How can I write and solve this equation? as a function??

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Cesar Cardenas 2022-2-10

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评论： Cesar Cardenas 2022-3-31

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Benjamin Thompson 2022-2-10

Solve the equation for what? Need more information, and the screenshot is a little difficult to read.

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Star Strider 2022-2-10

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Possibly, after first providing values for the constants:

T(z,z0,Tz0,Tinf) = Tinf - (Tinf - Tz0).*exp(z-z0);              % Function
zsol = fzero(@(z)T(z,z0,Tz0,Tinf), rand)                        % Correct Call To It (Here Using 'fzero', Can Be Any Function)

See the documentation on Anonymous Functions for details.

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Star Strider 2022-2-11

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As always, my pleasure!

The function I originally wrote was for the Symbolic Math Toolbox, although the fzero call was numeric.

The numeric version (written as such) produces some interesting results —

Tfcn = @(z,z0,Tz0,Tinf) Tinf - (Tinf - Tz0).*exp(z-z0); % Function

T = 200;

z0 = 120;

Tinf = 651;

Tz0 = 350;

[zsol,fv] = fzero(@(z)(T - Tfcn(z,z0,Tz0,Tinf)), 100) % Correct Call To It (Here Using 'fzero', Can Be Any Function)

zsol = 120.4044

fv = 6.2528e-13

[zsol,fv] = fsolve(@(z)(T - Tfcn(z,z0,Tz0,Tinf)), 100) % Correct Call To It (Here Using 'fzero', Can Be Any Function)

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.

zsol = 120.4044

fv = 6.2528e-13

zv = linspace(zsol - 1.5, zsol + 1.5, 500);

zi = interp1(Tfcn(zv,z0,Tz0,Tinf), zv, 0)

zi = 120.7714

figure

plot(zv, Tfcn(zv,z0,Tz0,Tinf))

hold on

plot(zsol, fv, 'cp', 'MarkerSize',10)

plot(zi, fv, 'mp', 'MarkerSize',10)

hold off

grid

xlabel('z')

ylabel('T(z)')

legend('Function Evaluation','Function Solution','Interpolation Solution', 'Location','best')

I leave you to explore the solution you prefer.

It might also be interesting to see what the Symbolic Math Toolbox solution is.

(Computer crashed — again — a few minutes ago, explaining much of the delay. I have no idea the reason both Windows 10 and Windows 11 do this so frequently.)

Star Strider 2022-2-12

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As always, my pleasure!

It would be necessary to do essentially everything in my earlier Comment with the appropriate parameters. If the parameters are different, just change them and run that code with them.

A more straightforward way would be to use ‘sfcn’ and ‘D1Tzfcn’ that I have introduced at the end of this Comment. They are anonymous functons that do not require the Symbolic Math Toolbox and can be run from base MATLAB. All the arguments of course must be present in the workspace when the functions are called.

syms s Tz T_inf T_z0 z z_0

Eqn = Tz == T_inf - (T_inf - T_z0) * exp(-s*(z-z_0))

Eqn =

D1Tz = diff(rhs(Eqn),z)

D1Tz =

Eqn = isolate(Eqn, s)

Eqn =

% Tz = 200;

% z_0 = 120;

% T_inf = 651;

% T_z0 = 350;

% z = 120.7714;

% Eqn = subs(Eqn)

Eqn = vpa(Eqn)

Eqn =

s = rhs(Eqn)

s =

D1Tz = subs(D1Tz)

D1Tz =

D1Tz = vpa(D1Tz)

D1Tz =

sfcn = matlabFunction(rhs(Eqn)) % Calculates 's'

sfcn = function_handle with value:

@(T_inf,T_z0,Tz,z,z_0)(log((T_inf-Tz.*1.0)./(T_inf-T_z0.*1.0)).*-1.0)./(z-z_0.*1.0)

D1Tzfcn = matlabFunction(D1Tz) % Calculates 'dT(z)/dz'

D1Tzfcn = function_handle with value:

@(T_inf,T_z0,Tz,z,z_0)(exp(log((T_inf-Tz.*1.0)./(T_inf-T_z0.*1.0)).*1.0).*log((T_inf-Tz.*1.0)./(T_inf-T_z0.*1.0)).*(T_inf-T_z0.*1.0).*-1.0)./(z-z_0.*1.0)

format longg

Tz = 200;

z_0 = 120;

T_inf = 651;

T_z0 = 350;

z = 120.7714;

s = sfcn(T_inf,T_z0,Tz,z,z_0) % Evaluate Function

s =

-0.524185992680584

dTdz = D1Tzfcn(T_inf,T_z0,Tz,z,z_0) % Evaluate Function

dTdz =

-236.407882698943

Using the anonymous functions to do the numeric calculations should make things easier. They are now independent of the Symbolic Math Toolbox. They should produce the same results with the same arguments, as demonostrated here, although with more limited precision (double instead of the symbolic extended precision, however that will likely not be a significant problem given their magnitudes).

Walter Roberson 2022-2-16

If I understood my earlier skimming properly, between 85 km and 115 km, the behaviour is dominated by mixing effects, so the temperature increase is largely unrelated to molecular weight, but above 115 km up to about 200 km, you get more and more stratification by molecular weight, and so you need a composite model that mixes between the 85/115 behaviour (that you have the data on) and other behaviour (which you do not have the data on.) The Bates-Walker model, if I understood correctly, is then considered valid between 85 and 115 but has to be ramped down above 115 km. And that the Bates-Walker model is old enough that those other effects were not known about, and the newer effects are still being studied so the newer models are still somewhat ad-hoc. If I understood the reading correctly, Bates-Walker is still used above 115 km by some people, but people are getting less and less satisfied with it.

All in all, it seems questionable to use the 88/113 data in the table to extrapolate up to 200 -- something that would have been acceptable in past times, and might be acceptable for a homework assignment these days, but not considered good enough for actual Science now ?

Cesar Cardenas 2022-2-20

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Thanks, basically, for q(h) I have this:

I do not have errors and it runs well when the other parameters are defined.

q_max = 0.5;
%h = 139.6;
h_max = 198.369;
H = 35.6757;
x = 15;
format long g
qh = @(h) q_max.* exp((1 - (h - h_max)./H - sec(x)).*exp(-(h-h_max)./H));
qh (198)

I have this other equation for q(h) as in this case it depends on h and x, at first I get some erros but now it seems to run well, but sure it would be well defined?

q_max = 0.5;
%h = 198;
h_max = 198.369;
H = 35.6757;
%x = 15;
format long g
qh = @(h,x) q_max.*exp((1 - (h - h_max)./H - sec(x)).*exp(-(h-h_max)./H));
qh (198,15)

and for q_max as you mention in this comment I defined like this for instance for h = 100 and get errors

format long g
q_max = @(h) q(h) ./exp((1 - (h - h_max)./H - sec(x)).*exp(-(h-h_max)./H))
q_max(100)

Index exceeds the number of array elements. Index must not exceed 1.

Thanks much.

Walter Roberson 2022-2-20

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q is a variable that is an array; it is not a function at that point.

We do not see what q is in your code.

qh = @(h,x) q_max.*exp((1 - (h - h_max)./H - sec(x)).*exp(-(h-h_max)./H));

Okay, you made qh a function of h and x. That potentially makes sense

q_max = @(h) q(h) ./exp((1 - (h - h_max)./H - sec(x)).*exp(-(h-h_max)./H))

You made q_max a function of h but not of x . That feels a bit inconsistent after the definition for qh

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Cesar Cardenas 2022-3-16

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Not sure what I'm not doing right as I'm not getting any change when the value is varied?? any help will be appreciated. Thanks.

net = 9.37e4;
T = 183;
mi= 5.31e-26;
k = 1.38e-23;
cs = 1.1304e-18;
format long g
Kz = @(z) cs*net*sqrt(8*k*T)/3.14*mi;
Kz(433)

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Walter Roberson 2022-3-16

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Kz = @(z) cs*net*sqrt(8*k*T)/3.14*mi;

That function accepts 0 or 1 parameters, and ignores the parameter and computes a constant.

Note: we recommend you use pi instead of 3.14 unless you have particular reason to approximate the value.

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Cesar Cardenas 2022-3-16

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Right thanks, how can I change it to accept any value? Thanks much

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Walter Roberson 2022-3-20

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I was curious about which power operation would be fastest. The three possibilities are ^(-1./2) (matrix power), .^(-1./2) (power), and 1/sqrt() (potentially using hardware square root).

Conclusions are a bit tricky.

If you look at the below code, notice I discard the first few samples. When you do not do that, then right at the beginning you get some timings that are much higher.

If you look at the graph below, notice there is an isolated large peak for ^ and another for .^ . In my tests up to N = 1000, it was not unusual to see those at isolated places, especially for ^ -- for some reason the ^ operator seems more "bursty" than the others, which I have no explanation for at the moment (because it should devolve to the same code as .^ for the case of scalars). If you use movmedian() with a mean of 3 then those peaks disappear.

You can see from the bottom half of the plot that the timings are pretty variable for the three different operations, with none of them being a clear winner. So we have to look at something like the mean() to talk about the statistical behaviour.

In my tests, most of the time ^ was slightly slower on average. That is not unexpected, as it has to do an extra step to determine whether to use scalar behaviour. Occasionally .^ shows up slower on average, which I suspect is due to those occasional large bursts.

In my tests, 1/sqrt() was almost always faster on average; with N = 1000, this becomes more obvious.

%initialize

N = 300;

t_mpower = zeros(N,1); t_power = zeros(N,1); t_sqrt = zeros(N,1);

%time

for K = 1 : N; d = rand(); start = tic; d^(-1./2); t = toc(start); t_mpower(K) = t; end

for K = 1 : N; d = rand(); start = tic; d.^(-1./2); t = toc(start); t_power(K) = t; end

for K = 1 : N; d = rand(); start = tic; 1./sqrt(d); t = toc(start); t_sqrt(K) = t; end

%discard "warm-up"

t_mpower(1:9) = []; t_power(1:9) = []; t_sqrt(1:9) = [];

%display

format long g

[mean(t_mpower), mean(t_power), mean(t_sqrt)]

ans = 1×3

1.0e+00 * 6.04810996563573e-07 5.60137457044673e-07 4.7766323024055e-07

plot(([t_mpower, t_power, t_sqrt]))

legend({'\^', '.\^', '1/sqrt'})

Cesar Cardenas 2022-3-21

Thanks much. Maybe I was not so clear, the idea is to use the contour plot in the coordinate range of [-50, 50] and I think the solution would be similar to the one in the previous comment. Not sure how to do it.

Thanks so much

Star Strider 2022-3-21

I do not understand ‘coordinate range of [-50, 50] ’

What coordinates does this refer to?

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Cesar Cardenas 2022-3-22

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Thanks much I think your approach here comment. is pretty similar to the below image. I think the coordinate range may refer to something like this graph...not sure though. Thanks much.

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Star Strider 2022-3-22

I’m still lost.

This is not an area of my expertise.

Perhaps re-defining ‘x’ and ‘y’ in the earlier code will do what you want.

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Cesar Cardenas 2022-3-27

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How can I convert this value 0.6667 to a fraction? Thanks

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Cesar Cardenas 2022-3-31

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How can I calculate the magnitude of this vector:

v = [2 30 0]*10^-3,, it is to 10^-3 not sure how to add it to the code

v = [2: 30: 0];
sv = v.* v;       %the vector with elements 
                  % as square of v's elements
dp = sum(sv);     % sum of squares -- the dot product
mag = sqrt(dp);   % magnitude
disp('Magnitude:'); 
disp(mag);

Cesar Cardenas 2022-3-31

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@Walter Roberson thank you regarding this comment. Basically here what I need to is find V perpendicular as shown here;

So first I did this:

%Magnitude of B
B = [2 30 0];%uniform magnetic field
norm(B*10^-3)

Then, this; to find V parallel,

dot(v,B)

So, would it be a nice approach? or how can I work this out? Thanks much

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