Solve non linear equation with vector

Dear,

I would like to solve a non linear equation, on the form:

yv = 1:1:19;
for k1 = 1:length(yv)
    y = yv(k1);
    f = @(x) exp(-0.5*((x-45)/dados.mass(k1))^2) == dados.par(k1) ;
    x(k1) = fzero(f, 0.5);
end

Send : x the variable

dados.mass(k1) vector with value, contain 19 lines

dados.par (k1) equal aspect to top.

I would like to solve each equation related a value of dados.mass(1) until (19) concomitantly at dados.par(1) until (19), obtaining the vector x with respectivly values.

Can help me, please?

Yours faithfully

Answer 1

[Torsten]

x(k1) = 45+dados.mass(k1)*sqrt(-2*log(dados.par(k1)))

or

x(k1) = 45-dados.mass(k1)*sqrt(-2*log(dados.par(k1)))

[Sam Chak]

@Torsten, 👍 That's a good analytical solution without using fzero()! Did I plot the visualization of the solutions correctly?

m      = linspace(0, 300, 31);
p      = linspace(0, 1,   31);  % assuming that Gaussian function has a range of 0 to 1
[M, P] = meshgrid(m, p);
X1     = 45 - sqrt(2)*M.*sqrt(log(1./P));
X2     = 45 + sqrt(2)*M.*sqrt(log(1./P));

tiledlayout(2,2);

nexttile
surf(M, P, X1), xlabel('mass'), ylabel('par'), 
title({'$x^{-}$'}, 'interpreter', 'latex', 'fontsize', 16)

nexttile
surf(M, P, X2), xlabel('mass'), ylabel('par'), 
title({'$x^{+}$'}, 'interpreter', 'latex', 'fontsize', 16)

nexttile
contourf(M, P, X1), xlabel('mass'), ylabel('par')

nexttile
contourf(M, P, X2), xlabel('mass'), ylabel('par')

Answer 2

[Star Strider]

It looks like you want to do a nonlinear regression.

Perhaps this —

dados = array2table(sortrows([10*randn(12,1)+35,rand(12,1)],1), 'VariableNames',{'mass','par'})

dados = 12×2 table
     mass       par  
    ______    _______

    20.408    0.66176
     21.47    0.37043
    24.976    0.14994
    28.527    0.72292
    29.961    0.91367
    33.376    0.67683
    33.456    0.64286
    34.064     0.2926
    34.807    0.49665
    37.383    0.91716
    39.025      0.984
    58.526    0.60612

f = @(x,m) exp(-0.5*((x-45)./m).^2) ;

x0 = rand;

mdl = fitnlm(dados, f, x0)

mdl = 
Nonlinear regression model:
    par ~ F(x,m)

Estimated Coefficients:
          Estimate      SE      tStat     pValue  
          ________    ______    _____    _________

    b1     15.791     3.7571    4.203    0.0014781


Number of observations: 12, Error degrees of freedom: 11
Root Mean Squared Error: 0.25
R-Squared: 0.0592,  Adjusted R-Squared 0.0592
F-statistic vs. zero model: 74.6, p-value = 3.13e-06

[y,yci] = predict(mdl, dados.mass);

figure
hp1 = plot(dados.mass, dados.par, 'b.', 'DisplayName','Data');
hold on
hp2 = plot(dados.mass, y, 'DisplayName','Regression');
hp3 = plot(dados.mass, yci, '--r', 'DisplayName','95% Confidence Intervals');
hold off
grid
xlabel('mass')
ylabel('par')
legend([hp1,hp2,hp3(1)], 'Location','best')

If you do not have the Statistics and Machine Learning Toolbox (for fitnlm), an alternative would be the fminsearch function —

xest = fminsearch(@(x) norm(dados.par - f(x,dados.mass)), x0)

xest = 15.7910

figure
hp1 = plot(dados.mass, dados.par, 'b.', 'DisplayName','Data');
hold on
hp2 = plot(dados.mass, f(xest,dados.mass), 'DisplayName','Regression');
hold off
grid
xlabel('mass')
ylabel('par')
legend([hp1,hp2], 'Location','best')

.

Answer 3

[Sam Chak]

Hi @Guilherme Lopes de Campos

I have fixed the function f in the code. Now it should be working correctly. The dados.mass and dados.par data are not provided. Thus I made up some value to test the code.

mass = linspace(110, 190,  19);
par  = linspace(0.1, 0.19, 19);

yv = 1:1:19;
for k1 = 1:length(yv)
    % y = yv(k1);       % unused
    f = @(x) exp(-0.5*((x - 45)/mass(k1)).^2) - par(k1);    % <-- fix it here
    x(k1) = fzero(f, 0.5);
end

% Solutions
x

x = 1×19

-191.0563 -197.9780 -204.7954 -211.5110 -218.1269 -224.6453 -231.0681 -237.3972 -243.6343 -249.7812 -255.8394 -261.8103 -267.6954 -273.4959 -279.2132 -284.8485 -290.4028 -295.8772 -301.2727

subplot(2,1,1)
plot(mass, x), xlabel('mass'), ylabel('x'), grid on
subplot(2,1,2)
plot(par,  x), xlabel('par'), ylabel('x'), grid on