MATLAB Answers

function handle producing different output

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www on 29 Feb 2016
Commented: Star Strider on 29 Feb 2016
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.


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www on 29 Feb 2016
I was wondering how do I alter the code above to use the bisection method where I converge the intersection of A and B from a upper and lower limit and achieve a tabled results similar to : @8.54mins @8.54mins
My case would be to match A=B.
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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