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







share|cite|improve this question
























  • Yes, thank you, I edited
    – Vwann
    yesterday
















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
























  • 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked yesterday









Vwann

225




225












  • 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






active

oldest

votes


















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062055%2frelative-roundoff-error-in-the-simple-precision%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









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














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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062055%2frelative-roundoff-error-in-the-simple-precision%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

Antonio Litta Visconti Arese

Image
Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
Read more