Relative roundoff error in the simple precision.












0














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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  • Yes, thank you, I edited
    – Vwann
    yesterday
















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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







share|cite|improve this question
























  • Yes, thank you, I edited
    – Vwann
    yesterday














0












0








0







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















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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked yesterday









Vwann

225




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  • Yes, thank you, I edited
    – Vwann
    yesterday


















  • Yes, thank you, I edited
    – Vwann
    yesterday
















Yes, thank you, I edited
– Vwann
yesterday




Yes, thank you, I edited
– Vwann
yesterday










1 Answer
1






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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























  • I have a slightly different answer
    – Vwann
    yesterday










  • For the absolute error I get 1/10×2^(-26)
    – Vwann
    yesterday










  • And relative is therefore 2^(-26)
    – Vwann
    yesterday










  • 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 $0$.
    – Ross Millikan
    yesterday












  • Well, not in this case. The question is about the precise value. Thank you anyway
    – Vwann
    yesterday











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1 Answer
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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























  • I have a slightly different answer
    – Vwann
    yesterday










  • For the absolute error I get 1/10×2^(-26)
    – Vwann
    yesterday










  • And relative is therefore 2^(-26)
    – Vwann
    yesterday










  • 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 $0$.
    – Ross Millikan
    yesterday












  • Well, not in this case. The question is about the precise value. Thank you anyway
    – Vwann
    yesterday
















0














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























  • I have a slightly different answer
    – Vwann
    yesterday










  • For the absolute error I get 1/10×2^(-26)
    – Vwann
    yesterday










  • And relative is therefore 2^(-26)
    – Vwann
    yesterday










  • 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 $0$.
    – Ross Millikan
    yesterday












  • Well, not in this case. The question is about the precise value. Thank you anyway
    – Vwann
    yesterday














0












0








0






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














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 yesterday

























answered yesterday









Ross Millikan

292k23197371




292k23197371












  • I have a slightly different answer
    – Vwann
    yesterday










  • For the absolute error I get 1/10×2^(-26)
    – Vwann
    yesterday










  • And relative is therefore 2^(-26)
    – Vwann
    yesterday










  • 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 $0$.
    – Ross Millikan
    yesterday












  • Well, not in this case. The question is about the precise value. Thank you anyway
    – Vwann
    yesterday


















  • I have a slightly different answer
    – Vwann
    yesterday










  • For the absolute error I get 1/10×2^(-26)
    – Vwann
    yesterday










  • And relative is therefore 2^(-26)
    – Vwann
    yesterday










  • 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 $0$.
    – Ross Millikan
    yesterday












  • Well, not in this case. The question is about the precise value. Thank you anyway
    – Vwann
    yesterday
















I have a slightly different answer
– Vwann
yesterday




I have a slightly different answer
– Vwann
yesterday












For the absolute error I get 1/10×2^(-26)
– Vwann
yesterday




For the absolute error I get 1/10×2^(-26)
– Vwann
yesterday












And relative is therefore 2^(-26)
– Vwann
yesterday




And relative is therefore 2^(-26)
– Vwann
yesterday












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 $0$.
– Ross Millikan
yesterday






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 $0$.
– Ross Millikan
yesterday














Well, not in this case. The question is about the precise value. Thank you anyway
– Vwann
yesterday




Well, not in this case. The question is about the precise value. Thank you anyway
– Vwann
yesterday


















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