# How a solution depends on a variable

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Amanda Hoxell on 17 Sep 2019
Commented: Star Strider on 17 Sep 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?

Star Strider on 17 Sep 2019
Edited: Star Strider on 17 Sep 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.

Star Strider on 17 Sep 2019
This works, however, ‘UE’ is uniformly 0 for all values of ‘gammav’.
gammav = linspace(0.5, 1, 10);
for k = 1:numel(gammav)
[UE(:,k),A]=Explicit(S,sigma,r,T,M,K,gammav(k),N);
end
figure
plot(gammav, UE, '.r')
This code runs without errors.
Amanda Hoxell on 17 Sep 2019
Thank you!!
Star Strider on 17 Sep 2019
My pleasure!

John D'Errico on 17 Sep 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 #### 1 Comment

Amanda Hoxell on 17 Sep 2019
The problem is that I am new to Matlab. I tried the following but it did not work..
gamma_all = 0.5:0.1:1;
y = zeros(1,length(gamma_all));
for i = 1:length(gamma_all)
gamma = gamma_all(i);
y(i) = Explicit(S,sigma,r,T,M,K,gamma,N);
end
plot(gamma_all,y)