How to find all solutions to a function?

I am trying to find all solutions to

f(y) = 1.1^2+y.^2-1-4.5*(sin(1.1.*y)).^2

However, I only get one of the solutions, which is

   -1.845486306009674675960636485178.

It's suppose to have four solutions, according to Wolframalpha.

y = -1.84
y = -0.21
y =  0.21 
y =  1.84

I have tried using the command vpasolve, like below,

sol = vpasolve(f,y).

But it does not work for some reason.

What other command can I use to find all four of the solutions to the function above?

Answer 1

[John D'Errico]

This could be a deflating experience. ;-)

x(1) = vpasolve(f)
x =
-1.8454863060096746759606364851789
>> x(2) = vpasolve(f./(y - x(1)))
x =
[ -1.8454863060096746759606364851789, 1.8454863060096746759606364851789]

x(2) = vpasolve(f./prod(y - x))
x =
[ -1.8454863060096746759606364851789, -0.21998207940626722662355300718635]

x(3) = vpasolve(f./prod(y - x))
x =
[ -1.8454863060096746759606364851789, -0.21998207940626722662355300718635, 0.21998207940626722662355300718635]

x(4) = vpasolve(f./prod(y - x))
x =
[ -1.8454863060096746759606364851789, -0.21998207940626722662355300718635, 0.21998207940626722662355300718635, 1.8454863060096746759606364851789]

Well, deflating, because the approach I used is often called deflation, where at each turn I kill off all of the other roots I found so far. So, if you plot f(x), we get this:

ezplot(f)
hold on
ezplot(f./(y - x(1)))

And now plot the deflated functions in turn. As you see, each one has one root "removed" from the previous curve.

ezplot(f./prod(y - x(1:2)))
ezplot(f./prod(y - x(1:3)))
h = refline(0,0);
h.Color = 'k';
legend('f(y)','1 root deflated','2 roots deflated','3 roots deflated','Zero')
axis([-6,6,-3,3])

At the very end, what would we see?

ezplot(f./prod(y - x))
refline(0,0)

Thus, a curve that has been thoroughly uprooted.

[wioqiiowqqiow]

Definitely, as I want to be able to take about 100 different points and find the y-values for every point.

[John D'Errico]

I showed how to find all 4 roots. It required only 4 calls to VPASOLVE. The graphics were merely to show you that it did indeed work.

vpa(x,5)
ans =
[ -1.8455, -0.21998, 0.21998, 1.8455]

Provably finding ALL roots of a completely general arbitrary function is impossible, since there may be infinitely many roots, and some may be very difficult to find.

However, I have no idea what you mean by "100 different points, finding the y values for each point." Sorry, but if you want more help than I have given you, you need to explain what you need more clearly than that.

[wioqiiowqqiow]

So, like I told madhan, I am trying to sketch an implicit function x.^2+y.^2=1+4.5.*(sin(x.*y).^2. If you want to see it, it looks like this , https://www.desmos.com/calculator/z0q9oodqtl.

And I am not allowed to use functions such as fimplicit,ezplot, contour etc. I cannot parameterize the implicit function either. Therefore, I am forced to use a numerical method, and put each x value in separately and get an output value, y, to then plot after those points.

And if I want to sketch the whole figure then there will be lots of points. Hence, the asking for a more simple method to get the root of a function.

Answer 2

[madhan ravi]

syms y

f = 1.1^2+y.^2-1-4.5*(sin(1.1.*y)).^2
fplot(f)
hold on
fplot(subs(y,0))
root1 = vpasolve(f,[-2 -1.5]) %see the graph and determine the interval of first root and the to the other roots 
root2 = vpasolve(f,[-0.5 0])
root3 = vpasolve(f,[1.5 2])
root4 = vpasolve(f,[0 0.5])

[madhan ravi]

see the edited answer and the attached screenshot of the graph

[wioqiiowqqiow]

Your way probably works. However, lets say that I want to find the solutions to 100 different x values. Is there a command that would give me the roots straight away, or would I have to use your way?

[madhan ravi]

have to use this way unfortunately (only for some equations - I guess for non-polynomial equations(not sure))

[wioqiiowqqiow]

Because I am trying to sketch this implicit function x.^2+y.^2=1+4.5.*(sin(x.*y).^2. And I am not allowed to use fimplicit. The graph looks like this, https://www.desmos.com/calculator/z0q9oodqtl, and you can see that it has several y-values for a certain x-value. Would you have any suggestions on how to approach the sketching of an implicit equation?

[madhan ravi]

use

ezplot(x.^2+y.^2-1-4.5.*(sin(x.*y).^2)
axis equal
hold on
ezplot(x.^2+y.^2-1)

[wioqiiowqqiow]

I dont think you are suppose to use any of matlabs commands to plot it for you. I think you are suppose to use a numerical method to solve for every x and y value and then plot it. Thats why I am trying to use vpasolve with a for-statement.

[Walter Roberson]

implicit solutions become possible if you convert to polar coordinates and solve for theta.

Answer 3

[Walter Roberson]

There is no method to find all roots of an arbitrary nonlinear function, even if it is not a "black box", and even if there are a finite number of roots . If there were such a method then solving for global minimum would just be a matter of solving for all roots of the derivative and choosing the one that gave the minimum.

If there were a general method to solve for all roots then there are a number of hard mathematical problems that would become easy by reexpressing them. For example it is not known if there is even one odd "perfect number" (a number equal to the sum of its factors including 1 but excluding itself, like 6 = 1+2+3). Create a function that runs over the odd integer and calculates number minus the sum of its factors, apply the hypothetical root finder, and out pops the complete list of them. Therefore such a hypothetical root finder would have to be a robust generalized existence prover for all kinds of mathematical problems. Gödel had something to say about that.

For functions for which there is a single y for each x, and which are not too strange, the technique of the following contribution can help.

https://www.mathworks.com/matlabcentral/fileexchange/55206-findroots