Partition?! On a step function Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is the space of uniformly left continuous functions on [0,1] complete?Partial derivative of a summation.Am I on the right track? Any hints on what to do next?Is there alternate definitions of an integral based on values of function on open or closed interval ?Taylor series and integrationUnderstanding the role of Partitioning in Riemann SumsRiemann and Darboux Integral of a product of two functionsProve a function is not Darboux-integrableEquivalent definition of the variation of the function on $[a,b]$How to prove that the limit of an upper sum equals the integral of a function with these limited definitions?
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Partition?! On a step function
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is the space of uniformly left continuous functions on [0,1] complete?Partial derivative of a summation.Am I on the right track? Any hints on what to do next?Is there alternate definitions of an integral based on values of function on open or closed interval ?Taylor series and integrationUnderstanding the role of Partitioning in Riemann SumsRiemann and Darboux Integral of a product of two functionsProve a function is not Darboux-integrableEquivalent definition of the variation of the function on $[a,b]$How to prove that the limit of an upper sum equals the integral of a function with these limited definitions?
$begingroup$
Let $s,r:[a,b]to mathbb R$ be step functions. Prove that there exists a partition $a=t_0<t_1<ldots<t_N=b$ such that $s$ and $r$ are both constant on $(t_i-1,t_i)$ for each $iin1,ldots,N$.
Yeah I have no idea could use a mini lesson explaining.
calculus
asked Apr 8 at 20:24
DoubleliftDoublelift
305
$endgroup$
add a comment |
$begingroup$
Let $s,r:[a,b]to mathbb R$ be step functions. Prove that there exists a partition $a=t_0<t_1<ldots<t_N=b$ such that $s$ and $r$ are both constant on $(t_i-1,t_i)$ for each $iin1,ldots,N$.
Yeah I have no idea could use a mini lesson explaining.
calculus
asked Apr 8 at 20:24
DoubleliftDoublelift
305
$endgroup$
1
$begingroup$
Do some drawing.
$endgroup$
– Phicar
Apr 8 at 20:25
add a comment |
$begingroup$
Let $s,r:[a,b]to mathbb R$ be step functions. Prove that there exists a partition $a=t_0<t_1<ldots<t_N=b$ such that $s$ and $r$ are both constant on $(t_i-1,t_i)$ for each $iin1,ldots,N$.
Yeah I have no idea could use a mini lesson explaining.
calculus
asked Apr 8 at 20:24
DoubleliftDoublelift
305
$endgroup$
Let $s,r:[a,b]to mathbb R$ be step functions. Prove that there exists a partition $a=t_0<t_1<ldots<t_N=b$ such that $s$ and $r$ are both constant on $(t_i-1,t_i)$ for each $iin1,ldots,N$.
Yeah I have no idea could use a mini lesson explaining.
calculus
calculus
asked Apr 8 at 20:24
DoubleliftDoublelift
305
asked Apr 8 at 20:24
DoubleliftDoublelift
305
asked Apr 8 at 20:24
DoubleliftDoublelift
305
asked Apr 8 at 20:24
DoubleliftDoublelift
305
asked Apr 8 at 20:24
DoubleliftDoublelift
305
305
1
$begingroup$
Do some drawing.
$endgroup$
– Phicar
Apr 8 at 20:25
add a comment |
1
$begingroup$
Do some drawing.
$endgroup$
– Phicar
Apr 8 at 20:25
1
1
$begingroup$
Do some drawing.
$endgroup$
– Phicar
Apr 8 at 20:25
$begingroup$
Do some drawing.
$endgroup$
– Phicar
Apr 8 at 20:25
add a comment |
1 Answer
1
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$begingroup$
Let's note 2 things about step functions:
(1) By definition the step function is peicewise constant. So each step function already has a "natural" partition $a=x_0 < x_1 < ... < x_N = b$ for which it is constant on $(x_i , x_i+1)$
(2) If we add any points to the natural partition of a step function, say we added the point $y$ to the above partition: $a = x_0 < ... < x_i < y < x_i+1 < ... < x_M = b $. Then the step function is still constant on every $(x_i , x_i+1)$. Namely it is constant on $(x_i , y)$ and $(y, x_i+1)$.
So now to answer your question. Take the natural partition on which $s$ is defined, and add to it all points of the natural partition on which $r$ is defined. Denote the combined partition by $t_0 , ... , t_N$. Then $s$ is still constant on $(t_i , t_i+1)$ by point (2).
Similarly, take the natural partition on which $r$ is defined, and add to it all points of the natural partition on which $s$ is defined. Denote the combined partition by $t_0 , ... , t_N$ (it is the same one as the previous paragraph). Then $r$ is still constant on $(t_i , t_i+1)$ by point (2).
This proves the result, since $t_0 , ... , t_N$ works for both $s$ and $r$
answered Apr 9 at 13:04
NazimJNazimJ
880110
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add a comment |
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$begingroup$
Let's note 2 things about step functions:
(1) By definition the step function is peicewise constant. So each step function already has a "natural" partition $a=x_0 < x_1 < ... < x_N = b$ for which it is constant on $(x_i , x_i+1)$
(2) If we add any points to the natural partition of a step function, say we added the point $y$ to the above partition: $a = x_0 < ... < x_i < y < x_i+1 < ... < x_M = b $. Then the step function is still constant on every $(x_i , x_i+1)$. Namely it is constant on $(x_i , y)$ and $(y, x_i+1)$.
So now to answer your question. Take the natural partition on which $s$ is defined, and add to it all points of the natural partition on which $r$ is defined. Denote the combined partition by $t_0 , ... , t_N$. Then $s$ is still constant on $(t_i , t_i+1)$ by point (2).
Similarly, take the natural partition on which $r$ is defined, and add to it all points of the natural partition on which $s$ is defined. Denote the combined partition by $t_0 , ... , t_N$ (it is the same one as the previous paragraph). Then $r$ is still constant on $(t_i , t_i+1)$ by point (2).
This proves the result, since $t_0 , ... , t_N$ works for both $s$ and $r$
answered Apr 9 at 13:04
NazimJNazimJ
880110
$endgroup$
add a comment |
$begingroup$
Let's note 2 things about step functions:
(1) By definition the step function is peicewise constant. So each step function already has a "natural" partition $a=x_0 < x_1 < ... < x_N = b$ for which it is constant on $(x_i , x_i+1)$
(2) If we add any points to the natural partition of a step function, say we added the point $y$ to the above partition: $a = x_0 < ... < x_i < y < x_i+1 < ... < x_M = b $. Then the step function is still constant on every $(x_i , x_i+1)$. Namely it is constant on $(x_i , y)$ and $(y, x_i+1)$.
So now to answer your question. Take the natural partition on which $s$ is defined, and add to it all points of the natural partition on which $r$ is defined. Denote the combined partition by $t_0 , ... , t_N$. Then $s$ is still constant on $(t_i , t_i+1)$ by point (2).
Similarly, take the natural partition on which $r$ is defined, and add to it all points of the natural partition on which $s$ is defined. Denote the combined partition by $t_0 , ... , t_N$ (it is the same one as the previous paragraph). Then $r$ is still constant on $(t_i , t_i+1)$ by point (2).
This proves the result, since $t_0 , ... , t_N$ works for both $s$ and $r$
answered Apr 9 at 13:04
NazimJNazimJ
880110
$endgroup$
add a comment |
$begingroup$
Let's note 2 things about step functions:
(1) By definition the step function is peicewise constant. So each step function already has a "natural" partition $a=x_0 < x_1 < ... < x_N = b$ for which it is constant on $(x_i , x_i+1)$
(2) If we add any points to the natural partition of a step function, say we added the point $y$ to the above partition: $a = x_0 < ... < x_i < y < x_i+1 < ... < x_M = b $. Then the step function is still constant on every $(x_i , x_i+1)$. Namely it is constant on $(x_i , y)$ and $(y, x_i+1)$.
So now to answer your question. Take the natural partition on which $s$ is defined, and add to it all points of the natural partition on which $r$ is defined. Denote the combined partition by $t_0 , ... , t_N$. Then $s$ is still constant on $(t_i , t_i+1)$ by point (2).
Similarly, take the natural partition on which $r$ is defined, and add to it all points of the natural partition on which $s$ is defined. Denote the combined partition by $t_0 , ... , t_N$ (it is the same one as the previous paragraph). Then $r$ is still constant on $(t_i , t_i+1)$ by point (2).
This proves the result, since $t_0 , ... , t_N$ works for both $s$ and $r$
answered Apr 9 at 13:04
NazimJNazimJ
880110
$endgroup$
Let's note 2 things about step functions:
(1) By definition the step function is peicewise constant. So each step function already has a "natural" partition $a=x_0 < x_1 < ... < x_N = b$ for which it is constant on $(x_i , x_i+1)$
(2) If we add any points to the natural partition of a step function, say we added the point $y$ to the above partition: $a = x_0 < ... < x_i < y < x_i+1 < ... < x_M = b $. Then the step function is still constant on every $(x_i , x_i+1)$. Namely it is constant on $(x_i , y)$ and $(y, x_i+1)$.
So now to answer your question. Take the natural partition on which $s$ is defined, and add to it all points of the natural partition on which $r$ is defined. Denote the combined partition by $t_0 , ... , t_N$. Then $s$ is still constant on $(t_i , t_i+1)$ by point (2).
Similarly, take the natural partition on which $r$ is defined, and add to it all points of the natural partition on which $s$ is defined. Denote the combined partition by $t_0 , ... , t_N$ (it is the same one as the previous paragraph). Then $r$ is still constant on $(t_i , t_i+1)$ by point (2).
This proves the result, since $t_0 , ... , t_N$ works for both $s$ and $r$
answered Apr 9 at 13:04
NazimJNazimJ
880110
answered Apr 9 at 13:04
NazimJNazimJ
880110
answered Apr 9 at 13:04
NazimJNazimJ
880110
answered Apr 9 at 13:04
NazimJNazimJ
880110
880110
add a comment |
add a comment |
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$begingroup$
Do some drawing.
$endgroup$
– Phicar
Apr 8 at 20:25