how to get polynomial fit for the set of data points????

i have set of data points and i want to get 2 nd degree piecewise polynomial equation for this point . i have used following matlab code for this purpose:

x = [x1 x2 ...];
y = [y1 y2 ...]; 
p = poly fit (x,y,2) 
f = polyval (p,x); 
plot (x,y,'o'x,f,'-')

but i unable to get exact polynomial equation . can any one explain me how to get correct value for this points ? and what is exact equation for this points

[Jan]

If you use polyfit (without space), the code should run. Then p contains the parameters of the polynomial. So what exacty is your problem? What does "piecewise" mean? How large should these pieces be?

[Deva Narayanan]

actually i want to get 2nd degree polynomial equation for the given points . but i unble to get exact fit

[madhan ravi]

eighth degree gives almost a perfect fit? is that ok?

[Matt J]

Since your data are not samples of a quadratic function, naturally you will not get an exact fit.

[John D'Errico]

polyfit(x,y,2) DOES give you the fit. However, there are many reasons why the resulting fit MAY be not to your satisfaction. These usually stem from various numerical issues, or mistakes you may be making. Since we cannot know which of MANY possible mistakes you MAY have made, we cannot answer your question as posed. (The only thing I do know is that you are making SOME mistake. It may be a mistake of code, it may be a floating point issue, it may be a mistake of understanding the mathematics. How do I know which?)

If you truly want an accurate, intelligent answer, then you need to show exactly what you have done. Attach the data you used, best in a .mat file attached to a comment here, so we can read it in and verfy what you did. Then explain why it is you think that polyfit did not give you the answer you expected.

[Bruno Luong]

"what is exact equation for this points?"

Likely the one who best know the correct model is you. I don't think anyone can guess the model solely from data value.

[Deva Narayanan]

but how can i findout the8th degree polynimial equations gives excat solution for this data points?

[John D'Errico]

What I think you do not understand is that an 8th degree polynomial is just a REALLY bad idea. High order polynomial fits are exactly that.

But, in order to do such a fit, you absolutely MUST use the centering/scaling options in polyfit for this problem.

[P,S,MU] = polyfit(x,y,8)
P =
  Columns 1 through 5
          325.005626152819         -1343.44789019254          1117.33013188067          1481.95173702699         -1769.64864034309
  Columns 6 through 9
         -710.281823181912          1180.13771869376         -722.974333148199          10279.1591621409
S = 
  struct with fields:

        R: [9×9 double]
       df: 9
    normr: 275.263953753075
MU =
       -0.0205173532222222
      0.000296418663882207
      
plot(x,y,'ro',xint,polyval(P,xint,[],MU),'b-')

So look carefully at that plot. The fit is actually quite "good", in one sense that the residual errors are tiny. However, the fit is in fact pure crapola. Look at the oscillations BETWEEN the data points. Look at the bumps, wiggles, etc. Sorry, but virtual crap. This is a fundamental characteristic of high order polynomial fits, andexactly what I would expect.

You need to decide what you want from this exercise. If your goal is merely to plot a smooth curve through the points, then a cubic interpolating spline is probably a great idea. For example, again the basic fitting tools gives this plot:

[Bruno Luong]

I tried it, even 8th order polynomial won't give exact solution. Not even close.

As John has noticed thet data seems to go to infinity of the left side of x.

No polynomial can have this behavior, sorry.

Is it hyperbolic or something else? Who know? (again only you who might know how the data are acquired/recorded/simulated).

Answer 1

[John D'Errico]

I see you have attached a .xlsx file.

xy = [-0.020830515	13256.65
-0.020827342	12829.94
-0.020817825	12434.39
-0.020801967	12069.29
-0.020779773	11733.98
-0.020751248	11427.76
-0.020716403	11149.94
-0.020675247	10899.84
-0.020627794	10676.78
-0.020574057	10480.08
-0.020514052	10309.03
-0.0204478	10162.97
-0.020375318	10041.2
-0.02029663	9943.04
-0.020211759	9867.804
-0.020120732	9814.806
-0.020023576	9783.362
-0.01992032	9772.784];
x = xy(:,1);
y = xy(:,2);

I can plot them. However, as I do, I see they do NOT look like a quadratic polynomial. That is simply not a quadratic I see plotted.

plot(x,y,'o')

I then used the basic fitting tools to plot a quadratic polynomial fit. As you see, I was correct. So asking for polyfit to produce THE quadratic polynomial exact fit is something that simply makes no sense. Sorry, but a basic quadratic will not fit those points exactly. It simply does not have the correct shape to do so. How you generated the points isan unknown to us. But it was not a basic quadratic polynomial, and trying to force it into that mold will always fail.

If I had to make a wild guess, I might wonder if it is possible that you have a hyperbolic relationship, which perhaps you have confused with a quadratic. After all, they are both conic sections.

ft = fittype('a + b./(1 + c*x)')
ft = 
     General model:
     ft(a,b,c,x) = a + b./(1 + c*x)

mdl = fit(x,y,ft)
Warning: Start point not provided, choosing random start point. 
> In curvefit.attention.Warning/throw (line 30)
  In fit>iFit (line 299)
  In fit (line 108) 
mdl = 
     General model:
     mdl(x) = a + b./(1 + c*x)
     Coefficients (with 95% confidence bounds):
       a =        9416  (9265, 9568)
       b =       17.07  (13.39, 20.74)
       c =       47.78  (47.73, 47.82)
plot(mdl)
hold on
plot(x,y,'o')

Note that this model is not in fact exact. However, it is far closer to a good fit than a quadratic polynomial.

Answer 2

[Bruno Luong]

You might fit parametric, so it meets yours question since it's polynomial.

xy=xlsread('E:\MATLAB\data points.xlsx',1)
x=xy(:,1);
y=xy(:,2);
t = (1:length(x))';
Px = polyfit(t,x,5);
Py = polyfit(t,y,5);
ti = linspace(min(t),max(t));
plot (x,y,'o',polyval(Px,ti),polyval(Py,ti),'-')