A bug? input: 1/0 output: -Inf

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qilin
qilin am 13 Feb. 2024
Kommentiert: Matt J am 13 Feb. 2024
if x=0, expected output of 1/x is Inf, but matlab output -Inf
load('x.mat')
x==0
Unrecognized function or variable 'x'.
1/0
1/x
  2 Kommentare
Matt J
Matt J am 13 Feb. 2024
Your attached file doesn't contain a variable called x.
Image Analyst
Image Analyst am 13 Feb. 2024
x==0 is a comparison, not an assignment. Since you didn't assign x yet, you can't see if it's equal to zero, hence the error. I think you meant x=0, not x==0.

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Stephen23
Stephen23 am 13 Feb. 2024
Bearbeitet: Stephen23 am 13 Feb. 2024
"A bug? input: 1/0 output: -Inf"
Not a bug. Because your value x is actually negative zero not positive zero:
Lets try it now:
x = -0
x = 0
1/x
ans = -Inf
num2hex(x) % yep, definitely negative zero
ans = '8000000000000000'
x = +0
x = 0
1/x
ans = Inf
num2hex(x) % yep, that is positive zero
ans = '0000000000000000'
So far everything is working exactly as expected. Note that a few simple arithmetic operations will convert negative zero to positive zero without affecting any other values, e.g. adding zero:
x = -0;
1/x
ans = -Inf
x = 0+x;
1/x
ans = Inf
  4 Kommentare
Dyuman Joshi
Dyuman Joshi am 13 Feb. 2024
How did you obtain that value?
Stephen23
Stephen23 am 13 Feb. 2024
"maybe the next step is to figure out why my ''deltaX'' is negative zero"
Most likely it makes zero(!) difference: note that by definition negative and positive zeros have the same value, so if you are happy with your array being zero-value then simply multiply your array by one (or add zero) and move on to more important tasks.

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Matt J
Matt J am 13 Feb. 2024
Bearbeitet: Matt J am 13 Feb. 2024
Theorem: 1/0 = Inf if and only if 1/0 is also -Inf.
Proof: Assume first that 1/0=Inf and let us deduce the implication that 1/0=-Inf. By multiplying the numerator and denominator by -1, we obtain,
1/0 = (-1)/(-0) = (-1)/0 = -(1/0) = -Inf
proving the implication Now assume that 1/0=-Inf. Proceeding similarly,
1/0 = (-1)/(-0) = (-1)/0 = -(1/0) = -(-Inf) = +Inf
and the reverse implication is also proved. Q.E.D.
  2 Kommentare
qilin
qilin am 13 Feb. 2024
Thank you for your answer, it's interesting, but: .
In other words, when we say , actually it's , here , so the second
''='' of ''1/0 = (-1)/(-0) =...'' should be modified.
Matt J
Matt J am 13 Feb. 2024
You took it way too seriously...

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