8a76faaa-1e52-40ed-96e4-fcc4ef3d96a2 50 overlord 4 5 false false 0 decimal

&[DATE] &[TIME]

underlying data

x 709011013015018022527532591 mat :

y 0.0024 0.0056 0.0051 0.00545 0.00565 0.00455 0.00302 0.00152 0.0019 91 mat :

x g1 70 0.0024 90 0.0056 110 0.0051 130 0.00545 150 0.00565 180 0.00455 225 0.00302 275 0.00152 325 0.0019 92 mat :

x g1

we introduce a formula for the approximation

a1 1002 π * sqrt * eval :

Wb h1 σ1 a1 * / (x μ1 - (2^2 σ1 2^* / - exp * h2 σ2 a1 * / (x μ2 - (2^- 2 σ2 2^* / exp * + h3 σ3 a1 * / (x μ3 - (2^- 2 σ3 2^* / exp * + :

initial values of 9 parameters

10 parameters that need to be calculated, we set arbitrary values for now

h1 0 :

h2 0 :

h3 0 :

μ1 50 :

μ2 100 :

μ3 200 :

Sqmin 1 :

hh1 0.15 :

hh2 0.7 :

hh3 0.40 :

σ1 1 :

σ2 1 :

σ3 1 :

μμ1 90 :

μμ2 160 :

μμ3 350 :

σσ1 19 :

σσ2 50 :

σσ3 80 :

The difference between the original data and those calculated by the formula

h1 h2 h3 μ1 μ2 μ3 σ1 σ2 σ3 Raz y Wb - eval :

the sum of the squares of the deviations of the initial data from the desired function

h1 h2 h3 μ1 μ2 μ3 σ1 σ2 σ3 Sq h1 h2 h3 μ1 μ2 μ3 σ1 σ2 σ3 Raz 2^vectorize (sum :

we iterate through the parameters to minimize the sum of the squared deviations

t0 0 time :

loop h1 11001 h2 11001 h3 11001 μ1 5120051 μ2 101350101 μ3 201500201 σ1 1801 σ2 11001 σ3 1150194 mat : i# 91 range Sqmin 1 : j# loop i# 4 el (loop i# 3 el (range tmp# loop 4 col : tmp# i# el j# : h1 tmp# 1 el : h2 tmp# 2 el : h3 tmp# 3 el : μ1 tmp# 4 el : μ2 tmp# 5 el : μ3 tmp# 6 el : σ1 tmp# 7 el : σ2 tmp# 8 el : σ3 tmp# 9 el : 33 mat Sq h1 h2 h3 μ1 μ2 μ3 σ1 σ2 σ3 Sq : loop i# 4 el j# : Sqmin Sq : 21 line Sq Sqmin < Sqmin Sqmin : cases 51 line for 21 line for h1 loop 14 el : hh1 h1 0.01 * : hh2 h2 0.01 * : hh3 h3 0.01 * : μμ1 μ1 : μμ2 μ2 : μμ3 μ3 : σσ1 σ1 : σσ2 σ2 : σσ3 σ3 : 33 mat 41 line

Here in the loop, Sq should be calculated and compared with Sqmin. The parameters under which the condition is met are saved

Dear alles,
Thank you for correcting my mistake.
But the desired result still did not work.
There, for some reason, the search of options stopped at the value of hh1 = 0.01, respectively, and hh2 = 001, hh3 = 0.01. That is, h1, h2, and h3 were only checked when the value =1.
Whereas I expected that the iteration of options should reach hh1 = hh2 = hh3 = 1, respectively h1 = h2 = h3 = 100.
For other parameters, it also seems that the search for options has not reached the end.
That is why the resulting graph shows a complete mismatch between the original data and the approximating formula.
I don't have an error in the If statement?
There I put continue in the else field.

loop h1 110020 h2 11008 h3 110093 μ1 51200131 μ2 101350135 μ3 201500201 σ1 18067 σ2 110011 σ3 11506294 mat

1 time t0 - 5 s *

Here the program should print the value of the smallest square of the deviation and the values of the 9 parameters at which this is achieved

h1 20

h2 8

h3 93

hh1 0.2

hh2 0.08

hh3 0.93

μ1 131

μ2 135

μ3 201

μμ1 131

μμ2 135

μμ3 201

σ1 67

σ2 11

σ3 62

σσ1 67

σσ2 11

σσ3 62

Sqmin h1 h2 h3 μ1 μ2 μ3 σ1 σ2 σ3 Sq : 0.000034847968104

x wh1 hh1 σσ1 2 π * sqrt * / x μμ1 - (2^2 σσ1 2^* / - exp * :

Draw the resulting graphs.
Three normal distributions and their sum with different weights

x wh2 hh2 σσ2 2 π * sqrt * / x μμ2 - (2^2 σσ2 2^* / - exp * :

x wh3 hh3 σσ3 2 π * sqrt * / x μμ3 - (2^2 σσ3 2^* / - exp * :

x WW x wh1 x wh2 + x wh3 + :

x g1 x WW x wh1 x wh2 x wh3 5 1 sys