Transfer Code from Mathematica to Matlab
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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;
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Walter Roberson
2016-1-9
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 ?
5 个评论
Walter Roberson
2016-1-10
Is the intention that zSolution should be the z such that z is a root of eqchem near 0.01 ?
Theon
2016-1-10
Walter Roberson
2016-1-10
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
2016-1-10
更多回答（1 个）
mohammad hodaei
2020-4-23
0 个投票
But, I have codes around 28000 line. how do i tranfer it?
1 个评论
Walter Roberson
2020-4-23
Mathematica has a function to translate Mathematica into MATLAB.
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