Relative roundoff error in the simple precision.
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I am struggling with a simple problem from the computer arithmetics. The goal is to find $|(fl(0.1)-0.1)/0.1|$. I have computed the binary representation of $0.1$, which is $(1.1001)_{2}times2^{-4}$, where the $1001$ after the decimal point is periodic. I know that $fl(0.1)$ is rounding the $0.1$ up, so we get $(1.1001...101)_{2}times2^{-4}$, where the last digit is on the $-23$rd position. But I don't know how to compute the relative error. Thank you for any help.
numerical-methods computer-arithmetic
asked Jan 4 at 20:16
VwannVwann
225
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add a comment |
$begingroup$
I am struggling with a simple problem from the computer arithmetics. The goal is to find $|(fl(0.1)-0.1)/0.1|$. I have computed the binary representation of $0.1$, which is $(1.1001)_{2}times2^{-4}$, where the $1001$ after the decimal point is periodic. I know that $fl(0.1)$ is rounding the $0.1$ up, so we get $(1.1001...101)_{2}times2^{-4}$, where the last digit is on the $-23$rd position. But I don't know how to compute the relative error. Thank you for any help.
numerical-methods computer-arithmetic
asked Jan 4 at 20:16
VwannVwann
225
$endgroup$
$begingroup$
Yes, thank you, I edited
$endgroup$
– Vwann
Jan 4 at 20:35
add a comment |
$begingroup$
I am struggling with a simple problem from the computer arithmetics. The goal is to find $|(fl(0.1)-0.1)/0.1|$. I have computed the binary representation of $0.1$, which is $(1.1001)_{2}times2^{-4}$, where the $1001$ after the decimal point is periodic. I know that $fl(0.1)$ is rounding the $0.1$ up, so we get $(1.1001...101)_{2}times2^{-4}$, where the last digit is on the $-23$rd position. But I don't know how to compute the relative error. Thank you for any help.
numerical-methods computer-arithmetic
asked Jan 4 at 20:16
VwannVwann
225
$endgroup$
I am struggling with a simple problem from the computer arithmetics. The goal is to find $|(fl(0.1)-0.1)/0.1|$. I have computed the binary representation of $0.1$, which is $(1.1001)_{2}times2^{-4}$, where the $1001$ after the decimal point is periodic. I know that $fl(0.1)$ is rounding the $0.1$ up, so we get $(1.1001...101)_{2}times2^{-4}$, where the last digit is on the $-23$rd position. But I don't know how to compute the relative error. Thank you for any help.
numerical-methods computer-arithmetic
numerical-methods computer-arithmetic
asked Jan 4 at 20:16
VwannVwann
225
asked Jan 4 at 20:16
VwannVwann
225
edited Jan 4 at 20:35
Vwann
asked Jan 4 at 20:16
VwannVwann
225
asked Jan 4 at 20:16
VwannVwann
225
asked Jan 4 at 20:16
VwannVwann
225
225
$begingroup$
Yes, thank you, I edited
$endgroup$
– Vwann
Jan 4 at 20:35
add a comment |
$begingroup$
Yes, thank you, I edited
$endgroup$
– Vwann
Jan 4 at 20:35
$begingroup$
Yes, thank you, I edited
$endgroup$
– Vwann
Jan 4 at 20:35
$begingroup$
Yes, thank you, I edited
$endgroup$
– Vwann
Jan 4 at 20:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just plug into the formula. You have $fl(0.1)$ and $0.1$ so now subtract. Starting from the last position in your $fl(0.1)$ the true value continues $011 001 100ldots$ so if we subtract we get (about) $001$ with the first $0$ matching the last $1$ in $fl(0.1)$. That is $22$ places to the right of the radix point, so the difference is $2^{-4}cdot 2^{-22}cdot 2^{-2}=2^{-28}$ where the first factor is from the exponent of $fl(0.1)$, the second is from all the places to the right of the radix point, and the third is the two places the $1$ in the error is to the right of the end of $fl(0.1)$. The relative error is then $frac {2^{-28}}{0.1}approx 3.7cdot 10^{-8}$
If you want the exact answer, you need the exact value of $fl(0.1)$. I asked Alpha and got $frac {6710885}{16cdot4194304}$. Then the relative error is $10(frac {6710885}{16cdot4194304}-frac 1{10})=-frac 7{33554432}$
answered Jan 4 at 20:42
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
I have a slightly different answer
$endgroup$
– Vwann
Jan 4 at 21:16
$begingroup$
For the absolute error I get 1/10×2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:17
$begingroup$
And relative is therefore 2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:18
$begingroup$
My computation of $fl(0.1)-0.1$ was not exact. Usually we are only interested in the magnitude of errors like this. $2^{-26}$ is not correct because that is the whole value of the last $1$ in your expression, but $0.1$ is greater than your expression with the last $1$ changed to a $0$.
$endgroup$
– Ross Millikan
Jan 4 at 21:18
$begingroup$
Well, not in this case. The question is about the precise value. Thank you anyway
$endgroup$
– Vwann
Jan 4 at 21:19
|
show 6 more comments
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1 Answer
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1 Answer
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$begingroup$
Just plug into the formula. You have $fl(0.1)$ and $0.1$ so now subtract. Starting from the last position in your $fl(0.1)$ the true value continues $011 001 100ldots$ so if we subtract we get (about) $001$ with the first $0$ matching the last $1$ in $fl(0.1)$. That is $22$ places to the right of the radix point, so the difference is $2^{-4}cdot 2^{-22}cdot 2^{-2}=2^{-28}$ where the first factor is from the exponent of $fl(0.1)$, the second is from all the places to the right of the radix point, and the third is the two places the $1$ in the error is to the right of the end of $fl(0.1)$. The relative error is then $frac {2^{-28}}{0.1}approx 3.7cdot 10^{-8}$
If you want the exact answer, you need the exact value of $fl(0.1)$. I asked Alpha and got $frac {6710885}{16cdot4194304}$. Then the relative error is $10(frac {6710885}{16cdot4194304}-frac 1{10})=-frac 7{33554432}$
answered Jan 4 at 20:42
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
I have a slightly different answer
$endgroup$
– Vwann
Jan 4 at 21:16
$begingroup$
For the absolute error I get 1/10×2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:17
$begingroup$
And relative is therefore 2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:18
$begingroup$
My computation of $fl(0.1)-0.1$ was not exact. Usually we are only interested in the magnitude of errors like this. $2^{-26}$ is not correct because that is the whole value of the last $1$ in your expression, but $0.1$ is greater than your expression with the last $1$ changed to a $0$.
$endgroup$
– Ross Millikan
Jan 4 at 21:18
$begingroup$
Well, not in this case. The question is about the precise value. Thank you anyway
$endgroup$
– Vwann
Jan 4 at 21:19
|
show 6 more comments
$begingroup$
Just plug into the formula. You have $fl(0.1)$ and $0.1$ so now subtract. Starting from the last position in your $fl(0.1)$ the true value continues $011 001 100ldots$ so if we subtract we get (about) $001$ with the first $0$ matching the last $1$ in $fl(0.1)$. That is $22$ places to the right of the radix point, so the difference is $2^{-4}cdot 2^{-22}cdot 2^{-2}=2^{-28}$ where the first factor is from the exponent of $fl(0.1)$, the second is from all the places to the right of the radix point, and the third is the two places the $1$ in the error is to the right of the end of $fl(0.1)$. The relative error is then $frac {2^{-28}}{0.1}approx 3.7cdot 10^{-8}$
If you want the exact answer, you need the exact value of $fl(0.1)$. I asked Alpha and got $frac {6710885}{16cdot4194304}$. Then the relative error is $10(frac {6710885}{16cdot4194304}-frac 1{10})=-frac 7{33554432}$
answered Jan 4 at 20:42
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
I have a slightly different answer
$endgroup$
– Vwann
Jan 4 at 21:16
$begingroup$
For the absolute error I get 1/10×2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:17
$begingroup$
And relative is therefore 2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:18
$begingroup$
My computation of $fl(0.1)-0.1$ was not exact. Usually we are only interested in the magnitude of errors like this. $2^{-26}$ is not correct because that is the whole value of the last $1$ in your expression, but $0.1$ is greater than your expression with the last $1$ changed to a $0$.
$endgroup$
– Ross Millikan
Jan 4 at 21:18
$begingroup$
Well, not in this case. The question is about the precise value. Thank you anyway
$endgroup$
– Vwann
Jan 4 at 21:19
|
show 6 more comments
$begingroup$
Just plug into the formula. You have $fl(0.1)$ and $0.1$ so now subtract. Starting from the last position in your $fl(0.1)$ the true value continues $011 001 100ldots$ so if we subtract we get (about) $001$ with the first $0$ matching the last $1$ in $fl(0.1)$. That is $22$ places to the right of the radix point, so the difference is $2^{-4}cdot 2^{-22}cdot 2^{-2}=2^{-28}$ where the first factor is from the exponent of $fl(0.1)$, the second is from all the places to the right of the radix point, and the third is the two places the $1$ in the error is to the right of the end of $fl(0.1)$. The relative error is then $frac {2^{-28}}{0.1}approx 3.7cdot 10^{-8}$
If you want the exact answer, you need the exact value of $fl(0.1)$. I asked Alpha and got $frac {6710885}{16cdot4194304}$. Then the relative error is $10(frac {6710885}{16cdot4194304}-frac 1{10})=-frac 7{33554432}$
answered Jan 4 at 20:42
Ross MillikanRoss Millikan
298k24200373
$endgroup$
Just plug into the formula. You have $fl(0.1)$ and $0.1$ so now subtract. Starting from the last position in your $fl(0.1)$ the true value continues $011 001 100ldots$ so if we subtract we get (about) $001$ with the first $0$ matching the last $1$ in $fl(0.1)$. That is $22$ places to the right of the radix point, so the difference is $2^{-4}cdot 2^{-22}cdot 2^{-2}=2^{-28}$ where the first factor is from the exponent of $fl(0.1)$, the second is from all the places to the right of the radix point, and the third is the two places the $1$ in the error is to the right of the end of $fl(0.1)$. The relative error is then $frac {2^{-28}}{0.1}approx 3.7cdot 10^{-8}$
If you want the exact answer, you need the exact value of $fl(0.1)$. I asked Alpha and got $frac {6710885}{16cdot4194304}$. Then the relative error is $10(frac {6710885}{16cdot4194304}-frac 1{10})=-frac 7{33554432}$
answered Jan 4 at 20:42
Ross MillikanRoss Millikan
298k24200373
edited Jan 4 at 21:29
answered Jan 4 at 20:42
Ross MillikanRoss Millikan
298k24200373
answered Jan 4 at 20:42
Ross MillikanRoss Millikan
298k24200373
answered Jan 4 at 20:42
Ross MillikanRoss Millikan
298k24200373
298k24200373
$begingroup$
I have a slightly different answer
$endgroup$
– Vwann
Jan 4 at 21:16
$begingroup$
For the absolute error I get 1/10×2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:17
$begingroup$
And relative is therefore 2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:18
$begingroup$
My computation of $fl(0.1)-0.1$ was not exact. Usually we are only interested in the magnitude of errors like this. $2^{-26}$ is not correct because that is the whole value of the last $1$ in your expression, but $0.1$ is greater than your expression with the last $1$ changed to a $0$.
$endgroup$
– Ross Millikan
Jan 4 at 21:18
$begingroup$
Well, not in this case. The question is about the precise value. Thank you anyway
$endgroup$
– Vwann
Jan 4 at 21:19
|
show 6 more comments
$begingroup$
I have a slightly different answer
$endgroup$
– Vwann
Jan 4 at 21:16
$begingroup$
For the absolute error I get 1/10×2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:17
$begingroup$
And relative is therefore 2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:18
$begingroup$
My computation of $fl(0.1)-0.1$ was not exact. Usually we are only interested in the magnitude of errors like this. $2^{-26}$ is not correct because that is the whole value of the last $1$ in your expression, but $0.1$ is greater than your expression with the last $1$ changed to a $0$.
$endgroup$
– Ross Millikan
Jan 4 at 21:18
$begingroup$
Well, not in this case. The question is about the precise value. Thank you anyway
$endgroup$
– Vwann
Jan 4 at 21:19
$begingroup$
I have a slightly different answer
$endgroup$
– Vwann
Jan 4 at 21:16
$begingroup$
I have a slightly different answer
$endgroup$
– Vwann
Jan 4 at 21:16
$begingroup$
For the absolute error I get 1/10×2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:17
$begingroup$
For the absolute error I get 1/10×2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:17
$begingroup$
And relative is therefore 2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:18
$begingroup$
And relative is therefore 2^(-26)
$endgroup$
– Vwann
Jan 4 at 21:18
$begingroup$
My computation of $fl(0.1)-0.1$ was not exact. Usually we are only interested in the magnitude of errors like this. $2^{-26}$ is not correct because that is the whole value of the last $1$ in your expression, but $0.1$ is greater than your expression with the last $1$ changed to a $0$.
$endgroup$
– Ross Millikan
Jan 4 at 21:18
$begingroup$
My computation of $fl(0.1)-0.1$ was not exact. Usually we are only interested in the magnitude of errors like this. $2^{-26}$ is not correct because that is the whole value of the last $1$ in your expression, but $0.1$ is greater than your expression with the last $1$ changed to a $0$.
$endgroup$
– Ross Millikan
Jan 4 at 21:18
$begingroup$
Well, not in this case. The question is about the precise value. Thank you anyway
$endgroup$
– Vwann
Jan 4 at 21:19
$begingroup$
Well, not in this case. The question is about the precise value. Thank you anyway
$endgroup$
– Vwann
Jan 4 at 21:19
|
show 6 more comments
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$begingroup$
Yes, thank you, I edited
$endgroup$
– Vwann
Jan 4 at 20:35