Error in re-arranging equation - matlab

Hey guys I want to solve this equation to rearrange for x but it's giving a message which is strange. Does anyone know what the problem is? I know that the solution is complex but it should still solve for it. I know that my syntax etc are correct. Would anyone know an alternate way to display complex roots?

      syms x y
      >> eqn = (x^2.5 - (1250*(1-(0.525*x^0.5))^2)/(y*(1-(0.3*x^0.5)))==0)

      eqn =

      x^(5/2) + (1250*((21*x^(1/2))/40 - 1)^2)/(y*((3*x^(1/2))/10 - 1)) == 0

      >> v_x=solve(eqn,x)
      Warning: The solutions are parametrized by the symbols:
      z1 = ({0.0} union Dom::ImageSet(1.0*x*I + y, [x, y],  [R_,Dom::Interval(0.0, RD_INF)]) union
      Dom::ImageSet(1.0*x*I, x, Dom::Interval(0.0, RD_INF))) intersect RootOf(48*y*z^6 - 160*y*z^5 + 55125*z^2 -
      210000*z + 200000, z)

   > In solve at 190 

   v_x =

   z1^2

Answer 1

[John D'Errico]

So you want to solve for x, given an unknown y? Why do you think an analytical solution exists, or even if it does, that it can be found?

First, get rid of those radicals. Substitute

x = u^2

simplify(subs(eqn,x,u^2))
ans =
48*y*u^6 + 55125*u^2 + 200000 == 80*(2*y*u^4 + 2625)*(u^2)^(½) & ~u in {-10/3, 10/3} & y ~= 0

So, subject to some constraints on y and u, this reduces to a 6th degree polynomial in u, where the coefficients are functions of y. While it is possible to solve some for the roots of SOME simple polynomials, they are relatively rare. This is surely not one of them. In general, such polynomials of degree 5 or higher are not solvable analytically.

If you choose to supply a value for y, then you could solve for the roots numerically, but that is as much as you can ask to do. Even there you will get only numerical values for those roots, and only if a root exists.

Having done that, now lets go back to your original equation, because the transformation may introduce spurious solutions. ezplot being the impressive tool that it is, we get:

ezplot(eqn,[0,10])
grid on

vpasolve(subs(eqn,y,5))
ans =
2.6155171172702278618796721666051

To solve for a point on the upper branch of that curve,

vpasolve(subs(eqn,y,5),5)
ans =
6.25

[sarah]

Thanks for that. That's really helpful.

So there is no way of obtaining an analytical solution besides graphical? I was hoping to get equation in terms of x and hence substitute into a=b*x^0.5.

If not, I know the range of x from graphical solution so i could try working with that.

[John D'Errico]

I am pretty confident there would be no solution that you can get from a formula. That does not make it impossible, just that we cannot get to it. As I pointed out, this reduces to what is fundamentally a 6th degree polynomial, and that was shown long ago to have no analytical solutions for general coefficients.

Still, there are always many ways to cole a problem. For example, for the lower branch of x as a function of y, do this...

ylist = logspace(-5,2,101);
xlist = nan(size(ylist));
for ii = 1:numel(ylist)
  xlist(ii) = double(vpasolve(subs(eqn,y,ylist(ii)),1));
end
semilogx(ylist,xlist)
xlabel 'Y'
ylabel 'X'
grid on

Now, we can use a spline to interpolate that relationship, modeling it as x(log(y)).

spl_lower_branch = pchip(log10(ylist),xlist);

Now interpolation is simple, to interpolate any point off from that curve.

fnval(spl_lower_branch,log10(1.25))
ans =
        3.02304946224984

(ppval would work too here.)

Thus, when y was 1.25, x is 3.023... on the lower branch.

You should be able to solve for the other branch by a careful choice of starting values.

ylist = logspace(-5,4,201);
xlist2 = nan(size(ylist));
for ii = 1:numel(ylist)
xlist2(ii) = double(vpasolve(subs(eqn,y,ylist(ii)),[3.626 12]));
end
[xlist2;ylist]
semilogx(ylist,xlist2)
grid on

Again, we can fit that with a spline as above and use it for interpolation for points off the upper branch of the curve.

spl_upper_branch = pchip(log10(ylist),xlist2);

fnval(spl_upper_branch,log10(1.25))
ans =
        4.61207127135728

I might have chosen a nonlinear model to fit these curves. A bit of effort would give us something there if you dislike using splines.

[sarah]

thank you so much that was really helpful :))

[sarah]

I have another question for you regarding vpasolve for upper and lower branch. At times the graph would show two solutions but vpasolve for upper and lower branch gives the same answer. Would you know possibly why?