What is meant by interval notation in $mathbb{R}^n$?Multivariable Mean Value Theorem With EqualitiesCan a...
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What is meant by interval notation in $mathbb{R}^n$?
Multivariable Mean Value Theorem With EqualitiesCan a value $c$ satisfying the Mean Value Theorem be on the interval $[a,b]$?What is an example for the mean-value theorem to fail to hold for differentiable functions $mathbb{R}^{n} to mathbb{R}^{m}$ if $m geq 2$?newton raphson method- root outside the intervalProving Mean Value Theorem for Harmonic functions in $mathbb{R}^n$What is meant by $d(xy)$?Why does the Mean Value Theorem require a closed interval for continuity and an open interval for differentiability?How do you prove that the “c” guaranteed by the Mean Value Theorem for a quadractic function is the midpoint of the interval [a,b]?Rolle's theorem and the closed intervalIncreasing and Decreasing functions using interval notation
$begingroup$
I am reading a version of the mean-value theorem and it goes as follows:
What is meant by, 'the interval $[x,x+s]$'?
calculus multivariable-calculus
asked Feb 26 at 12:05
foshofosho
4,7711033
$endgroup$
add a comment |
$begingroup$
I am reading a version of the mean-value theorem and it goes as follows:
What is meant by, 'the interval $[x,x+s]$'?
calculus multivariable-calculus
asked Feb 26 at 12:05
foshofosho
4,7711033
$endgroup$
$begingroup$
Try with a simple example : $x=1$ and $s=dfrac 1 2$.
$endgroup$
– Mauro ALLEGRANZA
Feb 26 at 12:11
1
$begingroup$
This doesn't help, since the OP's problem concerns dimensions greater than 1.
$endgroup$
– Mars Plastic
Feb 26 at 12:12
add a comment |
$begingroup$
I am reading a version of the mean-value theorem and it goes as follows:
What is meant by, 'the interval $[x,x+s]$'?
calculus multivariable-calculus
asked Feb 26 at 12:05
foshofosho
4,7711033
$endgroup$
I am reading a version of the mean-value theorem and it goes as follows:
What is meant by, 'the interval $[x,x+s]$'?
calculus multivariable-calculus
calculus multivariable-calculus
asked Feb 26 at 12:05
foshofosho
4,7711033
asked Feb 26 at 12:05
foshofosho
4,7711033
asked Feb 26 at 12:05
foshofosho
4,7711033
asked Feb 26 at 12:05
foshofosho
4,7711033
asked Feb 26 at 12:05
foshofosho
4,7711033
4,7711033
$begingroup$
Try with a simple example : $x=1$ and $s=dfrac 1 2$.
$endgroup$
– Mauro ALLEGRANZA
Feb 26 at 12:11
1
$begingroup$
This doesn't help, since the OP's problem concerns dimensions greater than 1.
$endgroup$
– Mars Plastic
Feb 26 at 12:12
add a comment |
$begingroup$
Try with a simple example : $x=1$ and $s=dfrac 1 2$.
$endgroup$
– Mauro ALLEGRANZA
Feb 26 at 12:11
1
$begingroup$
This doesn't help, since the OP's problem concerns dimensions greater than 1.
$endgroup$
– Mars Plastic
Feb 26 at 12:12
$begingroup$
Try with a simple example : $x=1$ and $s=dfrac 1 2$.
$endgroup$
– Mauro ALLEGRANZA
Feb 26 at 12:11
$begingroup$
Try with a simple example : $x=1$ and $s=dfrac 1 2$.
$endgroup$
– Mauro ALLEGRANZA
Feb 26 at 12:11
1
1
$begingroup$
This doesn't help, since the OP's problem concerns dimensions greater than 1.
$endgroup$
– Mars Plastic
Feb 26 at 12:12
$begingroup$
This doesn't help, since the OP's problem concerns dimensions greater than 1.
$endgroup$
– Mars Plastic
Feb 26 at 12:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Usually, this notation is meant to describe the set
$$ [x,x+s]:={x+lambda s:lambdain[0,1]}.$$
More general-looking: for any $x,yinmathbb R^n$, one sets
$$ [x,y]:={x+lambda (y-x):lambdain[0,1]},$$
which is the convex hull of the set ${x,y}$.
answered Feb 26 at 12:09
Mars PlasticMars Plastic
1,461121
$endgroup$
$begingroup$
Thanks. Ill accept this when it allows me to.
$endgroup$
– fosho
Feb 26 at 12:12
$begingroup$
But such thing cannot be a member of the open set S as asserted by the theorem.
$endgroup$
– William Elliot
Feb 26 at 12:12
2
$begingroup$
@WilliamElliot Yes, that's what I just thought as well. But this is probably an error in the notation and it should say $[x,x+s]subset mathcal S$. The rest wouldn't make sense otherwise, or am I missing something?
$endgroup$
– Mars Plastic
Feb 26 at 12:14
$begingroup$
Check the errata for the book. If it is not there point out the mistake to the author or publisher.
$endgroup$
– William Elliot
Feb 26 at 21:45
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Usually, this notation is meant to describe the set
$$ [x,x+s]:={x+lambda s:lambdain[0,1]}.$$
More general-looking: for any $x,yinmathbb R^n$, one sets
$$ [x,y]:={x+lambda (y-x):lambdain[0,1]},$$
which is the convex hull of the set ${x,y}$.
answered Feb 26 at 12:09
Mars PlasticMars Plastic
1,461121
$endgroup$
$begingroup$
Thanks. Ill accept this when it allows me to.
$endgroup$
– fosho
Feb 26 at 12:12
$begingroup$
But such thing cannot be a member of the open set S as asserted by the theorem.
$endgroup$
– William Elliot
Feb 26 at 12:12
2
$begingroup$
@WilliamElliot Yes, that's what I just thought as well. But this is probably an error in the notation and it should say $[x,x+s]subset mathcal S$. The rest wouldn't make sense otherwise, or am I missing something?
$endgroup$
– Mars Plastic
Feb 26 at 12:14
$begingroup$
Check the errata for the book. If it is not there point out the mistake to the author or publisher.
$endgroup$
– William Elliot
Feb 26 at 21:45
add a comment |
$begingroup$
Usually, this notation is meant to describe the set
$$ [x,x+s]:={x+lambda s:lambdain[0,1]}.$$
More general-looking: for any $x,yinmathbb R^n$, one sets
$$ [x,y]:={x+lambda (y-x):lambdain[0,1]},$$
which is the convex hull of the set ${x,y}$.
answered Feb 26 at 12:09
Mars PlasticMars Plastic
1,461121
$endgroup$
$begingroup$
Thanks. Ill accept this when it allows me to.
$endgroup$
– fosho
Feb 26 at 12:12
$begingroup$
But such thing cannot be a member of the open set S as asserted by the theorem.
$endgroup$
– William Elliot
Feb 26 at 12:12
2
$begingroup$
@WilliamElliot Yes, that's what I just thought as well. But this is probably an error in the notation and it should say $[x,x+s]subset mathcal S$. The rest wouldn't make sense otherwise, or am I missing something?
$endgroup$
– Mars Plastic
Feb 26 at 12:14
$begingroup$
Check the errata for the book. If it is not there point out the mistake to the author or publisher.
$endgroup$
– William Elliot
Feb 26 at 21:45
add a comment |
$begingroup$
Usually, this notation is meant to describe the set
$$ [x,x+s]:={x+lambda s:lambdain[0,1]}.$$
More general-looking: for any $x,yinmathbb R^n$, one sets
$$ [x,y]:={x+lambda (y-x):lambdain[0,1]},$$
which is the convex hull of the set ${x,y}$.
answered Feb 26 at 12:09
Mars PlasticMars Plastic
1,461121
$endgroup$
Usually, this notation is meant to describe the set
$$ [x,x+s]:={x+lambda s:lambdain[0,1]}.$$
More general-looking: for any $x,yinmathbb R^n$, one sets
$$ [x,y]:={x+lambda (y-x):lambdain[0,1]},$$
which is the convex hull of the set ${x,y}$.
answered Feb 26 at 12:09
Mars PlasticMars Plastic
1,461121
edited Mar 12 at 0:20
answered Feb 26 at 12:09
Mars PlasticMars Plastic
1,461121
answered Feb 26 at 12:09
Mars PlasticMars Plastic
1,461121
answered Feb 26 at 12:09
Mars PlasticMars Plastic
1,461121
1,461121
$begingroup$
Thanks. Ill accept this when it allows me to.
$endgroup$
– fosho
Feb 26 at 12:12
$begingroup$
But such thing cannot be a member of the open set S as asserted by the theorem.
$endgroup$
– William Elliot
Feb 26 at 12:12
2
$begingroup$
@WilliamElliot Yes, that's what I just thought as well. But this is probably an error in the notation and it should say $[x,x+s]subset mathcal S$. The rest wouldn't make sense otherwise, or am I missing something?
$endgroup$
– Mars Plastic
Feb 26 at 12:14
$begingroup$
Check the errata for the book. If it is not there point out the mistake to the author or publisher.
$endgroup$
– William Elliot
Feb 26 at 21:45
add a comment |
$begingroup$
Thanks. Ill accept this when it allows me to.
$endgroup$
– fosho
Feb 26 at 12:12
$begingroup$
But such thing cannot be a member of the open set S as asserted by the theorem.
$endgroup$
– William Elliot
Feb 26 at 12:12
2
$begingroup$
@WilliamElliot Yes, that's what I just thought as well. But this is probably an error in the notation and it should say $[x,x+s]subset mathcal S$. The rest wouldn't make sense otherwise, or am I missing something?
$endgroup$
– Mars Plastic
Feb 26 at 12:14
$begingroup$
Check the errata for the book. If it is not there point out the mistake to the author or publisher.
$endgroup$
– William Elliot
Feb 26 at 21:45
$begingroup$
Thanks. Ill accept this when it allows me to.
$endgroup$
– fosho
Feb 26 at 12:12
$begingroup$
Thanks. Ill accept this when it allows me to.
$endgroup$
– fosho
Feb 26 at 12:12
$begingroup$
But such thing cannot be a member of the open set S as asserted by the theorem.
$endgroup$
– William Elliot
Feb 26 at 12:12
$begingroup$
But such thing cannot be a member of the open set S as asserted by the theorem.
$endgroup$
– William Elliot
Feb 26 at 12:12
2
2
$begingroup$
@WilliamElliot Yes, that's what I just thought as well. But this is probably an error in the notation and it should say $[x,x+s]subset mathcal S$. The rest wouldn't make sense otherwise, or am I missing something?
$endgroup$
– Mars Plastic
Feb 26 at 12:14
$begingroup$
@WilliamElliot Yes, that's what I just thought as well. But this is probably an error in the notation and it should say $[x,x+s]subset mathcal S$. The rest wouldn't make sense otherwise, or am I missing something?
$endgroup$
– Mars Plastic
Feb 26 at 12:14
$begingroup$
Check the errata for the book. If it is not there point out the mistake to the author or publisher.
$endgroup$
– William Elliot
Feb 26 at 21:45
$begingroup$
Check the errata for the book. If it is not there point out the mistake to the author or publisher.
$endgroup$
– William Elliot
Feb 26 at 21:45
add a comment |
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$begingroup$
Try with a simple example : $x=1$ and $s=dfrac 1 2$.
$endgroup$
– Mauro ALLEGRANZA
Feb 26 at 12:11
1
$begingroup$
This doesn't help, since the OP's problem concerns dimensions greater than 1.
$endgroup$
– Mars Plastic
Feb 26 at 12:12