6 ms·

Double precision floats have a maximum value of 1.7976931348623158 E + 308

by blurbleblurble 4y ago

onion2k 4y ago
Aka not infinity
- PartiallyTyped 4y ago
  They meant maximum finite value. 0 11111111111 0000000000000000000000000000000000000000000000000000 (base 2) ≙ 7FF0 0000 0000 0000 (base 16) ≙ +∞ (positive infinity) 1 11111111111 0000000000000000000000000000000000000000000000000000 (base 2) ≙ FFF0 0000 0000 0000 (base 16) ≙ −∞ (negative infinity)
thaumasiotes 4y ago
That is not correct; they have a maximum value of positive infinity. See what you get when you square 1.7e+307.
- mjburgess 4y ago
  A computer is always just a function from {0,1}^N -> {0,1}^M The "Inf" interpretation, of, eg., 11111111111111111 isnt infinity. And, in general, almost every function isnt representable on the discrete domain above. floats are a hacky interpretation of discrete bit patterns
  - thaumasiotes 4y ago
    > A computer is always just a function from {0,1}^N -> {0,1}^M > The "Inf" interpretation, of, eg., 11111111111111111 isnt infinity. This is incoherent nonsense. If you want to say that the floating point value "infinity" isn't really infinity, you must also say that nothing else is really infinity either. That is true in a completely useless and uninformative sense, but it's false in every sense a person would ever use. And it fails to distinguish the correct mathematical proof demonstrating an infinite limit from the simulation suggesting, but not demonstrating, an infinite limit. Neither of them is really infinity. Each of them can represent infinity just as accurately as the other, though. A paper is just a function from a bounded countable subset of ℝ² to another bounded countable subset of ℝ². What would you conclude, from that, about the limitations of what you can represent on paper?
    - mjburgess 4y ago
      I'm not sure where your misunderstanding comes from, but at least, you might consider you're disagreeing with an article on quanta magazine which writes up a project by experts in their field. In any case, no. The idea that a finite number of bits in a particular state "must just be infinity!!!! because the IEEE ref docs say so" is strange to say the least. The issue is to demonstrate that a given function, say f, for given real-valued inputs, say x, has an output y which is not real-valued and goes to infinity. A computer cannot demonstrate such a thing, because real-valued functions aren't computable. No sequence of bit patterns on a computer can ever show the above, because the above is an issue of the limits of a continuous function. Relevant inputs to the function have an infinite precision, and relevant outputs have an infinte precision. Eg., x = pi, y = pi^100 A computer may very easily mistakenly conclude x = pi is an input which becomes infinite in y, because (1) a computer cannot represent pi; and (2) cannot represent pi^100 either. infinity isn't a bit pattern; and isn't here in any relevant sense even a number; the IEEE standard may as well have said "Overflow"
      - thaumasiotes 4y ago
        > The issue is to demonstrate that a given function, say f, for given real-valued inputs, say x, has an output y which is not real-valued and goes to infinity. Why would the output need to be not real? There's no difficulty with saying a real-valued function has a singularity. The issue is to demonstrate that this function has a singularity at some point, yes. Simulation is a bad way to do that, though conceivably you could get lucky. > A computer cannot demonstrate such a thing, because real-valued functions aren't computable. Obviously false; computers are fully capable of providing proofs that some function has an infinite limit somewhere. > The idea that a finite number of bits in a particular state "must just be infinity!!!! because the IEEE ref docs say so" is strange to say the least. That is the only way anything is ever infinity - by designation. As I pointed out elsewhere, IEEE infinity has all the correct mathematical properties of positive infinity in the extended reals, so it's difficult to see what you think you're saying. > I'm not sure where your misunderstanding comes from, but at least, you might consider you're disagreeing with an article on quanta magazine which writes up a project by experts in their field. Writing about an expert doesn't make you any smarter. The reason proffered by Quanta is nonsense. They are correct that the experiment they describe cannot achieve the goal sought; they are quite obviously wrong about why. > infinity isn't a bit pattern; and isn't here in any relevant sense even a number; the IEEE standard may as well have said "Overflow" That's what infinity is. In every sense. Overflowing is defined by exceeding a boundary; infinity is defined by exceeding all boundaries. I'm morbidly intrigued by your fetish for the idea of "bit patterns". Infinity is also not an image on paper. How do you expect a correct mathematical proof to represent infinity?
- Karliss 4y ago
  Just because some people decided to label floating point overflow conditiona as "infinity" doesn't mean it can actually represent values up to infinity. There are ways of getting floating point infinity which doesn't involve overflow like dividing by there. But with exception of most trivial cases you have the same problem. You can't know whether you had actual 0 or value closer to 0 than what floating point can represent. All of that of course depends on how floating point unit or calculation environment is configured. It's probably possible to configure it so that overflows/underflows report an error instead of simply returning "inf".
- gilbetron 4y ago
  1.7e+307 squared is 2.89e+614, not infinity even though a "computer" will say +inf
vardump 4y ago
You can combine two doubles to double precision, to 107 bits. https://en.wikipedia.org/wiki/Quadruple-precision_floating-point_format#Double-double_arithmetic https://en.wikipedia.org/wiki/Quadruple-precision_floating-p...

Double precision floats have a maximum value of 1.7976931348623158 E + 308

onion2k 4y ago

Aka not infinity

PartiallyTyped 4y ago
They meant maximum finite value. 0 11111111111 0000000000000000000000000000000000000000000000000000 (base 2) ≙ 7FF0 0000 0000 0000 (base 16) ≙ +∞ (positive infinity) 1 11111111111 0000000000000000000000000000000000000000000000000000 (base 2) ≙ FFF0 0000 0000 0000 (base 16) ≙ −∞ (negative infinity)

thaumasiotes 4y ago

That is not correct; they have a maximum value of positive infinity. See what you get when you square 1.7e+307.

mjburgess 4y ago
A computer is always just a function from {0,1}^N -> {0,1}^M The "Inf" interpretation, of, eg., 11111111111111111 isnt infinity. And, in general, almost every function isnt representable on the discrete domain above. floats are a hacky interpretation of discrete bit patterns
- thaumasiotes 4y ago
  > A computer is always just a function from {0,1}^N -> {0,1}^M > The "Inf" interpretation, of, eg., 11111111111111111 isnt infinity. This is incoherent nonsense. If you want to say that the floating point value "infinity" isn't really infinity, you must also say that nothing else is really infinity either. That is true in a completely useless and uninformative sense, but it's false in every sense a person would ever use. And it fails to distinguish the correct mathematical proof demonstrating an infinite limit from the simulation suggesting, but not demonstrating, an infinite limit. Neither of them is really infinity. Each of them can represent infinity just as accurately as the other, though. A paper is just a function from a bounded countable subset of ℝ² to another bounded countable subset of ℝ². What would you conclude, from that, about the limitations of what you can represent on paper?
  - mjburgess 4y ago
    I'm not sure where your misunderstanding comes from, but at least, you might consider you're disagreeing with an article on quanta magazine which writes up a project by experts in their field. In any case, no. The idea that a finite number of bits in a particular state "must just be infinity!!!! because the IEEE ref docs say so" is strange to say the least. The issue is to demonstrate that a given function, say f, for given real-valued inputs, say x, has an output y which is not real-valued and goes to infinity. A computer cannot demonstrate such a thing, because real-valued functions aren't computable. No sequence of bit patterns on a computer can ever show the above, because the above is an issue of the limits of a continuous function. Relevant inputs to the function have an infinite precision, and relevant outputs have an infinte precision. Eg., x = pi, y = pi^100 A computer may very easily mistakenly conclude x = pi is an input which becomes infinite in y, because (1) a computer cannot represent pi; and (2) cannot represent pi^100 either. infinity isn't a bit pattern; and isn't here in any relevant sense even a number; the IEEE standard may as well have said "Overflow"
    - thaumasiotes 4y ago
      > The issue is to demonstrate that a given function, say f, for given real-valued inputs, say x, has an output y which is not real-valued and goes to infinity. Why would the output need to be not real? There's no difficulty with saying a real-valued function has a singularity. The issue is to demonstrate that this function has a singularity at some point, yes. Simulation is a bad way to do that, though conceivably you could get lucky. > A computer cannot demonstrate such a thing, because real-valued functions aren't computable. Obviously false; computers are fully capable of providing proofs that some function has an infinite limit somewhere. > The idea that a finite number of bits in a particular state "must just be infinity!!!! because the IEEE ref docs say so" is strange to say the least. That is the only way anything is ever infinity - by designation. As I pointed out elsewhere, IEEE infinity has all the correct mathematical properties of positive infinity in the extended reals, so it's difficult to see what you think you're saying. > I'm not sure where your misunderstanding comes from, but at least, you might consider you're disagreeing with an article on quanta magazine which writes up a project by experts in their field. Writing about an expert doesn't make you any smarter. The reason proffered by Quanta is nonsense. They are correct that the experiment they describe cannot achieve the goal sought; they are quite obviously wrong about why. > infinity isn't a bit pattern; and isn't here in any relevant sense even a number; the IEEE standard may as well have said "Overflow" That's what infinity is. In every sense. Overflowing is defined by exceeding a boundary; infinity is defined by exceeding all boundaries. I'm morbidly intrigued by your fetish for the idea of "bit patterns". Infinity is also not an image on paper. How do you expect a correct mathematical proof to represent infinity?
Karliss 4y ago
Just because some people decided to label floating point overflow conditiona as "infinity" doesn't mean it can actually represent values up to infinity. There are ways of getting floating point infinity which doesn't involve overflow like dividing by there. But with exception of most trivial cases you have the same problem. You can't know whether you had actual 0 or value closer to 0 than what floating point can represent. All of that of course depends on how floating point unit or calculation environment is configured. It's probably possible to configure it so that overflows/underflows report an error instead of simply returning "inf".
gilbetron 4y ago
1.7e+307 squared is 2.89e+614, not infinity even though a "computer" will say +inf

vardump 4y ago

You can combine two doubles to double precision, to 107 bits. https://en.wikipedia.org/wiki/Quadruple-precision_floating-point_format#Double-double_arithmetic https://en.wikipedia.org/wiki/Quadruple-precision_floating-p...