Using Matlab symbolic toolkit to get the positive root

Using the Matlab symbolic toolkit, consider a square root formula like the following:

syms c; assume(c,'real'); assumeAlso(c > 0);

syms d; assume(d,'real'); assumeAlso(d > 0);

N = (d^2 + 8*c)^0.5 - d

So

And I would like to show if this is always positve.

isAlways(N > 0)

? No, this is ambiguous according to Matlab given these assumptions.

Matlab points out that this is ambiguous. And it is, because the root could be positive or negative. If d were 9 and c were 19/8, then the solutions are: -10 - 9 = -19 and +10-9 = 1.

Perhaps I can solve a problem by taking only the positive root? If I take the absolute value of the root, this doesn't fix the problem:

N2 = abs((d^2 + 8*c)^0.5) - d

isAlways(N2 > 0)

The result is still ambiguous. Therefore maybe the root is already positive and I'm misunderstanding what the problem is. If I ask:

isAlways((d^2 + 8*c)^0.5 > 0)

This is always true, so maybe I'm already getting the positive root. But if so, why isn't N2 always greater than 0? Given that c and d are assumed to be real numbers and positive, doesn't d^2 + 8*c have to be greater than d^2? And since the square root is a strictly increasing function, doesn't this imply that sqrt(d^2 + 8*c ) > d?

What I can't do is assume that that . In my more complex non-minimum-working example setting, what's inside the square root is changing but always greater than d^2 (so I want to show , where d itself is potentially a function of multiple inputs but positive and real).

For what it is worth, when I set this up in Mathematica (Mathematica, not Matlab code below), I don't get an ambiguous result, I get that N is always larger than 0.

cond3 = c > 0 && c \[Element] Reals && d > 0 && d \[Element] Reals

Numer = (d^2 + 8*c)^0.5 - d

inequality = ForAll[{b, c}, cond3, Numer > 0]

I fear that I am missing something very basic. Thank you for any help you can provide.

Answer 1

[Gayatri Rathod]

Hi Benjamin,

• The reason why isAlways(N2 > 0) is ambiguous because the absolute value function makes it possible for the expression to be zero or negative even when the argument of the square root is positive. For example, when d = 1 and c = 1/8, we have N2 = 0, which is not greater than zero.
• As for the original expression N = (d^2 + 8*c)^0.5 - d, it is not always positive because the square root could be negative.
• If you want to be absolutely sure that N is always positive, you could try using the simplify function in MATLAB.

                                    N = (d^2 + 8*c)^0.5 - d 
                                    isAlways(simplify(N > 0)) 

• The simplify function simplifies the expression inside the square root and eliminates any negative terms.

You can read more about the simplify and isAlways functions from the following documentations: simplify Function, isAlways Function.

Hope it helps!

Regards,

Gayatri Rathod