Using roots() and poly() for multivariable functions

How would I go about using poly() and roots() to solve the roots for this multivariable function?

This is my attempt

syms z1 z2
p = poly([1 -.5 -.25 -.25])

p = 1×5

1.0000 0 -0.6875 -0.2812 -0.0312

r = root(p,z1,z2)

Error using sym/root
First argument must be scalar.

[Star Strider]

I am not certain about using roots for that, however the correct result can be found symbolically —

syms z_1 z_2 
Eqn = 1 - 1/2*1/z_1 - 1/4*(1/z_1*1/z_2) - 1/4*1/z_2^2

Eqn = 

[Z_1,Z_2] = solve(Eqn)

Z_1 = 

Z_2 = 

EvalEqn1 = subs(Eqn, {z_1,z_2},{Z_1(1),Z_2(1)})             % Verify Following EDIT

EvalEqn1 = 

0

EvalEqn2 = subs(Eqn, {z_1,z_2},{Z_1(2),Z_2(2)})             % Verify Following EDIT

EvalEqn2 = 

0

sol1 = solve(Eqn, 'returnconditions',1)

sol1 = struct with fields:
           z_1: z_2/(2*z_2 - 1)
    parameters: [1×0 sym]
    conditions: z_2 ~= -1/2 & z_2 ~= 1/2

sol = solve(Eqn, z_1, 'returnconditions', true)             % Run Responding To Walter's Observation

sol = struct with fields:
           z_1: z_2/(2*z_2 - 1)
    parameters: [1×0 sym]
    conditions: z_2 ~= -1/2 & z_2 ~= 1/2

EDIT — (22 Feb 2023 at 17:12)

I solved for both, just to see what the correct result is, and returning the conditions there produces the same result as solving for ‘z_1’. It appears that if ‘z_2’ equals -1/2 then the denominator of ‘z_2/(2*z_2 - 1)’ would be 0 with an infinite result. Using subs with the appropriate root pairs gives the expected value for the function.

Perhaps solve is just being compulsive. (My apologies for the anthropomorphism.)

.

[Walter Roberson]

Star Strider, could you confirm something I am seeing?

sol = solve(Eqn, z_1, 'returnconditions', true)

I am being told that one of the conditions is z_2 ~= -1/2, but (A) -1/2 is one of the two specific values you had returned, and (B) I cannot see any reason why -1/2 should be a problem at all.

[Dyuman Joshi]

In my understanding -

syms z_1 z_2 
Eqn = 1 - 1/2*1/z_1 - 1/4*(1/z_1*1/z_2) - 1/4*1/z_2^2;
[Z_1,Z_2] = solve(Eqn)

Z_1 = 

Z_2 = 

Here, it has not been specified for which variable to solve the equation. And by default, solve determines the variable via symvar, which are

symvar(Eqn)

ans = 

So solve() solves for both the variable and we get explicit solutions.

In this case, a variable is specified to solve the equation for, and the solve returns the solution accordingly

sol = solve(Eqn, z_1, 'returnconditions', true)

sol = struct with fields:
           z_1: z_2/(2*z_2 - 1)
    parameters: [1×0 sym]
    conditions: z_2 ~= -1/2 & z_2 ~= 1/2

However, I do not know why does it repeat the same condition. Edit - my bad, they are not the same condition.

[Walter Roberson]

Odd. Notice that when you solve for z_2 that it singles out z2 = -1/2 as a specific solution that is always true, but that when you solve for z_1 that it singles out z2 = -1/2 as a specific case in which the z1 solution does not apply.... even though there is no mathematical problem with z2 = -1/2 ...

syms z_1 z_2 
Eqn = 1 - 1/2*1/z_1 - 1/4*(1/z_1*1/z_2) - 1/4*1/z_2^2;

sol1 = solve(Eqn, z_1, 'returnconditions', true)

sol1 = struct with fields:
           z_1: z_2/(2*z_2 - 1)
    parameters: [1×0 sym]
    conditions: z_2 ~= -1/2 & z_2 ~= 1/2

sol1.conditions

ans = 

sol2 = solve(Eqn, z_2, 'returnconditions', true)

sol2 = struct with fields:
           z_2: [2×1 sym]
    parameters: [1×0 sym]
    conditions: [2×1 sym]

sol2.z_2

ans = 

sol2.conditions

ans = 

simplify(sol2.conditions)

ans = 

[Dyuman Joshi]

Walter, for z_2 = -1/2, the equation becomes null, and thus we can't solve for z_1

syms z_1 z_2 
Eqn = 1 - 1/2*1/z_1 - 1/4*(1/z_1*1/z_2) - 1/4*1/z_2^2

Eqn = 

subs(Eqn,z_2,-1/2)

ans = 

0

Answer 1

[Walter Roberson]

[N D] = numden(Eqn)
sol = root(N, z_1)

You will get

root(4*z_1*z_2^2 - z_1 - z_2 - 2*z_2^2, z_1)

But you would have followed your requirements that root() specifically be involved in the process.