How a solution depends on a variable
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Amanda Hoxell
il 17 Set 2019
Commentato: Star Strider
il 17 Set 2019
I have a function UE=Explicit(S,sigma,r,T,M,K,gamma,N); and I want to plot how the solution depends on gamma where 0.5<=gamma<=1. All of the other parameters are constants. How can I do this?
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Star Strider
il 17 Set 2019
Modificato: Star Strider
il 17 Set 2019
One approach:
gammav = linspace(0.5, 1, 10);
for k = 1:numel(gammav)
UE{k} = Explicit(S,sigma,r,T,M,K,gammav(k),N);
end
I have no idea what ‘Explicit’ is or what it produces, so it could be possible to vectorise this (instead of using the loop) depending on how you wrote the function. I am also saving each iteration to a cell array for that reason.
EDIT —
Plotting it depends on what ‘UE’ is. If ‘Explicit’ produces a scalar, this works:
figure
plot(gammav, [UE{:}])
grid
If it produces vectors or matrices, it will be necessary to use other approaches.
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Star Strider
il 17 Set 2019
My pleasure!
If my Answer helped you solve your problem, please Accept it!
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John D'Errico
il 17 Set 2019
Using gamma as the name of a variable is a bad idea. Regardless,...
Where is the problem? Just substitute a range of values for gamma. Save the solution for each gamma. Then plot, with gamma on the x axis, and your solution on the y axis. WTP?
You can even use tools like fimplicit, which will do the hard work for you. For example,
f = @(gam,y) gam.^2 - y.^2 - 2*gam.*y.^3 + gam.*y + 1;
fimplicit(f)
xlabel 'gam'
ylabel 'y'
grid on
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