Root() in answer of solve function

Hello, I am solving a partial derivative equation using symbolic expression. After using f1= diff(f,Qs) and solve(f1,Qs), I got the following answer that have root(....). Anybody can explain to me how can I get rid of root()? Note: z is not defined initially

Thanks for your help

Example:ans= root(3*a^2*k^2*z^6 - 2*Qn*a^2*k^2*z^5 + 4*a*k*z^5 - 3*Qn^2*a^2*k^2*z^4 + 8*Qn*a*k*z^4,z,1)

[pepe]

I have had a similar version dependent exprience. My code is:

syms x

solve(x^3+x-1==0)

In Matlab 2019 it produces the useless root() output that you also mentioned. But in MATLAB 2014 it produces a nicer result comprised of fractions and sums of numbers;...so I recommend you also give it a try using a old version of MATLAB.

good luck

پویا پاکاریان

پویاپاکاریان

[Walter Roberson]

syms x
>> simplify(solve(x^3+x-1==0, 'maxdegree', 3),'steps',50)
 
ans =
 
                                                                                        -(12^(1/3)*((93^(1/2) - 9)^(1/3) - (93^(1/2) + 9)^(1/3)))/6
 (2^(2/3)*3^(1/3)*((93^(1/2) - 9)^(1/3) - (93^(1/2) + 9)^(1/3)))/12 - (2^(2/3)*(3^(5/6)*(93^(1/2) - 9)^(1/3) + 3^(5/6)*(93^(1/2) + 9)^(1/3))*1i)/12
 (2^(2/3)*(3^(5/6)*(93^(1/2) - 9)^(1/3) + 3^(5/6)*(93^(1/2) + 9)^(1/3))*1i)/12 + (2^(2/3)*3^(1/3)*((93^(1/2) - 9)^(1/3) - (93^(1/2) + 9)^(1/3)))/12

Answer 1

[Walter Roberson]

RootOf(f(z), z) represents the set of values, z, that satisfy f(z) == 0 -- the roots of the expression.

You have a polynomial of degree 6. There are not generally exact solutions to such polynomials, but if the polynomial is what is shown, that particular one can be factored to create four 0's and two exact roots.

You could try applying simple() to the output.

[Linh Dang]

Thank you for your answer. Actually I got 4 roots with the same root(...) in the answers. I tried the simplify() but it stays the same. So that means there is no way to find a proper expression of the root ? I tried the vpa() function but the answer is so long that Matlab cannot presents it all

[Walter Roberson]

If you got 4 roots, then you can use 'MaxDegree', 4 as options on the solve(). The output will be very long, though.

In cases like the one you show,

syms Qn a k z
sol = root(3*a^2*k^2*z^6 - 2*Qn*a^2*k^2*z^5 + 4*a*k*z^5 - 3*Qn^2*a^2*k^2*z^4 + 8*Qn*a*k*z^4, z);

orig_expression = sol;

term0 = arrayfun(@(V) isempty(symvar(V)) && double(V) == 0, exprc)

num_significant_leading = find(~term0, 1, 'last');
num_zero_roots = length(term0) - num_significant_leading;

if num_significant_leading == 0
 reduced_roots = sym([]);
elseif num_zero_roots ~= 0
  recreated_expr = poly2sym(exprc(1:end-num_zero_roots), z);
  reduced_roots = solve(recreated_expr,z);
else
  reduced_roots = orig_expression;
end

The roots are then reduced_roots together with num_zero_roots copies of 0. If there are no non-zero roots, then reduced_roots will be set empty -- which is not to be taken to mean there are no roots, just that there are no non-zero roots in that case.

[Linh Dang]

I got it. Thank you for your help.

[Ariel Chazan]

Walter, I have a question on the nature of an expression of the form root(...). Is the expression an integer?

For example, if I apply the solve function a polynomial and get an array x with three elements all in the form of root(...), i.e. x = [root(...), root(...), root(...)], can I now assign a variable to one of these elements, i.e a = x(1), and use a later in a piece of code just like any other variable? Many thanks for your help

[Walter Roberson]

The expression would seldom be an integer. It would also not necessarily be real-valued.

However, yes you can assign it to a variable and use it in symbolic expressions.

If the root() expression contains only constants and the variable z then MATLAB will understand the represented value as being a constant.

Answer 2

[Victor Prohorov]

Read Matlab help:

Try to get an explicit solution for such equations by calling the solver with 'MaxDegree'. The option specifies the maximum degree of polynomials for which the solver tries to return explicit solutions. The default value is 2. Increasing this value, you can get explicit solutions for higher order polynomials.Solve the same equations for explicit solutions by increasing the value of 'MaxDegree' to 3.

[Walter Roberson]

Already discussed https://www.mathworks.com/matlabcentral/answers/282911-root-in-answer-of-solve-function#comment_365402

Also, when you are working with partial derivatives then when matlab sees the diff() then if you called solve() then matlab will call dsolve() in order to work out the solution. However, dsolve does not support the MaxDegree option and will return RootOf (though the presentation interface will rewrite them as root() to show to the user.) There is no method provided by matlab to reduce a RootOf that has already been returned.