Estimate the difference between two quotient

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Suppose I know that $|a-b|leq epsilon_1$ and $|c-d|leq epsilon_2$ all of them are non zero, can I estimate $$left|fracac-fracbdright|,?$$ I tried some thing simple $$left|fracac-fracbc+fracbc-fracbdright|leqleft|fraca-bcright|+|b|left|fracd-ccdright|leq fracepsilon_1+|b|fracepsilon_2,.$$ Is there another better way to do that?




real-analysis inequality estimation




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  • To do what? This is unclear. Why are you unsatified with it?
    – greedoid
    Aug 3 at 10:57















up vote
2
down vote

favorite












Suppose I know that $|a-b|leq epsilon_1$ and $|c-d|leq epsilon_2$ all of them are non zero, can I estimate $$left|fracac-fracbdright|,?$$ I tried some thing simple $$left|fracac-fracbc+fracbc-fracbdright|leqleft|fraca-bcright|+|b|left|fracd-ccdright|leq fracepsilon_1+|b|fracepsilon_2,.$$ Is there another better way to do that?




real-analysis inequality estimation




share|cite|improve this question





















  • To do what? This is unclear. Why are you unsatified with it?
    – greedoid
    Aug 3 at 10:57













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Suppose I know that $|a-b|leq epsilon_1$ and $|c-d|leq epsilon_2$ all of them are non zero, can I estimate $$left|fracac-fracbdright|,?$$ I tried some thing simple $$left|fracac-fracbc+fracbc-fracbdright|leqleft|fraca-bcright|+|b|left|fracd-ccdright|leq fracepsilon_1+|b|fracepsilon_2,.$$ Is there another better way to do that?




real-analysis inequality estimation




share|cite|improve this question













Suppose I know that $|a-b|leq epsilon_1$ and $|c-d|leq epsilon_2$ all of them are non zero, can I estimate $$left|fracac-fracbdright|,?$$ I tried some thing simple $$left|fracac-fracbc+fracbc-fracbdright|leqleft|fraca-bcright|+|b|left|fracd-ccdright|leq fracepsilon_1+|b|fracepsilon_2,.$$ Is there another better way to do that?





real-analysis inequality estimation





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share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 10:37









Batominovski

22.6k22776




22.6k22776









asked Aug 3 at 9:41









quallenjäger

462419




462419











  • To do what? This is unclear. Why are you unsatified with it?
    – greedoid
    Aug 3 at 10:57

















  • To do what? This is unclear. Why are you unsatified with it?
    – greedoid
    Aug 3 at 10:57
















To do what? This is unclear. Why are you unsatified with it?
– greedoid
Aug 3 at 10:57





To do what? This is unclear. Why are you unsatified with it?
– greedoid
Aug 3 at 10:57











1 Answer
1






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up vote
3
down vote



accepted










You can do this as well:
$$left|fracac-fracbdright|=left|fracac-fracad+fracad-fracbdright|leq |a|,left|fracd-ccdright|+left|fraca-bdright|leq fracaepsilon_1+frac1epsilon_2,.$$
Thus,
$$left|fracac-fracbdright|leq minBigglfracaepsilon_1+frac1epsilon_2,frac1epsilon_1+fracepsilon_2Biggr,.$$
I do not think that you can improve this inequality by much, as it is sharp.



However, you can reduce everything into just $a$, $c$, $epsilon_1$, and $epsilon_2$, provided that $|c|>epsilon_2$. That is, from
$$|b|leq |a|+epsilon_1text and |d|geq |c|-epsilon_2,,$$
we obtain
$$beginalign
left|fracac-fracbdright|&leq minBigglfracaepsilon_1+frac1epsilon_2,frac1epsilon_1+fracepsilon_2Biggr
\&leq minBigglfraca-epsilon_2big)epsilon_1+frac1-epsilon_2big)epsilon_2,frac1epsilon_1+frac+epsilon_1-epsilon_2big)epsilon_2Biggr,.endalign$$
This inequality is also sharp.






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  • Thank you for clarifying up, that’s what I sought after.
    – quallenjäger
    Aug 3 at 10:58










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










You can do this as well:
$$left|fracac-fracbdright|=left|fracac-fracad+fracad-fracbdright|leq |a|,left|fracd-ccdright|+left|fraca-bdright|leq fracaepsilon_1+frac1epsilon_2,.$$
Thus,
$$left|fracac-fracbdright|leq minBigglfracaepsilon_1+frac1epsilon_2,frac1epsilon_1+fracepsilon_2Biggr,.$$
I do not think that you can improve this inequality by much, as it is sharp.



However, you can reduce everything into just $a$, $c$, $epsilon_1$, and $epsilon_2$, provided that $|c|>epsilon_2$. That is, from
$$|b|leq |a|+epsilon_1text and |d|geq |c|-epsilon_2,,$$
we obtain
$$beginalign
left|fracac-fracbdright|&leq minBigglfracaepsilon_1+frac1epsilon_2,frac1epsilon_1+fracepsilon_2Biggr
\&leq minBigglfraca-epsilon_2big)epsilon_1+frac1-epsilon_2big)epsilon_2,frac1epsilon_1+frac+epsilon_1-epsilon_2big)epsilon_2Biggr,.endalign$$
This inequality is also sharp.






share|cite|improve this answer























  • Thank you for clarifying up, that’s what I sought after.
    – quallenjäger
    Aug 3 at 10:58














up vote
3
down vote



accepted










You can do this as well:
$$left|fracac-fracbdright|=left|fracac-fracad+fracad-fracbdright|leq |a|,left|fracd-ccdright|+left|fraca-bdright|leq fracaepsilon_1+frac1epsilon_2,.$$
Thus,
$$left|fracac-fracbdright|leq minBigglfracaepsilon_1+frac1epsilon_2,frac1epsilon_1+fracepsilon_2Biggr,.$$
I do not think that you can improve this inequality by much, as it is sharp.



However, you can reduce everything into just $a$, $c$, $epsilon_1$, and $epsilon_2$, provided that $|c|>epsilon_2$. That is, from
$$|b|leq |a|+epsilon_1text and |d|geq |c|-epsilon_2,,$$
we obtain
$$beginalign
left|fracac-fracbdright|&leq minBigglfracaepsilon_1+frac1epsilon_2,frac1epsilon_1+fracepsilon_2Biggr
\&leq minBigglfraca-epsilon_2big)epsilon_1+frac1-epsilon_2big)epsilon_2,frac1epsilon_1+frac+epsilon_1-epsilon_2big)epsilon_2Biggr,.endalign$$
This inequality is also sharp.






share|cite|improve this answer























  • Thank you for clarifying up, that’s what I sought after.
    – quallenjäger
    Aug 3 at 10:58












up vote
3
down vote



accepted







up vote
3
down vote



accepted






You can do this as well:
$$left|fracac-fracbdright|=left|fracac-fracad+fracad-fracbdright|leq |a|,left|fracd-ccdright|+left|fraca-bdright|leq fracaepsilon_1+frac1epsilon_2,.$$
Thus,
$$left|fracac-fracbdright|leq minBigglfracaepsilon_1+frac1epsilon_2,frac1epsilon_1+fracepsilon_2Biggr,.$$
I do not think that you can improve this inequality by much, as it is sharp.



However, you can reduce everything into just $a$, $c$, $epsilon_1$, and $epsilon_2$, provided that $|c|>epsilon_2$. That is, from
$$|b|leq |a|+epsilon_1text and |d|geq |c|-epsilon_2,,$$
we obtain
$$beginalign
left|fracac-fracbdright|&leq minBigglfracaepsilon_1+frac1epsilon_2,frac1epsilon_1+fracepsilon_2Biggr
\&leq minBigglfraca-epsilon_2big)epsilon_1+frac1-epsilon_2big)epsilon_2,frac1epsilon_1+frac+epsilon_1-epsilon_2big)epsilon_2Biggr,.endalign$$
This inequality is also sharp.






share|cite|improve this answer















You can do this as well:
$$left|fracac-fracbdright|=left|fracac-fracad+fracad-fracbdright|leq |a|,left|fracd-ccdright|+left|fraca-bdright|leq fracaepsilon_1+frac1epsilon_2,.$$
Thus,
$$left|fracac-fracbdright|leq minBigglfracaepsilon_1+frac1epsilon_2,frac1epsilon_1+fracepsilon_2Biggr,.$$
I do not think that you can improve this inequality by much, as it is sharp.



However, you can reduce everything into just $a$, $c$, $epsilon_1$, and $epsilon_2$, provided that $|c|>epsilon_2$. That is, from
$$|b|leq |a|+epsilon_1text and |d|geq |c|-epsilon_2,,$$
we obtain
$$beginalign
left|fracac-fracbdright|&leq minBigglfracaepsilon_1+frac1epsilon_2,frac1epsilon_1+fracepsilon_2Biggr
\&leq minBigglfraca-epsilon_2big)epsilon_1+frac1-epsilon_2big)epsilon_2,frac1epsilon_1+frac+epsilon_1-epsilon_2big)epsilon_2Biggr,.endalign$$
This inequality is also sharp.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Aug 3 at 10:59


























answered Aug 3 at 9:55









Batominovski

22.6k22776




22.6k22776











  • Thank you for clarifying up, that’s what I sought after.
    – quallenjäger
    Aug 3 at 10:58
















  • Thank you for clarifying up, that’s what I sought after.
    – quallenjäger
    Aug 3 at 10:58















Thank you for clarifying up, that’s what I sought after.
– quallenjäger
Aug 3 at 10:58




Thank you for clarifying up, that’s what I sought after.
– quallenjäger
Aug 3 at 10:58












 

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