please help me to find out mathmatical function between X and Y

for example

x=[0.42  0.43  0.44  0.45  0.46  0.47  0.48  0.49  0.5  0.51  0.52  0.53  0.54  0.55  0.56  0.57  0.58  0.59  0.6  0.61  0.62  0.63  0.64  0.65  0.66  0.67  0.68 0.7  0.71  0.72  0.73  0.74  0.75  0.76  0.77  0.78  0.79  0.8  0.81  0.82  0.83  0.84  0.85  0.86  0.87  0.88  0.89  0.9  0.91  0.92  0.93  0.94  0.95  0.96  0.97  0.98  0.99  1]

y=[89.72308444  89.21117298  88.69692085  88.18020302  87.66088218  87.13880655  86.61380705  86.08569379  85.55425157  85.0192342  84.48035703  83.9372872  83.38963067  82.83691475  82.27856424  81.71386818  81.14193297  80.56161474  79.97142004  79.36935656  78.75270387  78.11765265  77.45872388  76.7678153  76.03261795  75.23400847  74.34196715 78.30969848  77.07114557  75.9480519  74.94218396  74.03486878  73.20106164  72.41872047  71.67135281  70.94735035  70.23865101  69.53960043  68.84614818  68.15531367  67.46483672  66.77294781  66.07821495  65.37943962  64.67558442  63.96572164  63.24899546  62.52459342  61.79172401  61.04959825  60.29741384  59.53434075  58.75950745  57.97198693  57.17078199  56.35480903  55.52287962  54.67367898]

[Morteza]

In fact there are infinite number of equation can be fitted between these data. But BASIC-FITTING can let you to find the equation according to the fitting function that you select....

Answer 1

[Morteza]

[rahman sajadi]

thanks a lot

Answer 2

[Walter Roberson]

Your data has a discontinuity at the 28th point. If you separate the two portions, you can get a pretty decent fit with a pair of cubic polynomials and an very good fit with a pair of quartic polynomials.

d = 3; N = 27; plot(x,y, 'b', x(1:N), polyval(polyfit(x(1:N),y(1:N), d), x(1:N)),'r', x(N+1:end), polyval(polyfit(x(N+1:end),y(N+1:end),d), x(N+1:end)), 'g')

and change to d = 4 for the quadratic.

If on the other hand you try to fit the whole thing with one polynomial, then

d = 9; plot(x,y, 'b', x, polyval(polyfit(x,y, d), x),'r')

and you can see that the portion near 0.7 is not satisfactory. If you go for d = 10 or higher you will get warnings about bad conditionining. By d = 25, you will still not be seeing a good transition near 0.7 but you will be starting to see notable errors at the right hand side. By d = 30 you get:

which is not satisfactory at the centre or the edges.

When you have a discontinuity like this, although mathematically you can fit it using a polynomial, the reality in the world of finite precision values is that you cannot do a good job of fitting it numerically.

You need to reconsider whether a polynomial is appropriate for your system.

[rahman sajadi]

there is a point between .68 and .7 (.69) and in this point my function is Discontinuous.so not importanat in this part of curve for me

thank u for your answers

[Walter Roberson]

The discontinuity is going to affect the coefficients for the polynomial, unless you break it up into two polynomials like I did here. If it is acceptable that the fit is not very good near 0.7 then why not just fit to a straight line?