I want to make curve fitting to these points

I want to make curve fitting to these points but when i made it the blue straight line appeared. What does it mean ?

[Umar]

Hi @noura,

If you don’t mind, can you share your code with us to help resolve your issue.

[dpb]

You don't need his code; the answer is obvious by inspection. Put any set of symmetric points in and overall OLS will always return the mean; it's inevitable conclusion from the basis of the method to minimize the overall error sum of squares.

Answer 1

[dpb]

The top and bottom are identical pairs above and below the mean; hence when/if you fit the whole dataset, the single line that best fits all the data is (drum roll, please...) the mean.

You need two fits; one for the upper and one for the lower. The slope of one will be exactly the negative of the other, but the two intercepts will be different.

X=[16 17.5 20 25 32.5 37 42.5 45 48].';
Y=[0  9   17 21 27   34 41   48  55 ];
DY=Y-Y(end);
Y=[Y;Y(end)-DY].';
scatter(X,Y,'filled')
box on, grid on
ylim([-5 118])
b1=polyfit(X,Y(:,1),3)

b1 = 1x4

0.0037 -0.3546 11.9781 -113.5101

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b2=polyfit(X,Y(:,2),3)

b2 = 1x4

-0.0037 0.3546 -11.9781 223.5101

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y1=polyval(b1,X);
y2=polyval(b2,X);
hold on
plot(X,y1,'b-')
plot(X,y2,'r-')

[noura]

Not possible that one polynomial define the all points ?

[Umar]

@noura,

Could you share code snippet or example to share with us by explaining what exactly you are looking to achieve by mentioning, “ I want to make curve fitting to these points” since I can’t archive points from attached png file. Hope it makes sense.

[dpb]

"...one polynomial define the all points ?"

You can observe it is double-valued function -- at each X, there are two possible y values. So, no there is no single polynomial that can define both of them.

How would you propose being able to tell which value of Y, the upper or lower, is supposed to be returned given a value of X?

The issue is as outlined before, when you combine the two Y at each X, you get a constant which gives you the flat line you had before. As you can see by inspection, the coefficients of the two are the negative of each other for the terms in X while the intercepts are different.

[dpb]

@Umar, the data I created matches the example closely enough although the actual data are totally immaterial -- any set that is symmetric about the overall mean will suffice to create the same scenario.

[noura]

Yes i understood thank you

[Umar]

@dpb, thanks for your contribution and sharing your thoughts, really appreciated. @noura, please go ahead accept @dpb answer and give him a vote. If there are still further questions or issues, please let us know, we will be more happy to help.

[dpb]

"...the coefficients of the two are the negative of each other for the terms in X while the intercepts are different."

Also, notice that

X=[16 17.5 20 25 32.5 37 42.5 45 48].';
Y=[0  9   17 21 27   34 41   48  55 ];
DY=Y-Y(end);
Y=[Y;Y(end)-DY].';
b1=polyfit(X,Y(:,1),3);
b2=polyfit(X,Y(:,2),3);
mean_intercept=mean([b1(end) b2(end)])

mean_intercept = 55.0000

mean_y=mean(Y,'all')

mean_y = 55

The average of the two intercepts is the mean of the overall data to within machine precision/roundoff in the fitting calculations...

That being so, it is in one sense only one polynomial, the second is completely known by only fitting the first; the issue still being there isn't any convenient way to return the double-valued function from only a f(x).