Find the roots of an expression

Is there a function that can find the root(s) of an expression, which is not a polynomial? My constants and my final expression (for which I want to find the root) are listed below.

%Declare constants
Dw = 0.0000076;
S = 0.000239;
rho = 1.1;
h = 0.0030;
Nm=20,000;
Wo = 0.0025;
c1 = Dw*S/rho/h;
c2 = (3*Wo/(4*Nm*3.14159*rho))^(1/3);
c3 = (3*Wwf/(4*Nm*3.14159*rho))^(1/3);
twf=20;

%Need to determine the value of Wwf in the following expression
((c2 - c3 - h*log((h + c2)/(h + c3)))/c1)-twf=0;

Answer 1

[Matt Fig]

FZERO can be used to find roots.

Dw = 0.0000076; 
S = 0.000239; 
rho = 1.1; 
h = 0.0030;
Nm=20000; 
Wo = 0.0025; 
c1 = Dw*S/rho/h; 
c2 = (3*Wo/(4*Nm*3.14159*rho))^(1/3); 
c3 = @(Wwf) (3*Wwf/(4*Nm*3.14159*rho)).^(1/3); 
twf=20;
G = @(Wwf) ((c2 - c3(Wwf) - h*log((h + c2)./(h + c3(Wwf))))/c1)-twf;

rt = fzero(G,.01)

[Deanna]

Thanks!

[John D'Errico]

Note that fzero finds A root, based on the starting value. It will not return all roots of a nonlinear expression, something quite difficult to do in general.

[Deanna]

I have another question. How can this code be written as part of a loop, where different values of twf were evaluated? I tried to do it (see code below), but I get the error message:

??? The following error occurred converting from function_handle to

double:

Error using ==> double

Conversion to double from function_handle is not possible.

Error in ==> fzero_1 at 25

c3(i) = @(Wwf) (3*Wwf/(4*Nm*3.14159*rho)).^(1/3);

Code:

clear all

Dw = 0.0000076;

S = 0.000239;

rho = 1.1;

h = 0.0030;

Nm=20000;

Wo = 0.0025;

c1 = Dw*S/rho/h;

c2 = (3*Wo/(4*Nm*3.14159*rho))^(1/3);

%initializations

twf(1)=1;

c3(1)=c2;

rt(1)=Wo;

i=1;

for i=2:10800

c3(i) = @(Wwf) (3*Wwf/(4*Nm*3.14159*rho)).^(1/3);

G(i) = @(Wwf) ((c2 - c3(Wwf) - h*log((h + c2)./(h + c3(Wwf))))/c1)-twf(i-1);

rt(i) = fzero(G(i),.01);

twf(i)=i;

end

Thanks!

[Matt Fig]

Try this:

Dw = 0.0000076;

S = 0.000239;

rho = 1.1;

h = 0.0030;

Nm=20000;

Wo = 0.0025;

c1 = Dw*S/rho/h;

c2 = (3*Wo/(4*Nm*3.14159*rho))^(1/3);

%initializations

twf(1)=1;

c3=@(Wwf) (3*Wwf/(4*Nm*3.14159*rho)).^(1/3);

rt = zeros(1,10800);

rt(1) = Wo;

for ii=2:10800

G = @(Wwf) ((c2 - c3(Wwf) - h*log((h + c2)./(h + c3(Wwf))))/c1)-(ii-1);

try

rt(ii) = fzero(G,rt(ii-1));

catch

disp('No value found, aborting....')

rt = rt(1:ii-1);

break

end

end

[Deanna]

Yes, thanks, that works as long as I end the loop early enough. For the code about 10800 is too many iterations.

Answer 2

[Deanna]

The following works. For the constants I gave I need to end the loop at 2600. If the loop runs longer an error occurs.

clear all
Dw = 0.0000076; 
S = 0.000239; 
rho = 1.1; 
h = 0.0030; 
Nm=8477.7;
Wo = 0.0025; 
c1 = Dw*S/rho/h; 
c2 = (3*Wo/(4*Nm*3.14159*rho))^(1/3); 
%initializations 
c3=@(Wwf) (3*Wwf/(4*Nm*3.14159*rho)).^(1/3); 
rt = zeros(1,2600); 
rt(1) = Wo;
time(1)=0;
for ii=2:2600 
G = @(Wwf) ((c2 - c3(Wwf) - h*log((h + c2)./(h + c3(Wwf))))/c1)-(ii-1); 
rt(ii) = fzero(G,[0 0.0025]);
time(ii)=time(ii-1)+ii;
end