Solving a nonlinear equation using fsolve. cant reach at perfect result. output showing some error written below.

Where M*_N = m - (a*sigma)

Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the

function tolerance, and the problem appears regular as measured by the gradient.

used code :

Result must be around 22-25. but itc coming .0029.

Is there anything wrong i am doing?

[Star Strider]

Since fsolve is a root-finding algorithm, plot the function to see if there are any zero-crossings. Since fsolve can find complex as well as real roots, it might be worth giving it complex initial parameter estimates to see what it finds.

Answer 1

[Matt J]

Result must be around 22-25. but itc coming .0029.

As the plot shows, your function has no roots in the range 22-25

p = 1;

pfun=@(sigma) arrayfun( @(z)fun(z,p) , sigma);

fplot(pfun,[0,25])
xlabel sigma; ylabel fun

function sigmares = fun(sigma,p)
m = 939;
ms = 550;
b = -7.25*10^(-3);
gs = 8.238;
p0 = .15*(197.3)^3;
k = (1.5*p*p0*pi^2)^(1/3);
gss = (1 + (b*sigma/2))*gs;
m1 = (m-(gss*sigma));
F = ((1+(b*sigma))*gs*m1*(1/(2*ms^2*pi^3))*log((sqrt(k^2 + m1^2)+ k))) - sigma;
sigmares = F ;

end

[pritha]

Thanks for this point . How can it be solved other way, if the function is F?

[Matt J]

Since you have only one unknown, you could solve it with fzero.

p = 1;
sigma0 = 1;
sigma = fzero(@(sigma)fun(sigma,p),sigma0)

sigma = 0.0029

function sigmares = fun(sigma,p)
m = 939;
ms = 550;
b = -7.25*10^(-3);
gs = 8.238;
p0 = .15*(197.3)^3;
k = (1.5*p*p0*pi^2)^(1/3);
gss = (1 + (b*sigma/2))*gs;
m1 = (m-(gss*sigma));
F = ((1+(b*sigma))*gs*m1*(1/(2*ms^2*pi^3))*log((sqrt(k^2 + m1^2)+ k))) - sigma;
sigmares = F ;

end

[pritha]

still it is coming the same. thanks for your concern. i think i'm doing mistakes on basic ideas.

Thanks so much.

[Matt J]

Naturally it's the same. We can see in the plot above that that's the only root.

[Walter Roberson]

It is the only real root. There might be several complex roots

[shiv gaur]

use muller method