How to say “become smaller/lower” in one word in mathematical context?
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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
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show 1 more comment
$begingroup$
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
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2
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There’s not a good word for it. Just like there’s no common single-word opposite to “exceed.”
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– Steve Kass
Dec 15 '18 at 15:48
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I agree with @SteveKass. You might say, "falling behind."
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– saulspatz
Dec 15 '18 at 15:51
2
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"is dominated by"? Not a single word, certainly.
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– Patrick Stevens
Dec 15 '18 at 15:52
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"Becomes progressively smaller than"?
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– timtfj
Dec 15 '18 at 16:03
2
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What's wrong with using two or three words if they get the point across? Also, words are sometimes clearer than formulas, e.g., $$lim_{x to infty} frac{x^2}{2^x} = 0.$$
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– Robert Soupe
Dec 15 '18 at 17:37
|
show 1 more comment
$begingroup$
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
$endgroup$
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
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
|
show 1 more comment
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
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
"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 $y>x$.
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add a comment |
$begingroup$
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.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
"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 $y>x$.
$endgroup$
add a comment |
$begingroup$
"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 $y>x$.
$endgroup$
add a comment |
$begingroup$
"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 $y>x$.
$endgroup$
"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 $y>x$.
answered Dec 15 '18 at 15:52
R. BurtonR. Burton
627110
627110
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Dec 15 '18 at 22:40
The Short OneThe Short One
7341625
7341625
add a comment |
add a comment |
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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