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Hello,

I am not sure if I am overlooking something very obvious or whether the MATLAB symbolic solver is giving me a wrong solution.

I have the following code:

syms p t

formula = 84*p^2*t^4 - 112*p^4*t^2 - 64*p^6 + 27*t^6 + (4*p^2 + t^2)^(1/2)*(4*p^2 + 9*t^2)^(1/2)*(16*p^4 - 88*p^2*t^2 + 9*t^4)

solutions = solve(formula == 0, t)

vpasolutions = vpa(solutions)

Now MATLAB 2019a gives me the following result:

formula =

84*p^2*t^4 - 112*p^4*t^2 - 64*p^6 + 27*t^6 + (4*p^2 + t^2)^(1/2)*(4*p^2 + 9*t^2)^(1/2)*(16*p^4 - 88*p^2*t^2 + 9*t^4)

solutions =

0

-p*2i

p*2i

-(p*2i)/3

(p*2i)/3

-(p*(272/27 - (64*13^(1/2))/27)^(1/2))/2

(p*(272/27 - (64*13^(1/2))/27)^(1/2))/2

-(p*((64*13^(1/2))/27 + 272/27)^(1/2))/2

(p*((64*13^(1/2))/27 + 272/27)^(1/2))/2

vpasolutions =

0

-p*2.0i

p*2.0i

-p*0.66666666666666666666666666666667i

p*0.66666666666666666666666666666667i

-0.61797697405792080417600009731734*p

0.61797697405792080417600009731734*p

-2.1575776918969228428670635119892*p

2.1575776918969228428670635119892*p

So altogether there seem to be 5 real and 4 complex solutions for any given 'p'. I am only interested in the real ones. Now I plot the graph for p = 1 using the following code:

t_values = linspace(-2.2,2.2,1000);

formula_values = subs(formula,{p,t},{1,t_values});

plot(t_values,formula_values)

grid on

I get the attached figure. Here one can see that my function should not be zero around -0.618 and +0.618, which seems to contradict the answers from the symbolic solver. Now I know that one cannot always trust a graph with discrete values but even if you decrease the step size, this does not seem to change, and wolfram alpha gives me all results except for those two ominous ones.

Furthermore, if I add the assumption that p must be positive, these ominous results also vanish:

syms p t

assume(p>0)

formula = 84*p^2*t^4 - 112*p^4*t^2 - 64*p^6 + 27*t^6 + (4*p^2 + t^2)^(1/2)*(4*p^2 + 9*t^2)^(1/2)*(16*p^4 - 88*p^2*t^2 + 9*t^4)

solutions = solve(formula == 0, t)

vpasolutions = vpa(solutions)

yields

formula =

84*p^2*t^4 - 112*p^4*t^2 - 64*p^6 + 27*t^6 + (4*p^2 + t^2)^(1/2)*(4*p^2 + 9*t^2)^(1/2)*(16*p^4 - 88*p^2*t^2 + 9*t^4)

solutions =

0

-p*2i

p*2i

-(p*2i)/3

(p*2i)/3

-(p*((64*13^(1/2))/27 + 272/27)^(1/2))/2

(p*((64*13^(1/2))/27 + 272/27)^(1/2))/2

vpasolutions =

0

-p*2.0i

p*2.0i

-p*0.66666666666666666666666666666667i

p*0.66666666666666666666666666666667i

-2.1575776918969228428670635119892*p

2.1575776918969228428670635119892*p

In my understanding, this should not happen since before I made an assumption about p, MATLAB should give me the general solution that is valid for all p (even the positive ones). Now that I made an assumption, the number of solutions should not change.

Can you spot a mistake? Did I not understand the behavior of 'solve' correctly? Or is the symbolic solver really faulty?

John D'Errico
on 18 Jun 2019

Edited: John D'Errico
on 18 Jun 2019

It seems a subtle thing. We need to "branch" out from our preconceptions. ;-)

pretty(solutions)

/ 0 \

| |

| -p 2i |

| |

| p 2i |

| |

| p 2i |

| - ---- |

| 3 |

| |

| p 2i |

| ---- |

| 3 |

| |

| -#1 |

| |

| #1 |

| |

| -#2 |

| |

\ #2 /

where

/ 272 64 sqrt(13) \

p sqrt| --- - ----------- |

\ 27 27 /

#1 == ---------------------------

2

/ 64 sqrt(13) 272 \

p sqrt| ----------- + --- |

\ 27 27 /

#2 == ---------------------------

2

So #1 contains the questionable solutions. Are they? Are they spurious? It looks like case #1 converts to case #2, if we take the negative branch of the sqrt for sqrt(13).

It looks like solutions [1 2 3 4 5 8 9] all trivially satisfy the original problem, with 6 and 7 a problem.

simplify(subs(formula,t,solutions))

ans =

0

0

0

0

0

-(32768*p^6*(25*13^(1/2) - 86))/729

-(32768*p^6*(25*13^(1/2) - 86))/729

0

0

But suppose we look carefully at formula?

formula

formula =

84*p^2*t^4 - 112*p^4*t^2 - 64*p^6 + 27*t^6 + (4*p^2 + t^2)^(1/2)*(4*p^2 + 9*t^2)^(1/2)*(16*p^4 - 88*p^2*t^2 + 9*t^4)

In there, I see two square roots. When you take a square root, you can take the positive or the negative branch. Either is equally valid. So for any values of p and t, if I write Q as

Q = (4*p^2 + t^2)^(1/2)

then we might have intended -Q also, and that expression raised to the 1/2 power becomes ambiguous.

Let me now change formula, swapping the sign in front of the square rooted terms. This allows either of those 1/2 powers to act as if we took the negative branch.

>> formulamod = 84*p^2*t^4 - 112*p^4*t^2 - 64*p^6 + 27*t^6 - (4*p^2 + t^2)^(1/2)*(4*p^2 + 9*t^2)^(1/2)*(16*p^4 - 88*p^2*t^2 + 9*t^4)

formulamod =

84*p^2*t^4 - 112*p^4*t^2 - 64*p^6 + 27*t^6 - (4*p^2 + t^2)^(1/2)*(4*p^2 + 9*t^2)^(1/2)*(16*p^4 - 88*p^2*t^2 + 9*t^4)

Now subs in the solutions.

>> simplify(subs(formulamod,t,solutions))

ans =

-128*p^6

0

0

0

0

0

0

(32768*p^6*(25*13^(1/2) + 86))/729

(32768*p^6*(25*13^(1/2) + 86))/729

So solutions #6 and #7 trivially solve formulamod. Effectively, solutions #6 and #7 solve the problem where EXACTLY one of the terms in formula (where a 1/2 power was taken) follows the negative branch of the square root. So all 9 solutions are technically solutions to the problem, as long as either branch in those squre roots is acceptable.

Does this get to the "root" of the matter? Perhaps. Don't just "woodenly" think a square root is always a positive number. Should I just "leaf" things alone? ;-) (Sorry about that.)

John D'Errico
on 19 Jun 2019

That does not mean your understanding of mathematics is complete.

"In this case, one could define that the square root function does indeed have two branches. I have, however, never heard of such a definition. "

You certainly need to take a complex analysis course if you have never even read about these concepts. At least, if you will be making arguments about mathematics, it helps to have a decent background.

The function y = sqrt(x) is a single valued thing, as it must be, otherwise it is not useful as a function. So we agree that sqrt(x) as a function returns only one value - the principal value.

Anyway, it ia a good idea to test the solutions returned, to verify they are truly meaningful and valid. Spurious solutions can arise. Here, that happens because there is some ambiguity that can creep into an implicit function with those fractional powers. There are multiple places where a square root was apparently necessary in the computation of that solution.

Unfortunately, I cannot trace through the logical path the symbolic toolbox must have taken, to know what was done, and if there was something done that created invalid solutions. All that I can do is as I did. That is, I showed that IF we assume the terms of the form:

(4*p^2 + t^2)^(1/2)

actually have two branches, then solutions #6 and #7 naturally arise, and are solutions. At the same time, they fall into a gray area.

You can feel free to decide they are not valid by your choices, that only the positive branch of the sqrt exists in your world view. That world will be a chaste, limited one.

As for why those solutions disappear when p was constrained to be positive? I'd need to take a guess at that. Constraining p to be a positive number does more than just exclude negative numbers! It also constrains p to be a REAL number. That is just a guess though, without much thought invested.

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