Should I trust solutions obtained for a nonlinear non-transcendental equation in matlab using 'fzero' function?

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Alireza Babaei 2020-3-16

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评论： Walter Roberson 2020-3-18

y = @(x)((10^-9)*x*(cosh(x)*sin(x)-sinh(x)*cos(x)) - ((10^-9)-1*(x^4)*(1+cosh(x)*cos(x))))

Suppose I have a kind of equation similar to the one above:

'y' is a function in variable 'x', it is a nonlinear, non-transcendental eqaution.

I have tried 'fzero', and I hae also plotted the function. So that plot gives me a shot regarding approximate location of the point whereas the function crosses y =0 line (corresponding x point is the answer). This is how I found the initial guess (eyeballing) and found following values for x:

0.149524, 1.875236, 4.694101

But wenenver I apply these numbers into the main equaiton to find the value of function at the mentioned points using 'subs' function in matlab, it is NOT equal to zero!!!!! (however we know value of function at the solution points should be equal to zero!!!!). Of course, using funciton 'round' yields zero. In short, I am a bit hessitant if my roots are correct or not.

Does anybody havea recommendation to find the roots? (logically snice I plot the function, I can estimate the first root should be pretty close to 0.15, and applying this initial guess into the 'fzero' function, yields 0.149524.) I don't know should trust it or not. (another reasin which makes me hesitant is the fact that physically the system obtaining such kind of nonlinear non-transcendental eqaution should have the first root close to 1.87).

Does anybody has a recommendation to solve such an enigma?

Thanks a lot

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Alireza Babaei 2020-3-16

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Can you please run my code in the following:

clc
clear all
close all
syms x laL
L = 100e-3;
rm = input('rm = ')
rs = rm
x = linspace(0,10,100)
y1 = (rm*rs*(L^3)*laL*(cosh(laL)*sin(laL) - sinh(laL)*cos(laL)) - (rs*(L^3) - rm*(laL^4)*(1+cosh(laL)*cos(laL))))
y2 = (rm.*rs.*(L.^3).*x.*(cosh(x).*sin(x) - sinh(x).*cos(x)) - (rs.*(L^3) - rm.*(x.^4).*(1+cosh(x).*cos(x))))
xi = linspace(0,10,11)
for i = 1 : length(xi)
    aa = vpasolve(y1 == 0, xi(i))
end 
y3 = 1 + cos(x).*cosh(x)
subplot(2,1,1)
plot(x,y2,'LineWidth',2)
grid on
ylim([-.000001 .000001])
hold on l
plot(x,y3,'--r','LineWidth',2)
subplot(2,1,2)
plot(x,y2,'LineWidth',2)
grid on
ylim([-400 400])
hold on l
plot(x,y3,'--r','LineWidth',2)
hold off

Running this code, shows that the main function (presented in 'y1' or 'y2' has another root pretty much close to zero (almost 0.14)). I also plotted y function to see if there is really another root close to zero and it weems that it is. However practically it does not match with reality.

This is the thing making me confused.

On the other hand you say that your root even for initial guess around 0.15 is NOT 0.14

Walter Roberson 2020-3-17

What value for rm?

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Matt J 2020-3-16

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编辑：Matt J 2020-3-16

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fzero is a numerical root finder. One doesn't expect y(x) to be exactly zero at the roots that it finds, but it should be pretty close. One way to evaluate the result is to plot the function in a tight interval around the supposed root and see how close to zero it is compared to its neighbors. The three roots you've indicated look pretty good to me.

fun = @(x)((10^-9)*x*(cosh(x)*sin(x)-sinh(x)*cos(x)) - ((10^-9)-1*(x^4)*(1+cosh(x)*cos(x))));
   
[x1,y1]=fzero(fun,[0,0.2])
[x2,y2]=fzero(fun,[1,2])
[x3,y3]=fzero(fun,[4.2,5])
close all 
figure;
fplot(fun, x1+[-1,1]*.0001);
hold on; plot(x1,y1,'ro','MarkerSize',6); hold off
figure;
fplot(fun, x2+[-1,1]*.0001);
hold on; plot(x2,y2,'ro','MarkerSize',6); hold off
figure;
fplot(fun, x3+[-1,1]*.0001);
hold on; plot(x3,y3,'ro','MarkerSize',6); hold off

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Steven Lord 2020-3-16

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For some functions, the difference between values that are displayed in the current display format and the values that are actually stored is negligible.

y = @(x) 2*x;
p = pi/2;
p2 = p + eps(p);
[p, y(p); p2, y(p2)]

In this case, p and p2 display the same (using the default format.) Their values are very close (but not the same, as you can see by computing p2-p) and the values computed by y are also very close.

y2 = @tan;
[p, y2(p); p2, y2(p2)]

A very small difference between p and p2 causes a huge difference in the value of y2. They even have different signs. Now this is a somewhat artificial example that I carefully concocted to have p and p2 end up on different sides of the tangent function's asymptote. But sometimes (especially when your function is steep) small differences get magnified.

If you plot some of the components of your function, when x gets large enough the cosh, sinh, and x.^4 functions start growing rapidly. So a very small change in x can result in a large change in y.

The fzero function is nice in that you can check if the answer is correct. Simply call the function you passed into fzero with the output of fzero as its input. If the answer is "close enough" to 0, the answer is a correct answer. If it's not, you may need to adjust the tolerances in your fzero call to define "close enough" how you want it to be defined.

Alireza Babaei 2020-3-17

Yes, I'm doing with 'format long', ...

here is my abstract: I will go agead with both cases (root1 = 0.14 and root1 = 1.87) to see what is the response of the system and then decide about which one to nominate as the rirst root

Walter Roberson 2020-3-18

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You are working with two different equations. You need to figure out which of the two equations is right.

I suspect that when you built

y = @(x)((10^-9)*x*(cosh(x)*sin(x)-sinh(x)*cos(x)) - ((10^-9)-1*(x^4)*(1+cosh(x)*cos(x)))) 

that you made a mistake.

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