What is the norm of the partition used for?
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I'm confused on what purpose the norm has. I know that it defines the maximum length of the widest subinterval in a partition.
But what use do we have from defining $||P||$ to be $max[Delta t_1, Delta t_2, … ,Delta t_n] $??
Because usually don't we just use the regular n-th partitions anyways? Idk, my textbook just defines the norm once, and then never uses it again so I'm a little confused.
calculus
asked Jan 15 at 21:48
mingming
4606
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add a comment |
$begingroup$
I'm confused on what purpose the norm has. I know that it defines the maximum length of the widest subinterval in a partition.
But what use do we have from defining $||P||$ to be $max[Delta t_1, Delta t_2, … ,Delta t_n] $??
Because usually don't we just use the regular n-th partitions anyways? Idk, my textbook just defines the norm once, and then never uses it again so I'm a little confused.
calculus
asked Jan 15 at 21:48
mingming
4606
$endgroup$
$begingroup$
In some sense it measures the coarseness of the partition, like distinguishing between boulders, rocks, pebbles, sand. Everything in the partition is no bigger than the norm.
$endgroup$
– MPW
Jan 15 at 21:52
$begingroup$
So if the norm was like 5, then the function has the potential to be "coarse" as you say, because each width could be 1,2,3, 4.23, etc.... but if the norm was like 1, the widths don't have as wide of a range?
$endgroup$
– ming
Jan 15 at 22:05
$begingroup$
Probably the main thing is that if the norm goes to zero, then so do the lengths of all intervals in the partition.
$endgroup$
– MPW
Jan 15 at 22:23
add a comment |
$begingroup$
I'm confused on what purpose the norm has. I know that it defines the maximum length of the widest subinterval in a partition.
But what use do we have from defining $||P||$ to be $max[Delta t_1, Delta t_2, … ,Delta t_n] $??
Because usually don't we just use the regular n-th partitions anyways? Idk, my textbook just defines the norm once, and then never uses it again so I'm a little confused.
calculus
asked Jan 15 at 21:48
mingming
4606
$endgroup$
I'm confused on what purpose the norm has. I know that it defines the maximum length of the widest subinterval in a partition.
But what use do we have from defining $||P||$ to be $max[Delta t_1, Delta t_2, … ,Delta t_n] $??
Because usually don't we just use the regular n-th partitions anyways? Idk, my textbook just defines the norm once, and then never uses it again so I'm a little confused.
calculus
calculus
asked Jan 15 at 21:48
mingming
4606
asked Jan 15 at 21:48
mingming
4606
asked Jan 15 at 21:48
mingming
4606
asked Jan 15 at 21:48
mingming
4606
asked Jan 15 at 21:48
mingming
4606
4606
$begingroup$
In some sense it measures the coarseness of the partition, like distinguishing between boulders, rocks, pebbles, sand. Everything in the partition is no bigger than the norm.
$endgroup$
– MPW
Jan 15 at 21:52
$begingroup$
So if the norm was like 5, then the function has the potential to be "coarse" as you say, because each width could be 1,2,3, 4.23, etc.... but if the norm was like 1, the widths don't have as wide of a range?
$endgroup$
– ming
Jan 15 at 22:05
$begingroup$
Probably the main thing is that if the norm goes to zero, then so do the lengths of all intervals in the partition.
$endgroup$
– MPW
Jan 15 at 22:23
add a comment |
$begingroup$
In some sense it measures the coarseness of the partition, like distinguishing between boulders, rocks, pebbles, sand. Everything in the partition is no bigger than the norm.
$endgroup$
– MPW
Jan 15 at 21:52
$begingroup$
So if the norm was like 5, then the function has the potential to be "coarse" as you say, because each width could be 1,2,3, 4.23, etc.... but if the norm was like 1, the widths don't have as wide of a range?
$endgroup$
– ming
Jan 15 at 22:05
$begingroup$
Probably the main thing is that if the norm goes to zero, then so do the lengths of all intervals in the partition.
$endgroup$
– MPW
Jan 15 at 22:23
$begingroup$
In some sense it measures the coarseness of the partition, like distinguishing between boulders, rocks, pebbles, sand. Everything in the partition is no bigger than the norm.
$endgroup$
– MPW
Jan 15 at 21:52
$begingroup$
In some sense it measures the coarseness of the partition, like distinguishing between boulders, rocks, pebbles, sand. Everything in the partition is no bigger than the norm.
$endgroup$
– MPW
Jan 15 at 21:52
$begingroup$
So if the norm was like 5, then the function has the potential to be "coarse" as you say, because each width could be 1,2,3, 4.23, etc.... but if the norm was like 1, the widths don't have as wide of a range?
$endgroup$
– ming
Jan 15 at 22:05
$begingroup$
So if the norm was like 5, then the function has the potential to be "coarse" as you say, because each width could be 1,2,3, 4.23, etc.... but if the norm was like 1, the widths don't have as wide of a range?
$endgroup$
– ming
Jan 15 at 22:05
$begingroup$
Probably the main thing is that if the norm goes to zero, then so do the lengths of all intervals in the partition.
$endgroup$
– MPW
Jan 15 at 22:23
$begingroup$
Probably the main thing is that if the norm goes to zero, then so do the lengths of all intervals in the partition.
$endgroup$
– MPW
Jan 15 at 22:23
add a comment |
1 Answer
1
active
oldest
votes
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If you know that a function is Riemann integrable, then you can limit yourself to use the regular partitions. But if you want to prove that a function is Riemann integrable, you need all partitions, not just the regular ones. It is in this process that the norm of the partition is useful.
answered Jan 15 at 22:40
Julián AguirreJulián Aguirre
69.5k24297
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you know that a function is Riemann integrable, then you can limit yourself to use the regular partitions. But if you want to prove that a function is Riemann integrable, you need all partitions, not just the regular ones. It is in this process that the norm of the partition is useful.
answered Jan 15 at 22:40
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
add a comment |
$begingroup$
If you know that a function is Riemann integrable, then you can limit yourself to use the regular partitions. But if you want to prove that a function is Riemann integrable, you need all partitions, not just the regular ones. It is in this process that the norm of the partition is useful.
answered Jan 15 at 22:40
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
add a comment |
$begingroup$
If you know that a function is Riemann integrable, then you can limit yourself to use the regular partitions. But if you want to prove that a function is Riemann integrable, you need all partitions, not just the regular ones. It is in this process that the norm of the partition is useful.
answered Jan 15 at 22:40
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
If you know that a function is Riemann integrable, then you can limit yourself to use the regular partitions. But if you want to prove that a function is Riemann integrable, you need all partitions, not just the regular ones. It is in this process that the norm of the partition is useful.
answered Jan 15 at 22:40
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 15 at 22:40
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 15 at 22:40
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 15 at 22:40
Julián AguirreJulián Aguirre
69.5k24297
69.5k24297
add a comment |
add a comment |
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$begingroup$
In some sense it measures the coarseness of the partition, like distinguishing between boulders, rocks, pebbles, sand. Everything in the partition is no bigger than the norm.
$endgroup$
– MPW
Jan 15 at 21:52
$begingroup$
So if the norm was like 5, then the function has the potential to be "coarse" as you say, because each width could be 1,2,3, 4.23, etc.... but if the norm was like 1, the widths don't have as wide of a range?
$endgroup$
– ming
Jan 15 at 22:05
$begingroup$
Probably the main thing is that if the norm goes to zero, then so do the lengths of all intervals in the partition.
$endgroup$
– MPW
Jan 15 at 22:23