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How to pass values to a vector inside an anonymous function for ezplot?

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Ezz El-din Abdullah
Ezz El-din Abdullah 2017 年 3 月 5 日
コメント済み: Guillaume 2017 年 3 月 6 日
I wonder how can I benefit from passing additional parameter for ezplot function.
I'm trying to plot implicit function that has a vector th inside which has the problem.
Here is the code:
delta = pi*[1 2 3];
m = 1:100;
th{1,1}(1:100) = 2*pi*m*delta(1);
th{2,1}(1:100) = 2*pi*m*delta(2);
th{3,1}(1:100) = 2*pi*m*delta(3);
myfun = @(x,y,k) x + y.*th{k,1}(1:100) + k;
for k = 1:3
ezplot(@(x,y)myfun(x,y,k));
hold on
end
  10 件のコメント
Ezz El-din Abdullah
Ezz El-din Abdullah 2017 年 3 月 6 日
What I mean by three lines is that the equation x + y*th + k is satisfied.
Let's consider the simplest case when th is a scalar. Since th has the vector delta which has three values in it. It will produce three lines. I know that the output of ezplot is just 1x1 so when th is a scalar the ezplot will not complain.
The problem comes when we consider the th as a vector like I wrote in the code.
So for each value of delta which corresponds to the first vector of the th 2D matrix (as you explained) I would like to "sort of" convert this vector into a just symbolic 1x1 that ezplot output can produce. So actually, I don't know what myfun(x,y,k) should produce.
It happens that for a scalar value of th it produced this figure:
For that case:
>> myfun(1,1,1)
ans =
2
I hope I'm clear now and thanks.
Guillaume
Guillaume 2017 年 3 月 6 日
I understand everything you wrote except the I would like to "sort of" convert this vector into a just symbolic 1x1 that ezplot output can produce. It doesn't sound that you actually know what you want.
ezplot plots where myfun(x, y) == 0. What does equal to 0 mean when myfun(x, y) is a vector of 100 values? Do you want the zeros of the function
f(x, y) = x + a*y + k
for 100 different scalar a and 3 different scalar k, resulting in 300 plots?

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