function handle producing different output
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Hi,
I am trying to write a code to find the intersection of the two curve:
A = @(m) -2.*(1/1.895.*((1+(1.4-1).*m.^2/2).^(1.4/(1.4-1)))-1)./(1.4.*m.^2);
B = @(m) (0.38./sqrt(1-m.^2));
m_lower = 0;
m_upper = 1;
m_mid = (m_lower+m_upper)/2;
{while abs(A(m_mid))-(B(m_mid))) > 0
if (A(m_mid))<(B(m_mid))
m_lower = m_mid
else
m_upper = m_mid
end
m_mid = (m_lower+m_upper)/2
end
However, when I tried to plot the curves, (i.e. fplot(A,[0 1]);) it gives me an incorrect curve. but when I tried to solve individually A(1) etc. etc., it produce the correct answer. Similarly when I tried to loop the equation in the while loop, it just goes to infinity because it using the wrong curve.
Many thanks in advance!
采纳的回答
Star Strider
2016-2-29
Your functions produce complex results, so the best you can hope for is to compare the absolute values of them:
A = @(m) -2.*(1/1.895.*((1+(1.4-1).*m.^2/2).^(1.4/(1.4-1)))-1)./(1.4.*m.^2);
B = @(m) (0.38./sqrt(1-m.^2));
AminusB = @(m) abs(A(m)) - abs(B(m));
IntSct(1) = fzero(AminusB, 0.5)
IntSct(2) = fzero(AminusB, 1.5)
x = linspace(0, 5, 100);
figure(1)
semilogy(x, abs(A(x)), x, abs(B(x)) )
hold on
plot(IntSct(1), abs(A(IntSct(1))), 'gp', 'MarkerSize',10)
plot(IntSct(2), abs(A(IntSct(2))), 'gp', 'MarkerSize',10)
hold off
grid
Producing:
IntSct =
754.2877e-003 1.3359e+000
I found the intersections by taking the differences between the absolute values of your functions, and then letting the fzero function find the zero crossings.
6 个评论
www
2016-2-29
Star Strider
2016-2-29
This is how I would do it:
A = @(m) -2.*(1/1.895.*((1+(1.4-1).*m.^2/2).^(1.4/(1.4-1)))-1)./(1.4.*m.^2);
B = @(m) (0.38./sqrt(1-m.^2));
AminusB = @(m) abs(A(m)) - abs(B(m));
dw = 1E-4;
w = dw;
Val = AminusB(w);
while Val > 0
Val = AminusB(w);
w = w + dw;
end
fprintf('\nA = B at %.4f with value = %.4f\n', w, Val)
A = B at 0.7544 with value = -0.0001
www
2016-2-29
Star Strider
2016-2-29
As always, my pleasure!
www
2016-2-29
编辑：Star Strider
2016-2-29
Star Strider
2016-2-29
The bisection method is a root-finding method, so I would use it with my derived ‘AminusB’ to find the root. That will be where A=B.
更多回答（1 个）
jgg
2016-2-29
Not sure what the deal is, but this works:
A_vals = A(0.1:0.01:1);
vals = 0.1:0.01:1;
B_vals = B(0.1:0.01:1);
plot(vals,A_vals);
hold on
plot(vals,B_vals);
You can see it works here too:
fplot(B,[0.1,0.99])
hold on
fplot(A,[0.1,0.99])
I think the asymptotes are causing it to look funny; I'm not convinced it's wrong though.
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