Does the Symbolic Toolbox Support 0- and 0+ ?

syms s
syms t real
f = int(exp(-s*t),t,-inf,0)

f = 

Is the second term in the second condition a notation for (though I'm not exactly sure what int is returning for that case)? I'm not sure because doesn't really make sense if s is complex, and if it really means then that condition would make 0 < real(s) unneccesary in the joint condition. If so, is there a way for the user to enter and in expressions, particluarly as one or both bounds of int?

Exploring a bit further, the second term in the second expression is

c = children(f);
c = children(c{2,2});
c(2)

ans = 1×1 cell array
    {[~0 < s]}

Looks like the NOT (~) symbol is involved.

c = c{2}

c = 

Interestingly, the children of c, which looks like an ordinary symbolic expression, can't be obtained

whos c

  Name      Size            Bytes  Class    Attributes

  c         1x1                 8  sym                

children(c)

ans = 1×1 cell array
    {[0 < s]}

I was expecting to be able to do this:

children(0 < s)

ans = 1×2 cell array
    {[0]}    {[s]}

Because it appear that ~0 is a symbolic thing for , I tried this:

nzero = ~sym(0)

nzero = 

But does that mean ?

This integral returns 1/2

int(dirac(t),0,1)

ans = 

Actually, I was shocked when I saw this. This result is different than what I see in 2022a

though I found nothing in the release notes (including bug fixes) to indicate this change in behavior. Having said that, I can kind of see the appeal in the 2023b result, though I'm not sure it's a rigorous result.

Anyway, the next thing I tried is

int(dirac(t),t,~0,1)

ans = 

0

I guess that makes sense because ~0 is logical true, which I guess is converted to numerical (or symbolic) 1 for int. But using nzero results in an error

try
    int(dirac(t),t,nzero,1)
catch ME
    ME.message
end

ans = 'Invalid argument.'

I was hoping this result would be 1.

What exactly is nzero and for what can it be used?

[Paul]

Link to related question

[Paul]

Complete misunderstanding on my part. Looks like it really is a logical NOT applied to the inequality. I guess I got thrown off by no parentheses.

syms s
~(sym(0) < s)

ans = 

Is that the same as?

sym(0) >= s

ans = 

Returning to the original problem

syms s
syms t real
f = int(exp(-s*t),t,-inf,0)

f = 

c = children(f)

c = 3x2 cell array
    {[Inf                              ]}    {[0 < s                                                         ]}
    {[limit(exp(-s*t), t, -Inf)/s - 1/s]}    {[0 < real(s) & ~0 < s                                          ]}
    {[-1/s                             ]}    {[s ~= 0 & angle(-s) in Dom::Interval(-pi/2, pi/2) | real(s) < 0]}

h(s) = c{2,2}

h(s) = 

simplify(h(s),10)

ans = 

Are there any values of s s.t. h(s) is true?

h([-1, 0, 1, 1i])

ans = 

simplify(ans)

ans = 

If not, why is that solution returned from int?

Further confusing things for me is that this expression

~(sym(0) < 5)

ans = 

is displayed without parentheses, suggesting that the SMT uses an order of operations where < is evaluated before ~. But this expression seems to indicate the opposite (though I'm not sure what this epxression means mathematically).

~sym(0) < 5

ans = 

Answer 1

[Walter Roberson]

posZ = sym(0);
negZ = sym(-0);
negZ2 = -sym(0);
1/posZ

ans = 

∞

1/negZ

ans = 

∞

1/negZ2

ans = 

∞

So, no, there is no negative 0 -- if there were then one of the answers would have been -inf

[Paul]

I didn't mean negative zero. I meant which loosely means a very, very small number that's less than 0, often used in Laplace transform stuff (example)