how to code equation this equation?
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not sure how to code this equation:
I have tried this:
nz = @(z) n(z0)*(Tz0/Tz)^(1+g).*exp(-uc(z))
but I think it is not right.
采纳的回答
Assuming T is a function:
n = @(z) n(z0)*(T(z0)/T(z))^(1+g).*exp(-u*c(z))
10 个评论
Walter Roberson
2022-2-12
编辑：Walter Roberson
2022-2-13
nz0 = some constant
n = @(z) nz0 .* (T(z0)./T(z)).^(1+g).*exp(-u*c(z))
and c(z0) must equal 0
However we cannot tell what the value of n(z0) is.
Star Strider
2022-2-12
This is derivative of How can I write and solve this equation? as a function?? the problem being that it is recursive and that according to the expression, calculation of ‘n(z0)’ depends on already knowing what ‘n(z0)’ is.
This problem needs a larger context, in which this might be explained.
Cesar Cardenas
2022-2-12
Thanks much. The idea of this problem is to calculate ni which is the particle number density of a gas (Ar) at z = 200 km, and z0 = 120 km.(Walker, J. Atmos. Sci, 22, 462, 1965)
The modified form of the above equation is:
Any help will be greatly appreciated thanks much.
Walter Roberson
2022-2-12
z_0 = 120
n_iz_0 = some constant
mu = s + 1./(R_0 + z_0)
Gamma = m_i .* g(z_0) ./ (mu .* T_inf)
n_i = @(z) n_iz_0 .* (T(z_0)./T(z)).^(1+Gamma).*exp(-mu*sigma(z))
n_i(200)
What we are given so far does not give enough information to determine what 
is, or what 
is, or what 
is, or 
or what 
is, or what the formulae are for 
or 
is, or what is, or what is, or or what is, or what the formulae are for or
Cesar Cardenas
2022-2-13
Thanks so much. For the variables you mention the problem has some constants and other equations, for instance, for this one: I changed "sigma" for "zeta", as seen in the previous image e is 
n_i = @(z) n_iz_0 .* (T(z_0)./T(z)).^(1+Gamma).*exp(-mu*zeta(z))
I changed "sigma" for "zeta", as seen in the previous image e is 
so, zeta (z) could be coded like this??
zeta = @(z)*(z - z0)*(Re + z0)/(R+z);
zeta(200)
Any always any comments and help will be greatly appreciated. Thanks
Walter Roberson
2022-2-13
That is not a zeta. Zeta always has a upper-left extension. That is a sigma; in particular the form of sigma used in word-final positions; https://en.wikipedia.org/wiki/Greek_alphabet#cite_note-nicholas_finalsigma-31
Walter Roberson
2022-2-13
编辑：Walter Roberson
2022-2-13
zeta = @(z) (z - z_0) .* (R_e + z_0) ./ (R_e + z);
Cesar Cardenas
2022-2-13
Walter Roberson
2022-2-14
No. Read the equation again. Read the middle term, which is written using 
. Read the last term, which is also written in terms of 
.
. Read the last term, which is also written in terms of . Now look at your proposed implementation. The middle term in your proposed implementation is written in terms of the variable Re so we know that where 
appears in the equation, it will be replaced with Re in the code. Now look at the last term, which in your code is written in terms of R. The R-type variable in the last term needs to be the same as the R-type variable in the middle term, and since you used Re for the middle term, it follows that you need to use Re in the last term to be consistent with the middle term.
appears in the equation, it will be replaced with Re in the code. Now look at the last term, which in your code is written in terms of R. The R-type variable in the last term needs to be the same as the R-type variable in the middle term, and since you used Re for the middle term, it follows that you need to use Re in the last term to be consistent with the middle term.Your proposed code has an * immediately after the @(z) part that introduces the anonymous function and indicates what the dummy parameter names are for the function. That * could only work in that position if MATLAB had a unary asterisk operator... which it does not have. Some programming languages do define a unary asterisk operator; for example in C and C++, a unary asterisk is used to indicate pointer indirection. In C, the expression *(z-z0) would be valid, and would indicate that z-z0 is to be calculated and the difference is to be used as a pointer and the content stored at that pointer was to be extracted and used in further calculation of the expression. But MATLAB does not define any such operator. The * there is a syntax error in MATLAB.
I would also point out that if you continue to use the / operator, then your expression is not vectorized -- and so could not be used to fplot or integrate or anything else that requires a vector result for a vector input.
The code I posted for you in https://www.mathworks.com/matlabcentral/answers/1648105-how-to-code-equation-this-equation#comment_1984045 already took care of all of those problems. I do not understand why you felt it was necessary to revert back to your older incorrect code.
Cesar Cardenas
2022-2-14
更多回答（1 个）
Cesar Cardenas
2022-3-15
2 个评论
Voss
2022-3-15
You have an extraneous * there which is causing the error. Change that line to:
a = exp(1./(2*Hn)*z);
exp() is a function. Doing exp*(something) is the same as exp()*(something), i.e., calling exp() with no inputs, which is what the error is saying.
Cesar Cardenas
2022-3-15
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