How to say “become smaller/lower” in one word in mathematical context?












5












$begingroup$


We can say e.g. "You can see $2^x$ outgrowing $x^2$ as x increases in Fig. 6.18.".



How can we express the opposite?



The corresponding example: "You can see $x^2$ ... $2^x$ as x increases". What is a good single word for the gap?






calculus arithmetic







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    There’s not a good word for it. Just like there’s no common single-word opposite to “exceed.”
    $endgroup$
    – Steve Kass
    Dec 15 '18 at 15:48










  • $begingroup$
    I agree with @SteveKass. You might say, "falling behind."
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:51






  • 2




    $begingroup$
    "is dominated by"? Not a single word, certainly.
    $endgroup$
    – Patrick Stevens
    Dec 15 '18 at 15:52












  • $begingroup$
    "Becomes progressively smaller than"?
    $endgroup$
    – timtfj
    Dec 15 '18 at 16:03






  • 2




    $begingroup$
    What's wrong with using two or three words if they get the point across? Also, words are sometimes clearer than formulas, e.g., $$lim_{x to infty} frac{x^2}{2^x} = 0.$$
    $endgroup$
    – Robert Soupe
    Dec 15 '18 at 17:37
















5












$begingroup$


We can say e.g. "You can see $2^x$ outgrowing $x^2$ as x increases in Fig. 6.18.".



How can we express the opposite?



The corresponding example: "You can see $x^2$ ... $2^x$ as x increases". What is a good single word for the gap?






calculus arithmetic







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    There’s not a good word for it. Just like there’s no common single-word opposite to “exceed.”
    $endgroup$
    – Steve Kass
    Dec 15 '18 at 15:48










  • $begingroup$
    I agree with @SteveKass. You might say, "falling behind."
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:51






  • 2




    $begingroup$
    "is dominated by"? Not a single word, certainly.
    $endgroup$
    – Patrick Stevens
    Dec 15 '18 at 15:52












  • $begingroup$
    "Becomes progressively smaller than"?
    $endgroup$
    – timtfj
    Dec 15 '18 at 16:03






  • 2




    $begingroup$
    What's wrong with using two or three words if they get the point across? Also, words are sometimes clearer than formulas, e.g., $$lim_{x to infty} frac{x^2}{2^x} = 0.$$
    $endgroup$
    – Robert Soupe
    Dec 15 '18 at 17:37














5












5








5





$begingroup$


We can say e.g. "You can see $2^x$ outgrowing $x^2$ as x increases in Fig. 6.18.".



How can we express the opposite?



The corresponding example: "You can see $x^2$ ... $2^x$ as x increases". What is a good single word for the gap?






calculus arithmetic







share|cite|improve this question









$endgroup$




We can say e.g. "You can see $2^x$ outgrowing $x^2$ as x increases in Fig. 6.18.".



How can we express the opposite?



The corresponding example: "You can see $x^2$ ... $2^x$ as x increases". What is a good single word for the gap?






calculus arithmetic




calculus arithmetic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 15:45









Konstantinos AndreadisKonstantinos Andreadis

283




283








  • 2




    $begingroup$
    There’s not a good word for it. Just like there’s no common single-word opposite to “exceed.”
    $endgroup$
    – Steve Kass
    Dec 15 '18 at 15:48










  • $begingroup$
    I agree with @SteveKass. You might say, "falling behind."
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:51






  • 2




    $begingroup$
    "is dominated by"? Not a single word, certainly.
    $endgroup$
    – Patrick Stevens
    Dec 15 '18 at 15:52












  • $begingroup$
    "Becomes progressively smaller than"?
    $endgroup$
    – timtfj
    Dec 15 '18 at 16:03






  • 2




    $begingroup$
    What's wrong with using two or three words if they get the point across? Also, words are sometimes clearer than formulas, e.g., $$lim_{x to infty} frac{x^2}{2^x} = 0.$$
    $endgroup$
    – Robert Soupe
    Dec 15 '18 at 17:37














  • 2




    $begingroup$
    There’s not a good word for it. Just like there’s no common single-word opposite to “exceed.”
    $endgroup$
    – Steve Kass
    Dec 15 '18 at 15:48










  • $begingroup$
    I agree with @SteveKass. You might say, "falling behind."
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:51






  • 2




    $begingroup$
    "is dominated by"? Not a single word, certainly.
    $endgroup$
    – Patrick Stevens
    Dec 15 '18 at 15:52












  • $begingroup$
    "Becomes progressively smaller than"?
    $endgroup$
    – timtfj
    Dec 15 '18 at 16:03






  • 2




    $begingroup$
    What's wrong with using two or three words if they get the point across? Also, words are sometimes clearer than formulas, e.g., $$lim_{x to infty} frac{x^2}{2^x} = 0.$$
    $endgroup$
    – Robert Soupe
    Dec 15 '18 at 17:37








2




2




$begingroup$
There’s not a good word for it. Just like there’s no common single-word opposite to “exceed.”
$endgroup$
– Steve Kass
Dec 15 '18 at 15:48




$begingroup$
There’s not a good word for it. Just like there’s no common single-word opposite to “exceed.”
$endgroup$
– Steve Kass
Dec 15 '18 at 15:48












$begingroup$
I agree with @SteveKass. You might say, "falling behind."
$endgroup$
– saulspatz
Dec 15 '18 at 15:51




$begingroup$
I agree with @SteveKass. You might say, "falling behind."
$endgroup$
– saulspatz
Dec 15 '18 at 15:51




2




2




$begingroup$
"is dominated by"? Not a single word, certainly.
$endgroup$
– Patrick Stevens
Dec 15 '18 at 15:52






$begingroup$
"is dominated by"? Not a single word, certainly.
$endgroup$
– Patrick Stevens
Dec 15 '18 at 15:52














$begingroup$
"Becomes progressively smaller than"?
$endgroup$
– timtfj
Dec 15 '18 at 16:03




$begingroup$
"Becomes progressively smaller than"?
$endgroup$
– timtfj
Dec 15 '18 at 16:03




2




2




$begingroup$
What's wrong with using two or three words if they get the point across? Also, words are sometimes clearer than formulas, e.g., $$lim_{x to infty} frac{x^2}{2^x} = 0.$$
$endgroup$
– Robert Soupe
Dec 15 '18 at 17:37




$begingroup$
What's wrong with using two or three words if they get the point across? Also, words are sometimes clearer than formulas, e.g., $$lim_{x to infty} frac{x^2}{2^x} = 0.$$
$endgroup$
– Robert Soupe
Dec 15 '18 at 17:37










2 Answers
2






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5












$begingroup$

"Outgrowing" is probably best interpreted as "the difference between $2^x$ and $x^2$ grows larger as $x$ increases". If you represent this difference as a formula $2^x-x^2=d$, then the converse, given by the formula $x^2-2^x=-d$ would be "the difference between $x^2$ and $2^x$ decreases as $x$ increases; i.e. $2^x-x^2>x^2-2^x$ for some $x$, and all $y>x$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    How about just "decreases"?



    Maybe what you really want to say here is that the increases or decreases are exponential. The difference between, say, $2^3$ and $3^2$ is famously small, but the difference between $2^G$ and $G^2$, where $G$ is a googolplex, are mind-boggling, at least for puny human minds.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

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      active

      oldest

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      5












      $begingroup$

      "Outgrowing" is probably best interpreted as "the difference between $2^x$ and $x^2$ grows larger as $x$ increases". If you represent this difference as a formula $2^x-x^2=d$, then the converse, given by the formula $x^2-2^x=-d$ would be "the difference between $x^2$ and $2^x$ decreases as $x$ increases; i.e. $2^x-x^2>x^2-2^x$ for some $x$, and all $y>x$.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        "Outgrowing" is probably best interpreted as "the difference between $2^x$ and $x^2$ grows larger as $x$ increases". If you represent this difference as a formula $2^x-x^2=d$, then the converse, given by the formula $x^2-2^x=-d$ would be "the difference between $x^2$ and $2^x$ decreases as $x$ increases; i.e. $2^x-x^2>x^2-2^x$ for some $x$, and all $y>x$.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          "Outgrowing" is probably best interpreted as "the difference between $2^x$ and $x^2$ grows larger as $x$ increases". If you represent this difference as a formula $2^x-x^2=d$, then the converse, given by the formula $x^2-2^x=-d$ would be "the difference between $x^2$ and $2^x$ decreases as $x$ increases; i.e. $2^x-x^2>x^2-2^x$ for some $x$, and all $y>x$.






          share|cite|improve this answer









          $endgroup$



          "Outgrowing" is probably best interpreted as "the difference between $2^x$ and $x^2$ grows larger as $x$ increases". If you represent this difference as a formula $2^x-x^2=d$, then the converse, given by the formula $x^2-2^x=-d$ would be "the difference between $x^2$ and $2^x$ decreases as $x$ increases; i.e. $2^x-x^2>x^2-2^x$ for some $x$, and all $y>x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 15:52









          R. BurtonR. Burton

          627110




          627110























              4












              $begingroup$

              How about just "decreases"?



              Maybe what you really want to say here is that the increases or decreases are exponential. The difference between, say, $2^3$ and $3^2$ is famously small, but the difference between $2^G$ and $G^2$, where $G$ is a googolplex, are mind-boggling, at least for puny human minds.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                How about just "decreases"?



                Maybe what you really want to say here is that the increases or decreases are exponential. The difference between, say, $2^3$ and $3^2$ is famously small, but the difference between $2^G$ and $G^2$, where $G$ is a googolplex, are mind-boggling, at least for puny human minds.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  How about just "decreases"?



                  Maybe what you really want to say here is that the increases or decreases are exponential. The difference between, say, $2^3$ and $3^2$ is famously small, but the difference between $2^G$ and $G^2$, where $G$ is a googolplex, are mind-boggling, at least for puny human minds.






                  share|cite|improve this answer









                  $endgroup$



                  How about just "decreases"?



                  Maybe what you really want to say here is that the increases or decreases are exponential. The difference between, say, $2^3$ and $3^2$ is famously small, but the difference between $2^G$ and $G^2$, where $G$ is a googolplex, are mind-boggling, at least for puny human minds.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 22:40









                  The Short OneThe Short One

                  7341625




                  7341625






























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