can help me to found empirical equation for data L1 vs T1

L1 = [0.784 0.804 0.8845 0.9707 1.062 1.156 1.256 1.36  1.469 1.579 1.696   1.815  1.937 2.063 2.191 2.323 2.457 2.595 2.735 2.876 3.022 3.167 3.316  3.618 3.773 3.927 4.085 4.242 4.405 4.566 4.726 4.891 5.057 5.224 5.236 5.255 5.274 5.242 5.209 5.19 4.634 4.254 3.337 1.257 0.2902 0.11 0.09331 0.02474 0.001703 0.00103 0 0 0];
T1 = [0.9379 1.002 1.25 1.50  1.751    2    2.251 2.502 2.75    3   3.252   3.50 3.752   4.001 4.25  4.501 4.751 5.001 5.253 5.50  5.754 6.002 6.253  6.752 7.003 7.252 7.503  7.75 8.004 8.254  8.5  8.751 9.004 9.256 9.276 9.309 9.375 9.414 9.434 9.44 9.50  9.52  9.546 9.58  9.6   9.609 9.61    9.621    9.643 9.647 10 11 12]; 
plot(T1, L1, '.-'), grid on
xlabel T_1, ylabel L_1

[Dyuman Joshi]

What does empirical equation mean in this context?

[Mushtaq Al-Jubbori]

Any equation to fitting L1 vs T1

[Image Analyst]

What do these measurements represent? And is there any theoretical model that represents the physical process that you're observing? That would be better than just pulling some arbitrary equation out of thin air that just coincidentally happens to fit well.

[Mushtaq Al-Jubbori]

Brag curve

[Sam Chak]

@Mushtaq Al-Jubbori, Thanks for the information. Apparently, the Bragg curve describes the stopping power acting on charged particles over the total path length traversed by them, and it is often represented by the Bethe–Bloch formula. It appears that @Alex Sha's formula is more compact. Do you have plans to publish Alex Sha's formula?

References:

https://en.wikipedia.org/wiki/Stopping_power_(particle_radiation)

https://en.wikipedia.org/wiki/Bethe_formula

Answer 1

[Alex Sha]

How about the one below, much simple and works well:

Sum Squared Error (SSE): 0.348116593172766
Root of Mean Square Error (RMSE): 0.0810446642724449
Correlation Coef. (R): 0.999051954884231

Parameter	Best Estimate        
---------	-------------        
p1       	1.93075962947385     
p2       	1.42899205819801     
p3       	6.09575438791542     
p4       	-3.221136445554E-15  
p5       	-4.32095917648694E-12
p6       	13.1725280684151

[Sam Chak]

Hi @Alex Sha,

This is impressive. Could you please share what led you to discover the proposed model? Did you extensively search the Origin Pro library?

[Mushtaq Al-Jubbori]

Very thanks

if you can provied me the code

did can litle the praeters to 4th parameters only?

[Alex Sha]

If you hope the parameter number not exceed 4, then just make a little modification like below, of couse, the effect will becomes a little worse:

Sum Squared Error (SSE): 0.82094649993158
Root of Mean Square Error (RMSE): 0.124457047219041
Correlation Coef. (R): 0.997672207622239
R-Square: 0.995349833861832

Parameter	Best Estimate        
---------	-------------        
p1       	1.82633074134699     
p2       	-5.58009896272397E-17
p3       	-9.82892599133683E-12
p4       	12.8417903882769   

[Alex Sha]

Another fitting function, slight different but more better result

Sum Squared Error (SSE): 0.317320862097167
Root of Mean Square Error (RMSE): 0.0773769091997414
Correlation Coef. (R): 0.999141448452466
R-Square: 0.998283634015693

Parameter	Best Estimate        
---------	-------------        
p1       	0.36398078591103     
p2       	-0.014145274526113   
p3       	1.68059813246038     
p4       	-2.31298883825941E-16
p5       	-6.41439068709134E-12
p6       	13.0135415721063 

[Mushtaq Al-Jubbori]

Very thanks Dear Alex Sha, if you can provide me code or m.file of the fitting

[Alex Sha]

@Sam Chak: not done in Origin Pro library, but in 1stOpt, there are over ten-thousand functions in its formula library.

@Mushtaq Al-Jubbori: once known the type of fitting function and corresponded data, it is easy to write the fitting code for Matlab like the below, however, the problem now is how to guess the start-values so ensure the convergence, even employ global optimization algorithm like Matlab GA, it is still very difficult to get the right results like ones I provided above. So the most important thing is the tools or algorithms which ensure the computation results are the best solution.

x = [0.9379,1.002,1.25,1.5,1.751,2,2.251,2.502,2.75,3,...
3.252,3.5,3.752,4.001,4.25,4.501,4.751,5.001,5.253,5.5,...
5.754,6.002,6.253,6.752,7.003,7.252,7.503,7.75,8.004,8.254,...
8.5,8.751,9.004,9.256,9.276,9.309,9.375,9.414,9.434,9.44,...
9.5,9.52,9.546,9.58,9.6,9.609,9.61,9.621,9.643,9.647,...
10,11,12];
y = [0.784,0.804,0.8845,0.9707,1.062,1.156,1.256,1.36,1.469,1.579,...
1.696,1.815,1.937,2.063,2.191,2.323,2.457,2.595,2.735,2.876,...
3.022,3.167,3.316,3.618,3.773,3.927,4.085,4.242,4.405,4.566,...
4.726,4.891,5.057,5.224,5.236,5.255,5.274,5.242,5.209,5.19,...
4.634,4.254,3.337,1.257,0.2902,0.11,0.09331,0.02474,0.001703,0.00103,...
0,0,0];

Fit_Fun = @(p,x) (exp(p(1).*x+p(2).*x.^2))./(p(3)-p(4).*exp(0-p(5).*x.^p(6)));
InitialGuess = ones(6,1);  %[1,1,1,1,1,1];
p = lsqcurvefit(Fit_Fun, InitialGuess, x, y);

plot(x, y,'pg');
hold on;
plot(x, Fit_Fun(p, x), '-r');
hold off;
grid;
xlabel('x');
ylabel('y');

Answer 2

[Alex Sha]

It is not easy to find a function to describe the curve. Refer to the fitting function and result below:

Sum Squared Error (SSE): 0.0168127809048645
Root of Mean Square Error (RMSE): 0.0178107349995405
Correlation Coef. (R): 0.999951618092273
R-Square: 0.999903238525356

Parameter	Best Estimate    
---------	-------------    
p1       	-113.045620847505
p2       	-1.98381297332703
p3       	0.199823219199342
p4       	-37.8432173816184
p5       	2.28764952811147 
p6       	1.80019545749072 
p7       	13.5735683684756 
p8       	3.19896435789107 
p9       	1082.4843757759  
p10      	21.6922662996556 
p11      	-1.93061546894078
p12      	360.858731186603 
p13      	-8.53668056447937

[Mushtaq Al-Jubbori]

No good, I want one equation for the total curve My be the equation type of tanh, 1/cos or exp

with litle coffetions

[John D'Errico]

Sorry, but just wanting something to be found, or even to exist is not sufficient. (If it was, then I'd have found a nice shiny Lamborghini in my garage wrapped up as a Christmas present.) As you can see, @Alex Sha gave you a shot using a sum of exponentials, but that fails miserably on the rght end, where it starts oscillating. There is no magic way to know what functional form MIGHT fit that set of points. And the very sharp transitions, then changing shape into smooth curves, or well-behaved sections means that any simple functions you choose (certainly NOT polynomials) will fail.

This is why tools like splines (and piecewise polynomials in general) were invented, to allow you to create curves that will represent such things. However, splines are not functions where you can write down the form either. So we can do this:

L1 = [0.784 0.804 0.8845 0.9707 1.062 1.156 1.256 1.36  1.469 1.579 1.696   1.815  1.937 2.063 2.191 2.323 2.457 2.595 2.735 2.876 3.022 3.167 3.316  3.618 3.773 3.927 4.085 4.242 4.405 4.566 4.726 4.891 5.057 5.224 5.236 5.255 5.274 5.242 5.209 5.19 4.634 4.254 3.337 1.257 0.2902 0.11 0.09331 0.02474 0.001703 0.00103 0 0 0];
T1 = [0.9379 1.002 1.25 1.50  1.751    2    2.251 2.502 2.75    3   3.252   3.50 3.752   4.001 4.25  4.501 4.751 5.001 5.253 5.50  5.754 6.002 6.253  6.752 7.003 7.252 7.503  7.75 8.004 8.254  8.5  8.751 9.004 9.256 9.276 9.309 9.375 9.414 9.434 9.44 9.50  9.52  9.546 9.58  9.6   9.609 9.61    9.621    9.643 9.647 10 11 12]; 
fun = pchip(T1,L1);
fnplt(fun)
hold on
plot(T1,L1,'ro')
grid on

As you can see, it very nicely fits the data. You can evaluate that function at any point, and get a nice smooth looking result. However, there is no simple function you can write down as a result.

fnval(fun,3.75)

ans = 1.9360

The function encapsulated inside fun is actually a set of many tiny segments of cubic polynomials. And that is likely to be the only way to turn that set of points into a function you can then use.

Answer 3

[Sulaymon Eshkabilov]

Just looking at your data, a first part of it looks like a nice polynomial fit (until it it drops). Here is one easy fit model with poly2, e.g.:

L1 = [0.784 0.804 0.8845 0.9707 1.062 1.156 1.256 1.36  1.469 1.579 1.696   1.815  1.937 2.063 2.191 2.323 2.457 2.595 2.735 2.876 3.022 3.167 3.316  3.618 3.773 3.927 4.085 4.242 4.405 4.566 4.726 4.891 5.057 5.224 5.236 5.255 5.274 5.242 5.209 5.19 4.634 4.254 3.337 1.257 0.2902 0.11 0.09331 0.02474 0.001703 0.00103 0 0 0];
T1 = [0.9379 1.002 1.25 1.50  1.751    2    2.251 2.502 2.75    3   3.252   3.50 3.752   4.001 4.25  4.501 4.751 5.001 5.253 5.50  5.754 6.002 6.253  6.752 7.003 7.252 7.503  7.75 8.004 8.254  8.5  8.751 9.004 9.256 9.276 9.309 9.375 9.414 9.434 9.44 9.50  9.52  9.546 9.58  9.6   9.609 9.61    9.621    9.643 9.647 10 11 12]; 
plot(T1, L1, '.-'), grid on
xlabel T_1, ylabel L_1
hold on
%% Partial fit because the other part causes the bad fit
IDX = T1<9.5;
T1_S =T1(IDX);
L1_S = L1(IDX);
% Fit Model:
Model = fittype('poly2');
FModel = fit(T1_S', L1_S', Model)

FModel = 
     Linear model Poly2:
     FModel(x) = p1*x^2 + p2*x + p3
     Coefficients (with 95% confidence bounds):
       p1 =     0.01782  (0.01579, 0.01986)
       p2 =      0.3546  (0.3318, 0.3773)
       p3 =      0.3858  (0.3325, 0.439)

plot(FModel, T1_S, L1_S)