function handle producing different output

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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
m_upper = m_mid
m_mid = (m_lower+m_upper)/2
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!
  1 Comment
www on 29 Feb 2016
What do I have to do if I want to continue the loop is A(m_mid) ~= B(m_mid)? Is there another function or neater way?
Pardon me, I am very new to MATLAB!

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Accepted Answer

Star Strider
Star Strider on 29 Feb 2016
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);
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
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.
Star Strider
Star Strider on 29 Feb 2016
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.

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More Answers (1)

jgg on 29 Feb 2016
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);
hold on
You can see it works here too:
hold on
I think the asymptotes are causing it to look funny; I'm not convinced it's wrong though.

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