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Groups > comp.soft-sys.math.mathematica > #1911 > unrolled thread
| Started by | Rafael Dunn <worthless.trash.junk@gmail.com> |
|---|---|
| First post | 2011-04-27 09:39 +0000 |
| Last post | 2011-04-28 10:36 +0000 |
| Articles | 3 — 3 participants |
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Output Precision Exploration Rafael Dunn <worthless.trash.junk@gmail.com> - 2011-04-27 09:39 +0000
Re: Output Precision Exploration Barak Shoshany <baraksh@gmail.com> - 2011-04-28 10:33 +0000
Re: Output Precision Exploration Joseph Gwinn <joegwinn@comcast.net> - 2011-04-28 10:36 +0000
| From | Rafael Dunn <worthless.trash.junk@gmail.com> |
|---|---|
| Date | 2011-04-27 09:39 +0000 |
| Subject | Output Precision Exploration |
| Message-ID | <ip8o8c$piq$1@smc.vnet.net> |
Mathematica 8.0.1.0, Mac OSX x86 In:= Log[173.5/173.5] Out:= -1.11022*10^-16 I expect an output of exactly 0. Although 10^-16 is small, it turned out to be the largest factor in a chemical equation I was attempting to compute. I discovered this is because Mathematica does not actually evaluate 173.5/173.5 = 1. The output is actually some number 0.9999999999... However, for most decimal constants x/x produces an exact output of 1. By entering a few decimals off the top of my head I also found 1733.5, 26.44, and 27.44 do not produce an output of 1 when divided by themselves. Why? I understand Mathematica's algorithms for working with decimals must make approximations, but why is there so much variance among decimal calculations? 173.49/173.49 = 1, while 173.5/173.5 != 1. Furthermore, I find: x=173.49999999999999 x/x = 173.5/173.5, with infinite precision. If you add or remove a single 9 to the end of x, this ceases to be true. Furthermore, this looks like a contradiction to me: In:= 173.5/173.5 = 1 Log[1] = 0 Log[173.5/173.5] = 0 Out:= True True False I have learned a lot about Mathematica's precision and approximation through the help documentation, but I still can not explain this or see how I can expect Log[x/x] = 0 for the sake of calculations on the 10^-16 scale.
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| From | Barak Shoshany <baraksh@gmail.com> |
|---|---|
| Date | 2011-04-28 10:33 +0000 |
| Message-ID | <ipbfou$aac$1@smc.vnet.net> |
| In reply to | #1911 |
On Apr 27, 12:39 pm, Rafael Dunn <worthless.trash.j...@gmail.com> wrote: > Mathematica 8.0.1.0, Mac OSX x86 > > In:= > Log[173.5/173.5] > > Out:= > -1.11022*10^-16 > > I expect an output of exactly 0. Although 10^-16 is small, it turned > out to be the largest factor in a chemical equation I was attempting to > compute. > > I discovered this is because Mathematica does not actually evaluate > 173.5/173.5 = 1. The output is actually some number 0.9999999999... > > However, for most decimal constants x/x produces an exact output of 1. > By entering a few decimals off the top of my head I also found 1733.5, > 26.44, and 27.44 do not produce an output of 1 when divided by > themselves. > > Why? I understand Mathematica's algorithms for working with decimals > must make approximations, but why is there so much variance among > decimal calculations? 173.49/173.49 = 1, while 173.5/173.5 != 1. > Furthermore, I find: > x=173.49999999999999 > x/x = 173.5/173.5, with infinite precision. If you add or remove a > single 9 to the end of x, this ceases to be true. > > Furthermore, this looks like a contradiction to me: > > In:= > 173.5/173.5 = 1 > Log[1] = 0 > Log[173.5/173.5] = 0 > > Out:= > True > True > False > > I have learned a lot about Mathematica's precision and approximation through the help documentation, but I still can not explain this or see how I can expect Log[x/x] = 0 for the sake of calculations on the 10^-16 scale. 173.5 is a machine precision number, not an exact number. So it's only know up to a certain precision. Try this: InputForm[173.5/173.5] 0.9999999999999999 So it's only *approximately* 1. Of course, you could specify the numbers to be of arbitrary precision, for example 100 digits: Log[173.5`100/173.5`100] 0.*10^-100 However, the answer will never be *exactly* 0 because it is only known to a certain precision. This is what Chop was made for: Chop@Log[173.5/173.5] 0 See tutorial/NumericalPrecision in the Mathematica documentation.
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| From | Joseph Gwinn <joegwinn@comcast.net> |
|---|---|
| Date | 2011-04-28 10:36 +0000 |
| Message-ID | <ipbfvs$ag8$1@smc.vnet.net> |
| In reply to | #1911 |
In article <ip8o8c$piq$1@smc.vnet.net>, Rafael Dunn <worthless.trash.junk@gmail.com> wrote: > Mathematica 8.0.1.0, Mac OSX x86 > > In:= > Log[173.5/173.5] > > Out:= > -1.11022*10^-16 > > I expect an output of exactly 0. Although 10^-16 is small, it turned > out to be the largest factor in a chemical equation I was attempting to > compute. > > I discovered this is because Mathematica does not actually evaluate > 173.5/173.5 = 1. The output is actually some number 0.9999999999... > > However, for most decimal constants x/x produces an exact output of 1. > By entering a few decimals off the top of my head I also found 1733.5, > 26.44, and 27.44 do not produce an output of 1 when divided by > themselves. > > Why? I understand Mathematica's algorithms for working with decimals > must make approximations, but why is there so much variance among > decimal calculations? 173.49/173.49 = 1, while 173.5/173.5 != 1. > Furthermore, I find: > x=173.49999999999999 > x/x = 173.5/173.5, with infinite precision. If you add or remove a > single 9 to the end of x, this ceases to be true. > > Furthermore, this looks like a contradiction to me: > > In:= > 173.5/173.5 = 1 > Log[1] = 0 > Log[173.5/173.5] = 0 > > Out:= > True > True > False > > I have learned a lot about Mathematica's precision and approximation through > the help documentation, but I still can not explain this or see how I can > expect Log[x/x] = 0 for the sake of calculations on the 10^-16 scale. The computations are performed using 64-bit double precision floating point numbers, as defined in IEEE Std 754. This is by definition a finite-precision computation, and errors of order 10^-16 are to be expected, and cannot be removed unless one goes to Mathematica's arbitrary precision arithmetic, which is orders of magnitude slower to compute. You may wish to reformulate your problem. With an explanation of what you are trying to solve and why, people will be able to suggest alternatives. Joe Gwinn
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