8a76faaa-1e52-40ed-96e4-fcc4ef3d96a2 27 4 5 false false 0 decimal

&[DATE] &[TIME] - &[FILENAME]

underlying data

xv 709011013015018022527532591 mat :

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

This is a more practical definition for g. Can use linterp too

Don't use x, leave it undefined. Have special mean for plots

x g1 xv yv x linterp :

x g1

we introduce a formula for the approximation

a1 1002 π * sqrt * :

x 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 * + :

x h1 h2 h3 μ1 μ2 μ3 σ1 σ2 σ3 Raz x Wb yv - :

Here , where all parameters are undefined, except a1. Also, introduce x.

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 :

σ1 1 :

σ2 1 :

σ3 1 :

Sqmin 1 :

hh1 0.15 :

hh2 0.7 :

hh3 0.40 :

μμ1 90 :

μμ2 160 :

μμ3 350 :

σσ1 19 :

σσ2 50 :

σσ3 80 :

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

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

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

introduce x here

This seems to be a very good value for the difference error.

xv h1 h2 h3 μ1 μ2 μ3 σ1 σ2 σ3 Sq 0.0002

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

h1 100 < h1 h1 1 + : h2 100 < h2 h2 1 + : h3 100 < h3 h3 1 + : μ1 200 < μ1 μ1 1 + : μ2 350 < μ2 μ2 1 + : μ3 500 < μ3 μ3 1 + : σ1 80 < σ1 σ1 1 + : σ2 100 < σ2 σ2 1 + : σ3 150 < σ3 σ3 1 + : Sq xv h1 h2 h3 μ1 μ2 μ3 σ1 σ2 σ3 Sq : Sq Sqmin < hh1 h1 100 / : hh2 h2 100 / : hh3 h3 100 / : μμ1 μ1 : μμ2 μ2 : μμ3 μ3 : σσ1 σ1 : σσ2 σ2 : σσ3 σ3 : Sqmin Sq : 101 line continue if 31 line while 21 line while 21 line while 21 line while 21 line while 21 line while 21 line while 21 line while 21 line while

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

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

Sqmin 0.0002

hh1 0.01

hh2 0.01

hh3 0.01

μμ1 51

μμ2 101

μμ3 201

σσ1 2

σσ2 2

σσ3 62

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

Above procedure have not reason for converge to a solution for the problem, try this other, but yours guess values seems to be a good solution for the problem.

hh1 hh2 hh3 μμ1 μμ2 μμ3 σσ1 σσ2 σσ3 91 mat 0.15 0.7 0.4 9016035019508091 mat :

sol xv h1a h2a h3a μ1a μ2a μ3a σ1a σ2a σ3a Sq hh1 hh2 hh3 μμ1 μμ2 μμ3 σσ1 σσ2 σσ3 91 mat NelderMead 11 line :

hh1 hh2 hh3 μμ1 μμ2 μμ3 σσ1 σσ2 σσ3 91 mat sol : 0.15 0.7 0.4 9016035019508091 mat

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

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

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

x WW x wh1 x wh2 + x wh3 + :

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