Curve fitting challenge: Find a better fit I have tried to use the method of deepest descent to find the $4$
parameter variables $v_1, \ldots, v_4$ of a chosen function $y(x,v_1,v_2,v_3,v_4)$
that most closely matches the $y$-values in the table below. 
The best fit values of the variables, I could find, are shown to 
the right of the table. 
They give a least squares error of $1.2264$ and the plot below 
the table shows an almost perfect fit.
Below this plot, I have given all the details of my method but 
I know there are many more sophisticated methods, which I, 
however don’t fully understand. 
I would therefore appreciate if someone could find a
better set of parameters using the same function and some 
first order method. 
Also please explain the algorithm using “old time” math notation,
so I am not interested in MatLab or Maple-like answers, since
their methods are probably not fully explained 
I should add, that my method seems to get stuck at many local 
minima so, while observing the resulting plot, I frequently had 
to adjust the variables to get more close to the most probable 
global minimum.
Here is the table and my “best fit” variables:

and here is the “best fit” plot obtained:

Here are all the details:

 A: It is surprising that you chose such a complicated function and method since a simple polynomial function leads to much better fit. 
See below the comparison, both with four adjustable parameters :

Even with a simple three parameters polynomial the fitting is better than with the mixed polynomial and sin function. 
Note: Your method of non-linear fitting isn't very good : Again with the function $y=(v_1+v_2x+v_3x^2)\sin(v_4x)$ and with $v_1=119.416327$ ; $v_2=-25.023188$ ; $v_3=1.82441$ ; $v_4=0.939$ the sum of square errors is $0.3829$ , lower than what you found (but still higher than with the simple polynomial).
A: I post this as an answer because it is to long to fit in a comment!
JJacuelin: Congratulations for your best fit parameters 
giving 0.3829 for the sum of the squared errors. Actually
inputting your parameters for v1,..,v4 to my model gives
me a teeny bit better 0,3828. You are right, of course, 
that my model could be a lot better as you might remember
I said I had to do a lot of manuel fiddling to get close 
to the global minimum and avoid getting stuck in what 
seemed to be many local minima. So exactly what 
method did you use? Or was it some standard math tool.
I might add, that I did the calculations using VBA in
Excel (double precision) where I also tried the Solver 
add-in. However, no matter which start parameters I used,
Solver never came even close to a good curve fit.
As to your polynomial of order 4 my model gives 0.019015
with your parameters, so my method gives accurate enough 
results but my problem of finding close enough start 
parameters remains and really bugs me. So again, exactly
how do you overcome this? Anyway, thanks for your answer
and your effort.  
