Relative roundoff error in the simple precision.












0












$begingroup$


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







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  • $begingroup$
    Yes, thank you, I edited
    $endgroup$
    – Vwann
    Jan 4 at 20:35
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, thank you, I edited
    $endgroup$
    – Vwann
    Jan 4 at 20:35














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 20:35







Vwann

















asked Jan 4 at 20:16









VwannVwann

225




225












  • $begingroup$
    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




$begingroup$
Yes, thank you, I edited
$endgroup$
– Vwann
Jan 4 at 20:35










1 Answer
1






active

oldest

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0












$begingroup$

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 $10(frac {6710885}{16cdot4194304}-frac 1{10})=-frac 7{33554432}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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 $10(frac {6710885}{16cdot4194304}-frac 1{10})=-frac 7{33554432}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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 $10(frac {6710885}{16cdot4194304}-frac 1{10})=-frac 7{33554432}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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 $10(frac {6710885}{16cdot4194304}-frac 1{10})=-frac 7{33554432}$






share|cite|improve this answer











$endgroup$



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 $10(frac {6710885}{16cdot4194304}-frac 1{10})=-frac 7{33554432}$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 21:29

























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


















  • $begingroup$
    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


















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