Anonymous function/ Loop

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Alyssa
Alyssa 2016-2-18
评论： Andrei Bobrov 2016-2-18
Hello,
I'm trying to solve an anonymous function with fzero but with one of the variable being a vector.
So if
w = linspace(0,1,500);
And my anonymous function is
anon = @(y) y*3-w*0.2;
Solving for each y when the equation is set to zero
answer = fzero(anon,1)
But I am getting the error:
Operands to the and && operators must be convertible to logical scalar values.
Error in fzero (line 308) elseif ~isfinite(fx) ~isreal(fx)
Error in Homework (line 39) answer = fzero(anon,1)
.
My guess would be a loop, but I'm unsure how to go about it. If someone were able to help me out, I would greatly appreciate it!!

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Andrei Bobrov
Andrei Bobrov 2016-2-18
编辑：Andrei Bobrov 2016-2-18
anon = @(y,w) y*3-w*0.2;
w = linspace(0,1,500);
answer = arrayfun(@(y)fzero(@(x)anon(x,y),1),w);
or
anon = @(y,w) y*3-w*0.2;
w = linspace(0,1,500);
answer = zeros(500,1);
for jj = 1:500
answer(jj) = fzero(@(x)anon(x,w(jj)),1);
end
  2 个评论
Alyssa
Alyssa 2016-2-18
Thank you for responding! My problem is a little trickier than that though, but I tried your method and am still getting an error.
B = 4;
a = 2*pi;
sigma = 0.1;
anon = @(y,rbar) y^2*8*(2/pi)* acos(exp(-((B/2)*((1-y)/rbar))))-sigma*a*(rbar*(0.3-0.2*rbar)-y);
rbar = linspace(0,1,500);
answer = arrayfun(@(y)fzero(@(x)anon(x,y),1),rbar);
Andrei Bobrov
Andrei Bobrov 2016-2-18
B = 4;
a = 2*pi;
sigma = 0.1;
anon = @(y,rbar) y^2*8*(2/pi)* acos(exp(-((B/2)*((1-y)/rbar))))-sigma*a*(rbar*(0.3-0.2*rbar)-y);
rbar = linspace( eps, 1,500);
answer = arrayfun(@(y)fzero(@(x)anon(x,y),1),rbar);

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