The meaning of $(forall M in mathbb{R} )( exists B in mathbb{R} )( forall x>B )( f(x)<M )$
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I'm trying to understand the meaning of this:
$$(forall M in mathbb{R} )( exists B in mathbb{R} )( forall x>B )( f(x)<M )$$
the only thing I could figure out that if $xrightarrow infty$ then $f(x)$ not going to $+infty$
contest-math
$endgroup$
add a comment |
$begingroup$
I'm trying to understand the meaning of this:
$$(forall M in mathbb{R} )( exists B in mathbb{R} )( forall x>B )( f(x)<M )$$
the only thing I could figure out that if $xrightarrow infty$ then $f(x)$ not going to $+infty$
contest-math
$endgroup$
$begingroup$
Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
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– астон вілла олоф мэллбэрг
Dec 12 '18 at 7:11
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What is "$R$"? The real numbers?
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– Eevee Trainer
Dec 12 '18 at 7:11
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R Mean real numbers(couldn't find the char sorry)
$endgroup$
– shay
Dec 12 '18 at 7:15
$begingroup$
I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
$endgroup$
– shay
Dec 12 '18 at 7:17
add a comment |
$begingroup$
I'm trying to understand the meaning of this:
$$(forall M in mathbb{R} )( exists B in mathbb{R} )( forall x>B )( f(x)<M )$$
the only thing I could figure out that if $xrightarrow infty$ then $f(x)$ not going to $+infty$
contest-math
$endgroup$
I'm trying to understand the meaning of this:
$$(forall M in mathbb{R} )( exists B in mathbb{R} )( forall x>B )( f(x)<M )$$
the only thing I could figure out that if $xrightarrow infty$ then $f(x)$ not going to $+infty$
contest-math
contest-math
edited Dec 12 '18 at 7:21
Eevee Trainer
5,7571936
5,7571936
asked Dec 12 '18 at 7:01
shayshay
103
103
$begingroup$
Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 7:11
$begingroup$
What is "$R$"? The real numbers?
$endgroup$
– Eevee Trainer
Dec 12 '18 at 7:11
$begingroup$
R Mean real numbers(couldn't find the char sorry)
$endgroup$
– shay
Dec 12 '18 at 7:15
$begingroup$
I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
$endgroup$
– shay
Dec 12 '18 at 7:17
add a comment |
$begingroup$
Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 7:11
$begingroup$
What is "$R$"? The real numbers?
$endgroup$
– Eevee Trainer
Dec 12 '18 at 7:11
$begingroup$
R Mean real numbers(couldn't find the char sorry)
$endgroup$
– shay
Dec 12 '18 at 7:15
$begingroup$
I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
$endgroup$
– shay
Dec 12 '18 at 7:17
$begingroup$
Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 7:11
$begingroup$
Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 7:11
$begingroup$
What is "$R$"? The real numbers?
$endgroup$
– Eevee Trainer
Dec 12 '18 at 7:11
$begingroup$
What is "$R$"? The real numbers?
$endgroup$
– Eevee Trainer
Dec 12 '18 at 7:11
$begingroup$
R Mean real numbers(couldn't find the char sorry)
$endgroup$
– shay
Dec 12 '18 at 7:15
$begingroup$
R Mean real numbers(couldn't find the char sorry)
$endgroup$
– shay
Dec 12 '18 at 7:15
$begingroup$
I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
$endgroup$
– shay
Dec 12 '18 at 7:17
$begingroup$
I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
$endgroup$
– shay
Dec 12 '18 at 7:17
add a comment |
3 Answers
3
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oldest
votes
$begingroup$
That's the definition of “$f(x) to -infty$ as $x to infty$”.
(At least for functions $f colon mathbf{R} to mathbf{R}$, i.e., functions defined for all $f in mathbf{R}$. If the domain is a subset of $mathbf{R}$, one has to modify the definition slightly.)
$endgroup$
add a comment |
$begingroup$
Graphically, think of $M$ as any horizontal line in the standard coordinate plane. Then your sentence in question is saying that there exists some value $B$ (think of $B$ has a vertical line) such that for all values to the right of $B$, if you evaluate at $f$, they will be below the horizontal line you originally drew ($M$). As the picture below shows, you're essentially saying that 'past $B$, $f$ lives in some quadrant'.
$endgroup$
add a comment |
$begingroup$
This suggests that, after a certain point, $f$ is bounded above by some number, and to any $M$ there is a corresponding "starting point" for this behavior $B$.
Thus, you can choose any upper bound $M in mathbb{R}$. Then we have a corresponding bound $B$ on the argument of $f$. For all $x$ beyond that point, then, $f(x)$ is less than this upper bound.
That is, for any upper bound $M$ on $f$, there is a corresponding lower bound $B$ on $x$, such that for any $x > B$, $f(x) < M$.
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That's the definition of “$f(x) to -infty$ as $x to infty$”.
(At least for functions $f colon mathbf{R} to mathbf{R}$, i.e., functions defined for all $f in mathbf{R}$. If the domain is a subset of $mathbf{R}$, one has to modify the definition slightly.)
$endgroup$
add a comment |
$begingroup$
That's the definition of “$f(x) to -infty$ as $x to infty$”.
(At least for functions $f colon mathbf{R} to mathbf{R}$, i.e., functions defined for all $f in mathbf{R}$. If the domain is a subset of $mathbf{R}$, one has to modify the definition slightly.)
$endgroup$
add a comment |
$begingroup$
That's the definition of “$f(x) to -infty$ as $x to infty$”.
(At least for functions $f colon mathbf{R} to mathbf{R}$, i.e., functions defined for all $f in mathbf{R}$. If the domain is a subset of $mathbf{R}$, one has to modify the definition slightly.)
$endgroup$
That's the definition of “$f(x) to -infty$ as $x to infty$”.
(At least for functions $f colon mathbf{R} to mathbf{R}$, i.e., functions defined for all $f in mathbf{R}$. If the domain is a subset of $mathbf{R}$, one has to modify the definition slightly.)
answered Dec 12 '18 at 7:49
Hans LundmarkHans Lundmark
35.5k564115
35.5k564115
add a comment |
add a comment |
$begingroup$
Graphically, think of $M$ as any horizontal line in the standard coordinate plane. Then your sentence in question is saying that there exists some value $B$ (think of $B$ has a vertical line) such that for all values to the right of $B$, if you evaluate at $f$, they will be below the horizontal line you originally drew ($M$). As the picture below shows, you're essentially saying that 'past $B$, $f$ lives in some quadrant'.
$endgroup$
add a comment |
$begingroup$
Graphically, think of $M$ as any horizontal line in the standard coordinate plane. Then your sentence in question is saying that there exists some value $B$ (think of $B$ has a vertical line) such that for all values to the right of $B$, if you evaluate at $f$, they will be below the horizontal line you originally drew ($M$). As the picture below shows, you're essentially saying that 'past $B$, $f$ lives in some quadrant'.
$endgroup$
add a comment |
$begingroup$
Graphically, think of $M$ as any horizontal line in the standard coordinate plane. Then your sentence in question is saying that there exists some value $B$ (think of $B$ has a vertical line) such that for all values to the right of $B$, if you evaluate at $f$, they will be below the horizontal line you originally drew ($M$). As the picture below shows, you're essentially saying that 'past $B$, $f$ lives in some quadrant'.
$endgroup$
Graphically, think of $M$ as any horizontal line in the standard coordinate plane. Then your sentence in question is saying that there exists some value $B$ (think of $B$ has a vertical line) such that for all values to the right of $B$, if you evaluate at $f$, they will be below the horizontal line you originally drew ($M$). As the picture below shows, you're essentially saying that 'past $B$, $f$ lives in some quadrant'.
edited Dec 12 '18 at 7:23
Eevee Trainer
5,7571936
5,7571936
answered Dec 12 '18 at 7:16
T. FoT. Fo
466311
466311
add a comment |
add a comment |
$begingroup$
This suggests that, after a certain point, $f$ is bounded above by some number, and to any $M$ there is a corresponding "starting point" for this behavior $B$.
Thus, you can choose any upper bound $M in mathbb{R}$. Then we have a corresponding bound $B$ on the argument of $f$. For all $x$ beyond that point, then, $f(x)$ is less than this upper bound.
That is, for any upper bound $M$ on $f$, there is a corresponding lower bound $B$ on $x$, such that for any $x > B$, $f(x) < M$.
$endgroup$
add a comment |
$begingroup$
This suggests that, after a certain point, $f$ is bounded above by some number, and to any $M$ there is a corresponding "starting point" for this behavior $B$.
Thus, you can choose any upper bound $M in mathbb{R}$. Then we have a corresponding bound $B$ on the argument of $f$. For all $x$ beyond that point, then, $f(x)$ is less than this upper bound.
That is, for any upper bound $M$ on $f$, there is a corresponding lower bound $B$ on $x$, such that for any $x > B$, $f(x) < M$.
$endgroup$
add a comment |
$begingroup$
This suggests that, after a certain point, $f$ is bounded above by some number, and to any $M$ there is a corresponding "starting point" for this behavior $B$.
Thus, you can choose any upper bound $M in mathbb{R}$. Then we have a corresponding bound $B$ on the argument of $f$. For all $x$ beyond that point, then, $f(x)$ is less than this upper bound.
That is, for any upper bound $M$ on $f$, there is a corresponding lower bound $B$ on $x$, such that for any $x > B$, $f(x) < M$.
$endgroup$
This suggests that, after a certain point, $f$ is bounded above by some number, and to any $M$ there is a corresponding "starting point" for this behavior $B$.
Thus, you can choose any upper bound $M in mathbb{R}$. Then we have a corresponding bound $B$ on the argument of $f$. For all $x$ beyond that point, then, $f(x)$ is less than this upper bound.
That is, for any upper bound $M$ on $f$, there is a corresponding lower bound $B$ on $x$, such that for any $x > B$, $f(x) < M$.
answered Dec 12 '18 at 7:19
Eevee TrainerEevee Trainer
5,7571936
5,7571936
add a comment |
add a comment |
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$begingroup$
Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 7:11
$begingroup$
What is "$R$"? The real numbers?
$endgroup$
– Eevee Trainer
Dec 12 '18 at 7:11
$begingroup$
R Mean real numbers(couldn't find the char sorry)
$endgroup$
– shay
Dec 12 '18 at 7:15
$begingroup$
I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
$endgroup$
– shay
Dec 12 '18 at 7:17