Curve fitting on dataset

Question

For my master's thesis I'm writing on a specific subject which requires curve fitting. In the first part I fixed everything with 12th degree polynomial fits. But when I derive the data from the place measures, to get the speeds, I get a curve which is hard to fit.

The curve looks like a sinewave, but they go much more pointy on the minimas. Does anyone have any idea about what a good polynomial or other function would be for this kind of curves?

My Curve

I have tried this already with 12th degree polynomial and with some sorts of sine wave. But maybe it might be a good idea to combine a sine wave and a triangle wave?

EDIT: as people are advising me to get into sine waves, the reason why I don't do this is because I need to fit a lot of datasets which are completely different to this dataset. I made another screenshot to show that sine waves are not a real option for me.

This dataset would be way harder with a sine wave

Answer 1

Here's a quick example of using a change point type method. The model.

\begin{equation} y(x) = \begin{cases} \alpha_0 + \alpha_1x + \alpha_2x^2 & x \leq c \\ \beta_0 + \beta_1x + \beta_2x^2 & x > c \end{cases} \end{equation}

We have 6 regression parameters, and a break-point parameter $c$. However, we want this model to be continuous at $c$, so we should impose the constraint: \begin{equation} \alpha_0 + \alpha_1c + \alpha_2c^2 = \beta_0 + \beta_1c + \beta_2c^2 \end{equation}

Or equivalently, $A \equiv \alpha_0 = \beta_0 + (\beta_1-\alpha_1)c + (\beta_2-\alpha_2)c^2$. So now we have 6 free parameters.

Here's a simulated data set which I will try this out on.

I'm using R, which is pretty easy to use. If you try to use the $\texttt{nls}$ function, it may complain about identifiability. But we can do this using $\texttt{optim}$

First, I write a simple model function, given a vector of parameters and the $x$ data. Here, par = $[\alpha_1, \alpha_2, \beta_0, \beta_1, \beta_2, c]$

 model <- function(par, x){
   n <- length(x)
   res <- rep(0,n)
   for(i in 1:n){
      A0 <- par[3] + (par[4]-par[1])*par[6] + (par[5]-par[2])*par[6]^2
      if(x[i] <= par[6]){
         res[i] <- A0 + par[1]*x[i] + par[2]*x[i]^2
      }else{
         res[i] <- par[3] + par[4]*x[i] +par[5]*x[i]^2
      }
   }
   return(res)
 }

Then we write a simple function to return the sum of squares. This is the function we want to minimize.

 sum_squares <- function(par, x, y){
    ss <- sum((y-model(par,x))^2)
    return(ss)
 }

Finally, we minimize the sum of squares with $\texttt{optim}$. Note that you will need to find good initial guesses to converge to the correct solution.

 #I found these initial values with a few minutes of guess and check.
 par0 <- c(7,-1,-395,70,-2.3,10)
 sol <- optim(par= par0, fn=sqerror, x=x, y=y)$par

When I plot my solution, this is what I get.

You may need to include third order terms for your data. But this certainly avoids having to consider 12th order polynomials! Plus the interpretability of the change-point is a nice feature.

Best of luck!