Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: tommya300 on 14/07/2010 21:51:27
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I will try to make this clear direct me if it is not.
In simple scientific solution, the equation contain individual numerical clusters separated by the math functions that will look like e.g. 1234.5 e^3 of defined digits that is 7 digits defining the value. Such as 1,234,500
If a particular constant value is introduced that has the fractional portion, which can extend infinitely longer then 7 digits; e.g. Pi or the square root of 2 or the reciprocal of the sin 45 deg
1.4142135623730950488016887242097
At which of these digits after the decimal point should the correct rounding out respectfully be 5th, 6th or 7th digit long?
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I will try to make this clear direct me if it is not.
In simple scientific solution, the equation contain individual numerical clusters separated by the math functions that will look like e.g. 1234.5 e^3 of defined digits that is 7 digits defining the value. Such as 1,234,500
If a particular constant value is introduced that has the fractional portion, which can extend infinitely longer then 7 digits; e.g. Pi or the square root of 2 or the reciprocal of the sin 45 deg
1.4142135623730950488016887242097
At which of these digits after the decimal point should the correct rounding out respectfully be 5th, 6th or 7th digit long?
Your original expression has 5 significan figures so after you add that to, say, pi, then round that to 5 significan figures as well.
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I will try to make this clear direct me if it is not.
In simple scientific solution, the equation contain individual numerical clusters separated by the math functions that will look like e.g. 1234.5 e^3 of defined digits that is 7 digits defining the value. Such as 1,234,500
If a particular constant value is introduced that has the fractional portion, which can extend infinitely longer then 7 digits; e.g. Pi or the square root of 2 or the reciprocal of the sin 45 deg
1.4142135623730950488016887242097
At which of these digits after the decimal point should the correct rounding out respectfully be 5th, 6th or 7th digit long?
Your original expression has 5 significan figures so after you add that to, say, pi, then round that to 5 significan figures as well.
Thanks for the reply Pmb!
Would I need to round out Pi = 3.14159 before the math manipulation?
Input to a calculator pi button will give a long readout.
Maintaining the rounding out each time through each step of the way??
Let's say I needed to document each step and reproducing the end results after.
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Would I need to round out Pi = 3.14159 before the math manipulation?
You're very welcome, and no. You don't need to round Pi
Rounding is only done at the end of a calculation, not for each step. But keep in mind this assumes that each term used is either exact or has more significan digits than the number input.
This topic is covered in text on numerical analysis. Usually in the first chapter. Would you like me to scan one in and post a link to it?
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Would I need to round out Pi = 3.14159 before the math manipulation?
You're very welcome, and no. You don't need to round Pi
Rounding is only done at the end of a calculation, not for each step. But keep in mind this assumes that each term used is either exact or has more significan digits than the number input.
This topic is covered in text on numerical analysis. Usually in the first chapter. Would you like me to scan one in and post a link to it?
Yes please! There may be other aspects that I may have no knowledge of to ask any questions.
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Yes please! There may be other aspects that I may have no knowledge of to ask any questions.
Okay. I'll do it tomorrow afternoon. Right now I'm off to bed and in the morning I have to go to the hospital. So tomorrow afternoon I'll be able to do that. See you tomorrow. :)
Please PM your e-mail address to me so I can e-mail the scans.
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Would I need to round out Pi = 3.14159 before the math manipulation?
You're very welcome, and no. You don't need to round Pi
It's a matter of definition but, if you want to multiply all of pi by two it will take you an infinitely long time.
You round before, or during the calculation.
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Would I need to round out Pi = 3.14159 before the math manipulation?
You're very welcome, and no. You don't need to round Pi
Rounding is only done at the end of a calculation, not for each step. But keep in mind this assumes that each term used is either exact or has more significan digits than the number input.
This topic is covered in text on numerical analysis. Usually in the first chapter. Would you like me to scan one in and post a link to it?
Yes please! There may be other aspects that I may have no knowledge of to ask any questions.
Sorry but I made a mistake above. Its not a text on numerical analysis that you want. I now recall where I first learned about the process of rounding. It was in a chemistry text. Since I have insomian :( I'm scanning this in now from the chemistry text I have in my bookcase. I'll see if I can post this on my website.
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Okay. I've scanned the text and have uploaded it to my website. See http://home.comcast.net/~peter.m.brown/rounding.pdf
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OK thanks guys
Let's see if I understand.
Rounding out to the most significant digit depends on... e.g.
1.2345 + 3.14159 = 4.37609 is incorrectly rounded?
should this be rounded during the aritmetic process 1.2345 + 3.1416 = 4.3761
or
1.23 + 3.1416 = 4.3716 is incorrectly rounded?
1.23 + 3.14 = 4.37 correctly rounded?
or
(6.3781e6)(1/.7071067) + 6.3781e6 ...needs correction?
rewrite to
(6.3781e6)(1.4142136) + (6.3781e6)= ... is this incorrectly rounding?
(6.3781e6)(1.4142) + (6.3781e6)= ... is this properly corrected rounding?
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Tommy
If you break down your first sum it can demonstrate where problems might arise
1.2345 + 3.14159 = 4.37609
But each of those numbers represents a range - as you have above mentioned
HI 1.234549
LOW 1.234450
hi 3.1415949
low 3.1415850
When you add the extremes of the range you find that the max potential error swamps the last digits you give
HIGH LOW
hi 4.376144 4.376045
low 4.376134 4.376035
your range is approximately 4.376035-4.376144 (of course this itself has uncertainties/errors), you cannot be certain of anything past the 3rd decimal place.
Personally, in all your examples I would try not to multiply out pi or root two (1.4142) until the very last moment in the hope that they will cancel or multiply out, and unless you need a plain number answer you can leave them in the actual answer. Hope that helped a bit - I would also say that the pages Peter put up are very informative and easily digested. Matthew
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Tommy
If you break down your first sum it can demonstrate where problems might arise
1.2345 + 3.14159 = 4.37609
But each of those numbers represents a range - as you have above mentioned
HI 1.234549
LOW 1.234450
hi 3.1415949
low 3.1415850
When you add the extremes of the range you find that the max potential error swamps the last digits you give
HIGH LOW
hi 4.376144 4.376045
low 4.376134 4.376035
your range is approximately 4.376035-4.376144 (of course this itself has uncertainties/errors), you cannot be certain of anything past the 3rd decimal place.
Personally, in all your examples I would try not to multiply out pi or root two (1.4142) until the very last moment in the hope that they will cancel or multiply out, and unless you need a plain number answer you can leave them in the actual answer. Hope that helped a bit - I would also say that the pages Peter put up are very informative and easily digested. Matthew
Helps a bit more Matthew.
Is that why tolerance values are highly observed?
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Is that why tolerance values are highly observed?
I don't think so. Manufacturing tolerances are more to do with cost than anything else.
Computational precision is more about knowing how precise your result needs to be. If you need an answer that's precise to 1% (one part in a hundred), the values that you use in your calculation should probably have a resolution of 0.1% (one part in a thousand).
So, it's really a question of knowing how much resolution (precision) you need in your result, and that very much depends on what you are going to do with the result.
A 1% error rate might not sound too bad, unless you are talking about midwives dropping babies on the floor when they are born.
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Is that why tolerance values are highly observed?
I don't think so. Manufacturing tolerances are more to do with cost than anything else.
Computational precision is more about knowing how precise your result needs to be. If you need an answer that's precise to 1% (one part in a hundred), the values that you use in your calculation should probably have a resolution of 0.1% (one part in a thousand).
So, it's really a question of knowing how much resolution (precision) you need in your result, and that very much depends on what you are going to do with the result.
A 1% error rate might not sound too bad, unless you are talking about midwives dropping babies on the floor when they are born.
Thanks Geezer I think I understand. Is it that the difference in hitting a targets bullseye vs. a manufacturing slip fit?