|
7
|
|
edited Dec 10 2010 at 19:07 |
Similarly, every numerical simulation in physics (or chemistry, biology, physiology, or medicine) always has to use finite precision representation of values, such that there is a limit to the largest and smallest integer represented by a fixed number of bits, and such that there is a limited amount of "floating-point-precision" available in dividing the bits of a floating-point representation of a real number into a set number of bits for the mantissa and a set number of bits for the exponent. For example, assuming that $d=64$-bits are used to represent "real numbers" as floating point numbers in computations, $m=48$ bits may be allocated to the mantissa, allowing the numerator to be $2^{48}$ yielding approximately $14$ digits of base-ten specificity to the numerator; this leaves $d-m=16$ bits to the exponent which may be signed (+/-) yielding a range of -32768 to +32767. In this case, the floating point number is in the range $n\times 2^{d-m}$, where $-(2^{47} \le n \le +(2^{47}-1)$, and ${-32768} \le d \le {32768}$.
|
|
|
|
|
6
|
|
edited Dec 10 2010 at 19:01 |
As Colin Tan said, "[using] only countably many symbols, then there will always be a countable models." Whereas the field $\mathbb{Q}$ of rational numbers uses a finite-number of symbols for an uncountable number of rational numbers. I misunderstood and misapplied a concept.
The field generated by rational numbers is quite different from the "approximation space" rendered by using a finite number of bits interpreted as a floating-point number low-precision approximation to real numbers. I'm editing my answer to point out my misunderstanding. @Hans-Stricker, I've fixed my error by pointing it out, but leaving it up (below the ruled line) so that some other bit-flipper like me will see why {0,1}$^n \times${0,1}$^n$ is not equivalent to $\mathbb{Q}$
below this is my original (erroneous) answer
If the total number of bits is $d$, the number of bits allocated to the exponent, $m$, may be decreased while simultaneously increasing the number of bits, $d-m$, allocated to the mantissa, increasing the "precision" of the numerator while decreasing the range over $\mathbb{R}$ spanned by this particular approximating set of {0,1}$^m \times ${0,1}$^{d-m}$ (which is not equivalent to $\mathbb{Q}$, as I erroneously stated originally)
Thus every numerical simulation is already, in a way, based on $\mathbb{Q}^d$ when models of $d$-dimensional systems are created and iterated using Euler or Runge-Kutta of whatever order.
|
|
|
|
|
5
|
|
edited Dec 10 2010 at 19:01 |
As Colin Tan said, "[using] only countably many symbols, then there will always be a countable models." Whereas the field $\mathbb{Q}$ of rational numbers uses a finite-number of symbols for an uncountable number of rational numbers. I misunderstood and misapplied a concept.
The field generated by rational numbers is quite different from the "approximation space" rendered by using a finite number of bits interpreted as a floating-point number low-precision approximation to real numbers. I'm editing my answer to point out my misunderstanding. @Hans-Stricker, I've fixed my error by pointing it out, but leaving it up (below the ruled line) so that some other bit-flipper like me will see why {0,1}$^n \times${0,1}$^n$ is not equivalent to $\mathbb{Q}$
below this is my original (erroneous) answer
If the total number of bits is $d$, the number of bits allocated to the exponent, $m$, may be decreased while simultaneously increasing the number of bits, $d-m$, allocated to the mantissa, increasing the "precision" of the numerator while decreasing the range over $\mathbb{R}$ spanned by this particular approximating set of {0,1}$^m \times ${0,1}$^{d-m}$ (which is not equivalent to $\mathbb{Q}$, as I erroneously stated originally)
Thus every numerical simulation is already, in a way, based on $\mathbb{Q}^d$ when models of $d$-dimensional systems are created and iterated using Euler or Runge-Kutta of whatever order.
|
|
|
|
4
|
|
edited Dec 10 2010 at 18:54 |
Similarly, every numerical simulation in physics (or chemistry, biology, physiology, or medicine) always has to use finite precision representation of values, such that there is a limit to If the largest and smallest integer represented by a fixed total number of bits , and such that there is a limited amount of "floating-point-precision" available in dividing the bits of a floating-point representation of a real number into a set number of bits for $d$, the mantissa and a set number of bits for the exponent. For example, assuming that $64$-bits are used to represent "real numbers" as floating point numbers in computations, $48$ bits may be allocated to the mantissa, allowing the numerator to be $2^{48}$ yielding approximately $14$ digits of base-ten specificity to the numerator; this leaves $16$ bits to the exponentwhich may be signed (+/-) yielding a range of -32768 to +32767. In this case, the floating point number is in the range $n\times 2^d$, where $-(2^{47} \le n \le +(2^{47}-1)$, and ${-32768} \le d \le {32768}$. The number of bits allocated to the exponent m$, may be decreased while simultaneously increasing the number of bits, $d-m$, allocated to the mantissa, increasing the "precision" of the numerator while decreasing the range over $\mathbb{R}$ spanned by this particular approximating set of {0,1}$^m \times $\mathbb{Q}$.{0,1}$^{d-m}$ (which is not equivalent to $\mathbb{Q}$, as I erroneously stated originally)
|
|
|
|
|
3
|
|
edited Dec 10 2010 at 18:46 |
As Colin Tan said, "[using] only countably many symbols, then there will always be a countable models." Whereas the field $\mathbb{Q}$ of rational numbers uses a finite-number of symbols for an uncountable number of rational numbers. I misunderstood and misapplied a concept.
The field generated by rational numbers is quite different from the "approximation space" rendered by using a finite number of bits interpreted as a floating-point number low-precision approximation to real numbers. I'm editing my answer to point out my misunderstanding. @Hans-Stricker, I've fixed my error by pointing it out, but leaving it up (below the ruled line) so that some other bit-flipper like me will see why {0,1}$^n \times${0,1}$^n$ is not equivalent to $\mathbb{Q}$
below this is my original (erroneous) answer
Similarly, every numerical simulation in physics (or chemistry, biology, physiology, or medicine) always has to use finite precision representation of values, such that there is a limit to the largest and smallest integer represented by a fixed number of bits, and such that there is a limited amount of "floating-point-precision" available in dividing the bits of a floating-point representation of a real number into a set number of bits for the mantissa and a set number of bits for the exponent.
For example, assuming that $64$-bits are used to represent "real numbers" as floating point numbers in computations, $48$ bits may be allocated to the mantissa, allowing the numerator to be $2^{48}$ yielding approximately $14$ digits of base-ten specificity to the numerator; this leaves $16$ bits to the exponent which may be signed (+/-) yielding a range of -32768 to +32767.
In this case, the floating point number is in the range $n\times 2^d$, where $-(2^{47} \le n \le +(2^{47}-1)$, and ${-32768} \le d \le {32768}$.
The number of bits allocated to the exponent may be decreased while simultaneously increasing the number of bits allocated to the mantissa, increasing the "precision" of the numerator while decreasing the range over $\mathbb{R}$ spanned by this particular set of $\mathbb{Q}$.
Thus every numerical simulation is already, in a way, based on $\mathbb{Q}^d$ when models of $d$-dimensional systems are created and iterated using Euler or Runge-Kutta of whatever order.
|
|
|
|
|
2
|
|
edited Dec 10 2010 at 13:24 |
As Colin Tan said, "[using] only countably many symbols, then there will always be a countable models." Similarly, every numerical simulation in physics (or chemistry, biology, physiology, or medicine) always has to use finite precision representation of values, such that there is a limit to the largest and smallest integer represented by a fixed number of bits, and such that there is a limited amount of "floating-point-precision" available in dividing the bits of a floating-point representation of a real number into a set number of bits for the mantissa and a set number of bits for the exponent.
For example, assuming that $64$-bits are used to represent "real numbers" as floating point numbers in computations, $48$ bits may be allocated to the mantissa, allowing the numerator to be $2^{48}$ yielding approximately $14$ digits of base-ten specificity to the numerator; this leaves $16$ bits to the exponent which may be signed (+/-) yielding a range of -32768 to +32767.
In this case, the floating point number is in the range $n\times 2^d$, where $-(2^{47} \le n \le +(2^{47}-1)$, and ${-32768} \le d \le {32768}$.
The number of bits allocated to the exponent may be decreased while simultaneously increasing the number of bits allocated to the mantissa, increasing the "precision" of the numerator while decreasing the range over $\mathbb{R}$ spanned by this particular set of $\mathbb{Q}$.
Thus every numerical simulation is already, in a way, based on $\mathbb{Q}^d$ when models of $d$-dimensional systems are created and iterated using Euler or Runge-Kutta of whatever order.
|
|
|
|
|
1
|
|
answered Dec 10 2010 at 13:15 |
As Colin Tan said, "[using] only countably many symbols, then there will always be a countable models." Similarly, every numerical simulation in physics (or chemistry, biology, physiology, or medicine) always has to use finite precision representation of values, such that there is a limit to the largest and smallest integer represented by a fixed number of bits, and such that there is a limited amount of "floating-point-precision" available in dividing the bits of a floating-point representation of a real number into a set number of bits for the mantissa and a set number of bits for the exponent.
Thus every numerical simulation is already, in a way, based on $\mathbb{Q}^d$ when models of $d$-dimensional systems are created and iterated using Euler or Runge-Kutta of whatever order.
|
|
|
