Transfer Code from Mathematica to Matlab

I'm trying to transfer the following code from Mathematica to Matlab but I'm having trouble. Can anyone help?

n[x_, z_, T_] = 1/(Exp[x/T]*z + 1)

f[x_] = (3/2)*Sqrt[x]

eqchem[z_, T_] = 
 1 == Integrate[f[x]*n[x, z, T], {x, 0, \[Infinity]}, Assumptions -> {T > 0}]

zSolution[T_] := z /. FindRoot[Evaluate[eqchem[z, T]], {z, .01}]

points = Prepend[Table[{T, -T Log[zSolution[T]]}, {T, 0.1, 5, 0.1}], {0., 1.}] // Re;

Answer 1

[Walter Roberson]

n = @(x, z, T) 1./(exp(x./T).*z + 1);
f = @(x) (3/2)*sqrt(x);
eqchem = @(z, T) integral(@(x) f(x)*n(x, z, T), 0, inf) - 1;

At this point the Mathematica code provided breaks down. It uses z without z being a parameter. We could add it as a parameter to zSolution, an assumption that it should have been written as

zSolution[T_, z_] := ....

but then we need to pass in two parameters in the reference to zSolution that appears in points. We can see that there is explicit range of T values being iterated over in the Table construction, but there is no match for z there. Is the table built to be symbolic in z? Or is the {0., 1.} intended to be a z range, or is that {0., 1.} intended to be a final (or leading) point in the table being constructed?

Perhaps I am misunderstanding the meaning of the zSolution code. The /. operator looks like you might have intended a rule replacement, but the right side of the ./ does not appear to be in the form of a Mathematica rule. Is the intention that zSolution should be the z such that z is a root of eqchem near 0.01 ?

[Theon]

I'm trying to use "points" to plot a figure with this command:

ListPlot[points, GridLines -> Automatic, Frame -> True]

In Mathematica, it works..

{0., 1.} is intended to be a leading point in the table being constructed.

[Walter Roberson]

Is the intention that zSolution should be the z such that z is a root of eqchem near 0.01 ?

[Theon]

Yes that is the intention.

[Walter Roberson]

zSolution = @(T) fzero(@(Z) -1 + integral(@(x) (3/2)*x^(1/2)/(exp(x/T)*Z+1), 0,inf), [eps(realmin),20]);

t = 0.1 : 0.1 : 5;
points = arrayfun(zSolution, t);
scatter([0,t], [1,points])

The 20 upper bound is arbitrary, needed to get the fzero to work for t = 5. The necessary upper bound goes up fairly quickly: for example by t = 17.87 you need slightly more than 100.

If you do not put in a range, just an initial point, then fzero will head to negatives and the integral will start to fail.

[Theon]

Thank you so much.

Answer 2

[mohammad hodaei]

But, I have codes around 28000 line. how do i tranfer it?

[Walter Roberson]

Mathematica has a function to translate Mathematica into MATLAB.