Mixing
When should I resolve significant figures?
$begingroup$
Background:
For a long time I have had some issues with significant figures. More specifically, I never know when to resolve them. In my high-school physics class the examples in the textbook usually resolved the significant figures at the end of an equation.
A real example (this is the last step of a longer problem); this is the equation in the course:
$-w cdot sin(theta) = m cdot a$
$-(290N) cdot sin(23°) = 30.1kg cdot a $
$ a = -3.8 frac{ m} {sec ^2} $
With no rounding in between steps here you get -3.764519.... when rounded -3.8.
If one broke the problem into smaller steps however, like
$-w cdot sin(theta) = m cdot a$
$-(290N) cdot sin(23°) = 30.1kg cdot a $
$ -110N = 30.1kg cdot a$
$ a = frac{-110N}{30.1kg} = -3.7$
Rounding at each step ends up 3.654485... 3.7 when rounded. There is enough difference there to make an answer like this one "wrong" against the textbook.
I personally think it makes more sense to do it like the second example, but which one is right?
significant-figures
$endgroup$
add a comment |
$begingroup$
Background:
For a long time I have had some issues with significant figures. More specifically, I never know when to resolve them. In my high-school physics class the examples in the textbook usually resolved the significant figures at the end of an equation.
A real example (this is the last step of a longer problem); this is the equation in the course:
$-w cdot sin(theta) = m cdot a$
$-(290N) cdot sin(23°) = 30.1kg cdot a $
$ a = -3.8 frac{ m} {sec ^2} $
With no rounding in between steps here you get -3.764519.... when rounded -3.8.
If one broke the problem into smaller steps however, like
$-w cdot sin(theta) = m cdot a$
$-(290N) cdot sin(23°) = 30.1kg cdot a $
$ -110N = 30.1kg cdot a$
$ a = frac{-110N}{30.1kg} = -3.7$
Rounding at each step ends up 3.654485... 3.7 when rounded. There is enough difference there to make an answer like this one "wrong" against the textbook.
I personally think it makes more sense to do it like the second example, but which one is right?
significant-figures
$endgroup$
add a comment |
$begingroup$
Background:
For a long time I have had some issues with significant figures. More specifically, I never know when to resolve them. In my high-school physics class the examples in the textbook usually resolved the significant figures at the end of an equation.
A real example (this is the last step of a longer problem); this is the equation in the course:
$-w cdot sin(theta) = m cdot a$
$-(290N) cdot sin(23°) = 30.1kg cdot a $
$ a = -3.8 frac{ m} {sec ^2} $
With no rounding in between steps here you get -3.764519.... when rounded -3.8.
If one broke the problem into smaller steps however, like
$-w cdot sin(theta) = m cdot a$
$-(290N) cdot sin(23°) = 30.1kg cdot a $
$ -110N = 30.1kg cdot a$
$ a = frac{-110N}{30.1kg} = -3.7$
Rounding at each step ends up 3.654485... 3.7 when rounded. There is enough difference there to make an answer like this one "wrong" against the textbook.
I personally think it makes more sense to do it like the second example, but which one is right?
significant-figures
$endgroup$
Background:
For a long time I have had some issues with significant figures. More specifically, I never know when to resolve them. In my high-school physics class the examples in the textbook usually resolved the significant figures at the end of an equation.
A real example (this is the last step of a longer problem); this is the equation in the course:
$-w cdot sin(theta) = m cdot a$
$-(290N) cdot sin(23°) = 30.1kg cdot a $
$ a = -3.8 frac{ m} {sec ^2} $
With no rounding in between steps here you get -3.764519.... when rounded -3.8.
If one broke the problem into smaller steps however, like
$-w cdot sin(theta) = m cdot a$
$-(290N) cdot sin(23°) = 30.1kg cdot a $
$ -110N = 30.1kg cdot a$
$ a = frac{-110N}{30.1kg} = -3.7$
Rounding at each step ends up 3.654485... 3.7 when rounded. There is enough difference there to make an answer like this one "wrong" against the textbook.
I personally think it makes more sense to do it like the second example, but which one is right?
significant-figures
significant-figures
asked Jan 16 at 13:27
user635303
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2 Answers
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$begingroup$
Do not round until you absolutely have to round at the end. Rounding along the way gives errors that multiply every time you do it, so what you write down at the end can end up very far further from the correct answer than rounding at the end. For an extreme example, consider multiplying $14$ by itself $100$ times, then rounding to 1 significant figure. If we do the rounding at the end, we get an answer of $4times 10^{114}$, very close to the exact answer of $4.100186...times 10^{114}$. If, however, we round at the start, we get $10^{100}$, which is 400 trillion times too small.
answered Jan 16 at 13:36
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
There are two sources of error here - the error due to imprecise measurements of data, and the error due to imprecise arithmetic along the way. We can't do anything about the former, and we'll report the final result to an accuracy based on the data's accuracy, but we can adjust the latter source of error. Specifically, we'd like it to be smaller than the error from imprecise measurements, so we aren't reporting incorrect digits.
Naively, we might say to calculate the intermediate steps without any rounding at all - but that isn't really an option. If nothing else, we'll be limited by the precision of whatever calculator/computer system we use. There are ways to get arbitrarily high precision (usually requiring writing a program), but even those require us to specify just how much precision we want, at a cost in the time it takes to calculate.
So then, what we really want is just enough precision in the calculations that the roundoff error there will be smaller than the error from the measurements. To do that, run all the calculations to one or two more digits of accuracy than the measurements, before rounding at the end. Or, if it's in a calculator or computer, don't bother with extra rounding when ordinary floating-point arithmetic is that good.
answered Jan 16 at 14:13
jmerryjmerry
9,8481225
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2 Answers
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active
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2 Answers
2
active
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active
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$begingroup$
Do not round until you absolutely have to round at the end. Rounding along the way gives errors that multiply every time you do it, so what you write down at the end can end up very far further from the correct answer than rounding at the end. For an extreme example, consider multiplying $14$ by itself $100$ times, then rounding to 1 significant figure. If we do the rounding at the end, we get an answer of $4times 10^{114}$, very close to the exact answer of $4.100186...times 10^{114}$. If, however, we round at the start, we get $10^{100}$, which is 400 trillion times too small.
answered Jan 16 at 13:36
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
Do not round until you absolutely have to round at the end. Rounding along the way gives errors that multiply every time you do it, so what you write down at the end can end up very far further from the correct answer than rounding at the end. For an extreme example, consider multiplying $14$ by itself $100$ times, then rounding to 1 significant figure. If we do the rounding at the end, we get an answer of $4times 10^{114}$, very close to the exact answer of $4.100186...times 10^{114}$. If, however, we round at the start, we get $10^{100}$, which is 400 trillion times too small.
answered Jan 16 at 13:36
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
Do not round until you absolutely have to round at the end. Rounding along the way gives errors that multiply every time you do it, so what you write down at the end can end up very far further from the correct answer than rounding at the end. For an extreme example, consider multiplying $14$ by itself $100$ times, then rounding to 1 significant figure. If we do the rounding at the end, we get an answer of $4times 10^{114}$, very close to the exact answer of $4.100186...times 10^{114}$. If, however, we round at the start, we get $10^{100}$, which is 400 trillion times too small.
answered Jan 16 at 13:36
user3482749user3482749
4,266919
$endgroup$
Do not round until you absolutely have to round at the end. Rounding along the way gives errors that multiply every time you do it, so what you write down at the end can end up very far further from the correct answer than rounding at the end. For an extreme example, consider multiplying $14$ by itself $100$ times, then rounding to 1 significant figure. If we do the rounding at the end, we get an answer of $4times 10^{114}$, very close to the exact answer of $4.100186...times 10^{114}$. If, however, we round at the start, we get $10^{100}$, which is 400 trillion times too small.
answered Jan 16 at 13:36
user3482749user3482749
4,266919
answered Jan 16 at 13:36
user3482749user3482749
4,266919
answered Jan 16 at 13:36
user3482749user3482749
4,266919
answered Jan 16 at 13:36
user3482749user3482749
4,266919
4,266919
add a comment |
add a comment |
$begingroup$
There are two sources of error here - the error due to imprecise measurements of data, and the error due to imprecise arithmetic along the way. We can't do anything about the former, and we'll report the final result to an accuracy based on the data's accuracy, but we can adjust the latter source of error. Specifically, we'd like it to be smaller than the error from imprecise measurements, so we aren't reporting incorrect digits.
Naively, we might say to calculate the intermediate steps without any rounding at all - but that isn't really an option. If nothing else, we'll be limited by the precision of whatever calculator/computer system we use. There are ways to get arbitrarily high precision (usually requiring writing a program), but even those require us to specify just how much precision we want, at a cost in the time it takes to calculate.
So then, what we really want is just enough precision in the calculations that the roundoff error there will be smaller than the error from the measurements. To do that, run all the calculations to one or two more digits of accuracy than the measurements, before rounding at the end. Or, if it's in a calculator or computer, don't bother with extra rounding when ordinary floating-point arithmetic is that good.
answered Jan 16 at 14:13
jmerryjmerry
9,8481225
$endgroup$
add a comment |
$begingroup$
There are two sources of error here - the error due to imprecise measurements of data, and the error due to imprecise arithmetic along the way. We can't do anything about the former, and we'll report the final result to an accuracy based on the data's accuracy, but we can adjust the latter source of error. Specifically, we'd like it to be smaller than the error from imprecise measurements, so we aren't reporting incorrect digits.
Naively, we might say to calculate the intermediate steps without any rounding at all - but that isn't really an option. If nothing else, we'll be limited by the precision of whatever calculator/computer system we use. There are ways to get arbitrarily high precision (usually requiring writing a program), but even those require us to specify just how much precision we want, at a cost in the time it takes to calculate.
So then, what we really want is just enough precision in the calculations that the roundoff error there will be smaller than the error from the measurements. To do that, run all the calculations to one or two more digits of accuracy than the measurements, before rounding at the end. Or, if it's in a calculator or computer, don't bother with extra rounding when ordinary floating-point arithmetic is that good.
answered Jan 16 at 14:13
jmerryjmerry
9,8481225
$endgroup$
add a comment |
$begingroup$
There are two sources of error here - the error due to imprecise measurements of data, and the error due to imprecise arithmetic along the way. We can't do anything about the former, and we'll report the final result to an accuracy based on the data's accuracy, but we can adjust the latter source of error. Specifically, we'd like it to be smaller than the error from imprecise measurements, so we aren't reporting incorrect digits.
Naively, we might say to calculate the intermediate steps without any rounding at all - but that isn't really an option. If nothing else, we'll be limited by the precision of whatever calculator/computer system we use. There are ways to get arbitrarily high precision (usually requiring writing a program), but even those require us to specify just how much precision we want, at a cost in the time it takes to calculate.
So then, what we really want is just enough precision in the calculations that the roundoff error there will be smaller than the error from the measurements. To do that, run all the calculations to one or two more digits of accuracy than the measurements, before rounding at the end. Or, if it's in a calculator or computer, don't bother with extra rounding when ordinary floating-point arithmetic is that good.
answered Jan 16 at 14:13
jmerryjmerry
9,8481225
$endgroup$
There are two sources of error here - the error due to imprecise measurements of data, and the error due to imprecise arithmetic along the way. We can't do anything about the former, and we'll report the final result to an accuracy based on the data's accuracy, but we can adjust the latter source of error. Specifically, we'd like it to be smaller than the error from imprecise measurements, so we aren't reporting incorrect digits.
Naively, we might say to calculate the intermediate steps without any rounding at all - but that isn't really an option. If nothing else, we'll be limited by the precision of whatever calculator/computer system we use. There are ways to get arbitrarily high precision (usually requiring writing a program), but even those require us to specify just how much precision we want, at a cost in the time it takes to calculate.
So then, what we really want is just enough precision in the calculations that the roundoff error there will be smaller than the error from the measurements. To do that, run all the calculations to one or two more digits of accuracy than the measurements, before rounding at the end. Or, if it's in a calculator or computer, don't bother with extra rounding when ordinary floating-point arithmetic is that good.
answered Jan 16 at 14:13
jmerryjmerry
9,8481225
answered Jan 16 at 14:13
jmerryjmerry
9,8481225
answered Jan 16 at 14:13
jmerryjmerry
9,8481225
answered Jan 16 at 14:13
jmerryjmerry
9,8481225
9,8481225
add a comment |
add a comment |
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