The appearance of infinity in the problem with finding the minimum of a function

I want to find the so-called "efficiency" of a function, defined as the ratio of the difference of the absolute values of the maximum and minimum to their sum. My very simple code is as follows

function z=cur_phi_T_0
x=-2*pi:0.001:2*pi;

r1=1;r2=1; 

phi0=1.90132; theta0=1*pi/2;


I1=cos(x/2).*atanh(sin(x/2));

I2=cos((x+2*phi0)/2).*atanh(sin((x+2*phi0)/2));

I3=cos((x+2*theta0)/2).*atanh(sin((x+2*theta0)/2));

I_sum=(I1+r1*I2+r2*I3);

I_min=min(I_sum)
I_max=max(I_sum)

eff=(I_max-abs(I_min))/(I_max+abs(I_min))


plot(x,I_sum,'LineWidth',3)
grid on
set(gca,'FontName','Times New Roman','FontSize',34)
xlabel('\phi','FontName','Times New Roman','fontsize',34,'fontweight','b');
ylabel('I','FontName','Times New Roman','fontsize',34,'fontweight','b');
set(gca,'XTick',-pi:pi/2:pi)
set(gca,'XTickLabel',{'-\pi','-\pi/2','0','\pi/2','\pi'})

end

However, for theta0=pi/2 , I suddenly get infinity for the minimum of the function, even though the plot does not contain infinity.

At the same time for a very close value theta0=0.99999999*pi/2 everything works fine.

What is the reason for this strange behavior?

Update. And how to avoid singularity in calculations?

Answer 1

[Sam Chak]

Here is what I discovered. When ,

and

atanh(1)

ans = Inf

[Sam Chak]

This is how looks like:

x   = -5:0.0001:5;
y   = atanh(x);
z   =  tanh(x);
plot(x, y, x, z), hold on

Warning: Imaginary parts of complex X and/or Y arguments ignored.

plot(x, x, '--'), grid on, xlabel x, ylabel y
xticks(-5:5), axis square
legend('atanh(x)', 'tanh(x)', '', 'location', 'eastoutside')

[Yuriy Yerin]

I agree. Yes. I just thought that since Matlab nevertheless plots the function, it could output the minimum value as well ignoring infinity and can find the finite limit because cos(pi/2)=0

[Yuriy Yerin]

Yes, thanks, I realized that while plotting the graph Matlab avoids infinity (because of the chosen interval for x) and it is looking for the minimum because it can't find the limit in its algorithm.

[Sam Chak]

When , and .

In pure mathematics, zero times infinity is undefined. But it appears to smoothly join the discontinuities if is very small.

dx  = 0.01;
x   = -2:dx:2;
y   = cos(x*pi/2).*atanh(sin(x*pi/2));
plot(x, y), grid on, xlabel x, ylabel y

Answer 2

[Ganesh]

Hi @Yuriy Yerin,

You are encountering this issue, as the value of "I_sum(1)" when theta0 = pi/2 is indeed "-Inf". You can try to manually calculate "I1", "I2" and "I3", and you will find "I3" to be "-Inf" when x=-2*pi.

You are not able to see the value on the graph as the plot is interpolated. You would find that at x=0, you will attain a value of "Inf" for "I3". However, your array does not contain x=0.

Mathematically, this should be treated as a discontinuity. The only workaround for the same is to modify the array "x" so that the domain is valid for the mathematical equations.

I hope this helps!

[Yuriy Yerin]

Many thanks! I just realized that.