Mixing
How can I be sure that a partial sum is accurate to some number of decimal places?
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In a homework assignment, I am asked how many terms of a series are needed to obtain four decimal places (chopped) of accuracy.
There are certain tricks that I am supposed to use, where I can determine an upper limit of the error and select the number of terms to make it small enough. However, it seems to me like there are some special cases where this will not work.
If the number that the series approaches is, for example, $0.123400005$, the error term being less than $10^{-4}$ is not enough to guarantee that the first four digits are correct. If a finite sum gives $0.12339999$, the error is $1.5*10^{-8}$, which is way more accurate than my textbook tells me I need. But, the first four digits are not correct.
So, if I am not given what the number that the series converges to is, how can I be sure that this kind of scenario will not happen?
sequences-and-series truncation-error
asked Jan 24 at 22:44
PolygonPolygon
628211
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add a comment |
$begingroup$
In a homework assignment, I am asked how many terms of a series are needed to obtain four decimal places (chopped) of accuracy.
There are certain tricks that I am supposed to use, where I can determine an upper limit of the error and select the number of terms to make it small enough. However, it seems to me like there are some special cases where this will not work.
If the number that the series approaches is, for example, $0.123400005$, the error term being less than $10^{-4}$ is not enough to guarantee that the first four digits are correct. If a finite sum gives $0.12339999$, the error is $1.5*10^{-8}$, which is way more accurate than my textbook tells me I need. But, the first four digits are not correct.
So, if I am not given what the number that the series converges to is, how can I be sure that this kind of scenario will not happen?
sequences-and-series truncation-error
asked Jan 24 at 22:44
PolygonPolygon
628211
$endgroup$
1
$begingroup$
The problem may in your use of the word "chopped". If you're just going to truncate digits, then the problem you demonstrate can happen. But usually, "4 decimal places" means "rounded", not "chopped". Both your numbers round to 4 correct decimal places.
$endgroup$
– B. Goddard
Jan 24 at 22:49
1
$begingroup$
Further to @B.Goddard's comment: there is a difference between "chopping" the number and "chopping" the series. Generally I think one would "chop" the series, so one is left with a finite computation, then round the value that gives.
$endgroup$
– Will R
Jan 24 at 23:43
$begingroup$
The problem point isn:t $1.234000005$. It's $1.234999999 $ (round down to $1.23$) versus $1.235000000$ (round up to $1.24$).
$endgroup$
– timtfj
Jan 25 at 0:14
add a comment |
$begingroup$
In a homework assignment, I am asked how many terms of a series are needed to obtain four decimal places (chopped) of accuracy.
There are certain tricks that I am supposed to use, where I can determine an upper limit of the error and select the number of terms to make it small enough. However, it seems to me like there are some special cases where this will not work.
If the number that the series approaches is, for example, $0.123400005$, the error term being less than $10^{-4}$ is not enough to guarantee that the first four digits are correct. If a finite sum gives $0.12339999$, the error is $1.5*10^{-8}$, which is way more accurate than my textbook tells me I need. But, the first four digits are not correct.
So, if I am not given what the number that the series converges to is, how can I be sure that this kind of scenario will not happen?
sequences-and-series truncation-error
asked Jan 24 at 22:44
PolygonPolygon
628211
$endgroup$
In a homework assignment, I am asked how many terms of a series are needed to obtain four decimal places (chopped) of accuracy.
There are certain tricks that I am supposed to use, where I can determine an upper limit of the error and select the number of terms to make it small enough. However, it seems to me like there are some special cases where this will not work.
If the number that the series approaches is, for example, $0.123400005$, the error term being less than $10^{-4}$ is not enough to guarantee that the first four digits are correct. If a finite sum gives $0.12339999$, the error is $1.5*10^{-8}$, which is way more accurate than my textbook tells me I need. But, the first four digits are not correct.
So, if I am not given what the number that the series converges to is, how can I be sure that this kind of scenario will not happen?
sequences-and-series truncation-error
sequences-and-series truncation-error
asked Jan 24 at 22:44
PolygonPolygon
628211
asked Jan 24 at 22:44
PolygonPolygon
628211
asked Jan 24 at 22:44
PolygonPolygon
628211
asked Jan 24 at 22:44
PolygonPolygon
628211
asked Jan 24 at 22:44
PolygonPolygon
628211
628211
1
$begingroup$
The problem may in your use of the word "chopped". If you're just going to truncate digits, then the problem you demonstrate can happen. But usually, "4 decimal places" means "rounded", not "chopped". Both your numbers round to 4 correct decimal places.
$endgroup$
– B. Goddard
Jan 24 at 22:49
1
$begingroup$
Further to @B.Goddard's comment: there is a difference between "chopping" the number and "chopping" the series. Generally I think one would "chop" the series, so one is left with a finite computation, then round the value that gives.
$endgroup$
– Will R
Jan 24 at 23:43
$begingroup$
The problem point isn:t $1.234000005$. It's $1.234999999 $ (round down to $1.23$) versus $1.235000000$ (round up to $1.24$).
$endgroup$
– timtfj
Jan 25 at 0:14
add a comment |
1
$begingroup$
The problem may in your use of the word "chopped". If you're just going to truncate digits, then the problem you demonstrate can happen. But usually, "4 decimal places" means "rounded", not "chopped". Both your numbers round to 4 correct decimal places.
$endgroup$
– B. Goddard
Jan 24 at 22:49
1
$begingroup$
Further to @B.Goddard's comment: there is a difference between "chopping" the number and "chopping" the series. Generally I think one would "chop" the series, so one is left with a finite computation, then round the value that gives.
$endgroup$
– Will R
Jan 24 at 23:43
$begingroup$
The problem point isn:t $1.234000005$. It's $1.234999999 $ (round down to $1.23$) versus $1.235000000$ (round up to $1.24$).
$endgroup$
– timtfj
Jan 25 at 0:14
1
1
$begingroup$
The problem may in your use of the word "chopped". If you're just going to truncate digits, then the problem you demonstrate can happen. But usually, "4 decimal places" means "rounded", not "chopped". Both your numbers round to 4 correct decimal places.
$endgroup$
– B. Goddard
Jan 24 at 22:49
$begingroup$
The problem may in your use of the word "chopped". If you're just going to truncate digits, then the problem you demonstrate can happen. But usually, "4 decimal places" means "rounded", not "chopped". Both your numbers round to 4 correct decimal places.
$endgroup$
– B. Goddard
Jan 24 at 22:49
1
1
$begingroup$
Further to @B.Goddard's comment: there is a difference between "chopping" the number and "chopping" the series. Generally I think one would "chop" the series, so one is left with a finite computation, then round the value that gives.
$endgroup$
– Will R
Jan 24 at 23:43
$begingroup$
Further to @B.Goddard's comment: there is a difference between "chopping" the number and "chopping" the series. Generally I think one would "chop" the series, so one is left with a finite computation, then round the value that gives.
$endgroup$
– Will R
Jan 24 at 23:43
$begingroup$
The problem point isn:t $1.234000005$. It's $1.234999999 $ (round down to $1.23$) versus $1.235000000$ (round up to $1.24$).
$endgroup$
– timtfj
Jan 25 at 0:14
$begingroup$
The problem point isn:t $1.234000005$. It's $1.234999999 $ (round down to $1.23$) versus $1.235000000$ (round up to $1.24$).
$endgroup$
– timtfj
Jan 25 at 0:14
add a comment |
1 Answer
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I think the way to handle this is to treat "accurate to $n$ decimal places" as meaning "differing from the correct value by no more than $0.5 × 10^{-n}$", or "having an error of no more than $5$ in the $n+1$th decimal place".
Otherwise there will be tuncation/rounding issues at certain points. For example, consider $1.234999999$ and $1.235000000$. If we round them to two decimal places, we get 1.23 and 1.24. If we truncate them to three decimal places, we get $1.234$ and $1.235$.
So you're right that in some instances, getting the specified number of correct digits can need much greater precision in the calculation. I don't see that you can guarantee it doesn't happen.
So I say the correct course of action is to define a maximum permitted error, and say you're using that as the criterion.
answered Jan 25 at 0:45
timtfjtimtfj
2,468420
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add a comment |
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$begingroup$
I think the way to handle this is to treat "accurate to $n$ decimal places" as meaning "differing from the correct value by no more than $0.5 × 10^{-n}$", or "having an error of no more than $5$ in the $n+1$th decimal place".
Otherwise there will be tuncation/rounding issues at certain points. For example, consider $1.234999999$ and $1.235000000$. If we round them to two decimal places, we get 1.23 and 1.24. If we truncate them to three decimal places, we get $1.234$ and $1.235$.
So you're right that in some instances, getting the specified number of correct digits can need much greater precision in the calculation. I don't see that you can guarantee it doesn't happen.
So I say the correct course of action is to define a maximum permitted error, and say you're using that as the criterion.
answered Jan 25 at 0:45
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
I think the way to handle this is to treat "accurate to $n$ decimal places" as meaning "differing from the correct value by no more than $0.5 × 10^{-n}$", or "having an error of no more than $5$ in the $n+1$th decimal place".
Otherwise there will be tuncation/rounding issues at certain points. For example, consider $1.234999999$ and $1.235000000$. If we round them to two decimal places, we get 1.23 and 1.24. If we truncate them to three decimal places, we get $1.234$ and $1.235$.
So you're right that in some instances, getting the specified number of correct digits can need much greater precision in the calculation. I don't see that you can guarantee it doesn't happen.
So I say the correct course of action is to define a maximum permitted error, and say you're using that as the criterion.
answered Jan 25 at 0:45
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
I think the way to handle this is to treat "accurate to $n$ decimal places" as meaning "differing from the correct value by no more than $0.5 × 10^{-n}$", or "having an error of no more than $5$ in the $n+1$th decimal place".
Otherwise there will be tuncation/rounding issues at certain points. For example, consider $1.234999999$ and $1.235000000$. If we round them to two decimal places, we get 1.23 and 1.24. If we truncate them to three decimal places, we get $1.234$ and $1.235$.
So you're right that in some instances, getting the specified number of correct digits can need much greater precision in the calculation. I don't see that you can guarantee it doesn't happen.
So I say the correct course of action is to define a maximum permitted error, and say you're using that as the criterion.
answered Jan 25 at 0:45
timtfjtimtfj
2,468420
$endgroup$
I think the way to handle this is to treat "accurate to $n$ decimal places" as meaning "differing from the correct value by no more than $0.5 × 10^{-n}$", or "having an error of no more than $5$ in the $n+1$th decimal place".
Otherwise there will be tuncation/rounding issues at certain points. For example, consider $1.234999999$ and $1.235000000$. If we round them to two decimal places, we get 1.23 and 1.24. If we truncate them to three decimal places, we get $1.234$ and $1.235$.
So you're right that in some instances, getting the specified number of correct digits can need much greater precision in the calculation. I don't see that you can guarantee it doesn't happen.
So I say the correct course of action is to define a maximum permitted error, and say you're using that as the criterion.
answered Jan 25 at 0:45
timtfjtimtfj
2,468420
edited Jan 25 at 0:56
answered Jan 25 at 0:45
timtfjtimtfj
2,468420
answered Jan 25 at 0:45
timtfjtimtfj
2,468420
answered Jan 25 at 0:45
timtfjtimtfj
2,468420
2,468420
add a comment |
add a comment |
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$begingroup$
The problem may in your use of the word "chopped". If you're just going to truncate digits, then the problem you demonstrate can happen. But usually, "4 decimal places" means "rounded", not "chopped". Both your numbers round to 4 correct decimal places.
$endgroup$
– B. Goddard
Jan 24 at 22:49
1
$begingroup$
Further to @B.Goddard's comment: there is a difference between "chopping" the number and "chopping" the series. Generally I think one would "chop" the series, so one is left with a finite computation, then round the value that gives.
$endgroup$
– Will R
Jan 24 at 23:43
$begingroup$
The problem point isn:t $1.234000005$. It's $1.234999999 $ (round down to $1.23$) versus $1.235000000$ (round up to $1.24$).
$endgroup$
– timtfj
Jan 25 at 0:14