Fitting a data with the best fit

Hi guys,

Please, I need a best fitting method for the data below. It is also attached in a text file.

I have tried tried few fitting methods but they are not giving good result.

M = [-40	242.118600000000
-39	242.717400000000
-38	239.597400000000
-37	236.149700000000
-36	241.291400000000
-35	243.789200000000
-34	244.742700000000
-33	237.290600000000
-32	239.856600000000
-31	236.112800000000
-30	233.345900000000
-29	235.154400000000
-28	233.210300000000
-27	235.488400000000
-26	230.774800000000
-25	237.738900000000
-24	237.066600000000
-23	242.997300000000
-22	241.418500000000
-21	243.022000000000
-20	251.080400000000
-19	255.720000000000
-18	257.327500000000
-17	255.298700000000
-16	257.262300000000
-15	259.756300000000
-14	260.172100000000
-13	261.738600000000
-12	262.423100000000
-11	269.295300000000
-10	266.707900000000
-9	263.961600000000
-8	263.291300000000
-7	267.552400000000
-6	266.540300000000
-5	272.394900000000
-4	266.560100000000
-3	265.945800000000
-2	269.381300000000
-1	269.702600000000
0	276.367300000000
1	277.915600000000
2	274.844200000000
3	274.439700000000
4	274.654000000000
5	277.024300000000
6	277.782500000000
7	278.889500000000
8	280.760300000000
9	281.655000000000
10	279.313700000000
11	278.863900000000
12	277.457200000000
13	275.902800000000
14	274.092300000000
15	273.431400000000
16	263.025100000000
17	255.005200000000
18	254.097200000000
19	255.055100000000
20	243.951200000000
21	236.658300000000
22	244.101800000000
23	240.492600000000
24	245.866900000000
25	239.299800000000
26	233.284500000000
27	234.925100000000
28	246.933500000000
29	231.857000000000
30	238.856600000000
31	234.920200000000
32	232.561800000000
33	242.456800000000
34	237.296900000000
35	235.602100000000
36	246.238300000000
37	240.722800000000
38	256.414100000000
39	243.881000000000];
plot(M(:,1),M(:,2))

Thanks

[Matt J]

What curve model would you be trying to fit to?

[dpb]

"Fit" for what purpose? There's nothing going to model that closely. What detail are you willing to sacrifice for a model of some sort?

One of the smoothing or interpolating splines or somesuch is about all I can imagine doing with the above.

Answer 1

[David Hill]

M = [-40	242.118600000000
-39	242.717400000000
-38	239.597400000000
-37	236.149700000000
-36	241.291400000000
-35	243.789200000000
-34	244.742700000000
-33	237.290600000000
-32	239.856600000000
-31	236.112800000000
-30	233.345900000000
-29	235.154400000000
-28	233.210300000000
-27	235.488400000000
-26	230.774800000000
-25	237.738900000000
-24	237.066600000000
-23	242.997300000000
-22	241.418500000000
-21	243.022000000000
-20	251.080400000000
-19	255.720000000000
-18	257.327500000000
-17	255.298700000000
-16	257.262300000000
-15	259.756300000000
-14	260.172100000000
-13	261.738600000000
-12	262.423100000000
-11	269.295300000000
-10	266.707900000000
-9	263.961600000000
-8	263.291300000000
-7	267.552400000000
-6	266.540300000000
-5	272.394900000000
-4	266.560100000000
-3	265.945800000000
-2	269.381300000000
-1	269.702600000000
0	276.367300000000
1	277.915600000000
2	274.844200000000
3	274.439700000000
4	274.654000000000
5	277.024300000000
6	277.782500000000
7	278.889500000000
8	280.760300000000
9	281.655000000000
10	279.313700000000
11	278.863900000000
12	277.457200000000
13	275.902800000000
14	274.092300000000
15	273.431400000000
16	263.025100000000
17	255.005200000000
18	254.097200000000
19	255.055100000000
20	243.951200000000
21	236.658300000000
22	244.101800000000
23	240.492600000000
24	245.866900000000
25	239.299800000000
26	233.284500000000
27	234.925100000000
28	246.933500000000
29	231.857000000000
30	238.856600000000
31	234.920200000000
32	232.561800000000
33	242.456800000000
34	237.296900000000
35	235.602100000000
36	246.238300000000
37	240.722800000000
38	256.414100000000
39	243.881000000000];
p=polyfit(M(:,1),M(:,2),5);%change to desired degree
plot(M(:,1),polyval(p,M(:,1)))
hold on;
plot(M(:,1),M(:,2))

[TTA]

Thanks David

Answer 2

[Torsten]

Try "fit" with fitType

gaussEqn = 'a*exp(-((x-b)/c)^2)+d'

https://de.mathworks.com/help/curvefit/fit.html#d124e31956

Answer 3

[John D'Errico]

As others have said, there is no magical way to know what is the "best" fitting model for such a problem. If you lack any model based on physical principles, then there is little magic you can do.

These are typically problems where a spine model of some sort may be best, since a spline makes very little in the way of assumptions, except that the curve be relatively smooth.

For example, using my SLM toolbox, there is relatively no information to provide, beyond a rough degree of fit in terms of how many knots to use, so I would do something like this:

slm = slmengine(x,y,'knots',7,'plot','on')

In the end, the result seems reasonable, but it depends on how much you want to chase what may well be noise.

[Alex Sha]

If want to have an explicit function expression，refer to one below:

y = p4*(1/(1+exp(p1*x+p8)))+p5*(1/(1+exp(p2*x+p9)))+p6*(1/(1+exp(p3*x+p10)))+p11+p7*x;

Sum Squared Error (SSE): 931.576981042263
Root of Mean Square Error (RMSE): 3.41243494634378
Correlation Coef. (R): 0.976919416755453
R-Square: 0.954371546833815

Parameter	Best Estimate     	Std. Deviation    	Confidence Bounds[95%] 
---------	-------------     	--------------    	--------------------------------
p1       	2.17071637227502  	2.00375241483378  	[-1.82666032070769, 6.16809306525773]
p2       	-0.866846695995501	0.502259597641453 	[-1.86882717750392, 0.135133785512921]
p3       	0.357015983521599 	0.0432728726246512	[0.270688964680532, 0.443343002362667]
p4       	-13.1100496826724 	2.55813476865172  	[-18.2133889106205, -8.00671045472424]
p5       	-16.4803026879968 	2.88984006155108  	[-22.2453758691813, -10.7152295068122]
p6       	64.8893226472405  	4.37934503986771  	[56.1527683387838, 73.6258769556972]
p7       	0.948998342347286 	0.0993732061413023	[0.750754220371189, 1.14724246432338]
p8       	44.370594281755   	40.924919037562   	[-37.2723853158642, 126.013573879374]
p9       	-27.8000726945408 	16.0846620989817  	[-59.8880956024535, 4.2879502133719]
p10      	-6.25791814028238 	0.774805804682342 	[-7.80361342793189, -4.71222285263287]
p11      	225.224588689396  	3.53354087170702  	[218.17536752829, 232.273809850502]