to find out.

I have a function $$f_\text{EC}(t,t_i) = \phi\cdot\left(\frac{t-t_i}{\beta+t-t_i}\right)^{0.3},\ (t>t_i)$$ that delivers “measurement”-data over the time, starting at $t_i$ and $\phi$ and $\beta$ are constants.

This data should be approximated by the following function $$ f_\text{KV}(t, t_i) = \sum_{\mu=1}^N\frac{1}{E_\mu(t_i)}\cdot\left(1-\exp\left(\frac{-(t-t_i)}{\tau_\mu}\right)\right), $$ where $\tau_\mu$ are constants (ranging from 1 to 1000) and $E_\mu$ are parameters that can be found by Regression for each $t_i$. In my case, $N = 4$.

I now determinend these Parameters $E_\mu$ for several Points $t_i$ which are those points in time marked in the graph above. When I then plotted the values of $E_\mu$ against the corresponding $t_i$, I found out that there seems to be a functional connection behind ($E_1$, $E_2$ and $E_3$ changed, while $E_4$ remained nearly constant). And this connection is what I am trying to find, because I need the values of $E_\mu$ at nearly each Point in time, and a linear-Interpolation between my determined $t_i$ is not sufficient enough, especially between $t = 1, \ldots, 21$.

Maybe now someone has an idea what I can try? - It would also be helpful for me just to find a procedure for a (discrete) determination of the $E_\mu$ at each $t_i$ without using the (costly) regression.

$\endgroup$
4
  • $\begingroup$ If this is real data from a real system, then you should try using a mathematical model to derive a model curve to fit. Have you done that ? $\endgroup$ Aug 18, 2017 at 20:26
  • $\begingroup$ What kind of system/value are you measuring? $\endgroup$
    – Nick ODell
    Aug 18, 2017 at 22:16
  • $\begingroup$ It is the elastic modulus in a spring-dashpot-system (4-Kelvin-Chain-Model) that is fitted to resulting creep-curves starting from each time-step marked in the graphs above. The shown graphs are the elastic modules of the springs that lead to the considered creep curve (one of these springs has a constant module). I just realized that Bazant/Xi did something like that and presented a dirichlet-series-approximation when the given creep-curve(s) is differentiable. I hope that I can work it out by this weekend. If one is interested, I can post the solution or the way I used this Bazant-algorithm. $\endgroup$ Aug 19, 2017 at 0:24
  • $\begingroup$ ... but anyway, it would be of interest to me if there exist some (simple?) functions that show such behaviour ... $\endgroup$ Aug 19, 2017 at 7:45

1 Answer 1

1
$\begingroup$

I don't understand the problem that you're trying to model, but...

Two functions come to mind: $-c_1xe^{-c_2x}$, and $c_1e^{-c_2x} + c_3x$

Plot 1 - Plot 2

$\endgroup$
3
  • $\begingroup$ thank you! - It is indeed something like "-c1*x*e^(-c2*x)+c3"; that helped me a lot! $\endgroup$ Aug 20, 2017 at 15:34
  • $\begingroup$ hm ... maybe I replied too early; at the Moment I didn't manage it to fit this function of type "-c1*x*e^(-c2*x)+c3" to my data. Ist not possible for me to Stretch it somehow though the plots in Wolfram Alpha do look like my data ... but not really in log-scale :/ $\endgroup$ Aug 20, 2017 at 17:12
  • $\begingroup$ ... I edited my post above and added a few more details what I should have done right from the beginning, I think. $\endgroup$ Aug 21, 2017 at 0:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .