2019-05-07
May 8 In-Class Exercise.
Please post your solutions to the May 8 In-Class Exercise to this thread.
Best,
Chris
Please post your solutions to the May 8 In-Class Exercise to this thread.
Best,
Chris
`L= langle 50,51,55,70,71,82,83,84,99 rangle`
`delta = 0.1`
`y/(1+delta) leq z leq y`
`L' = langle 50, 70,82,99 rangle`
(Edited: 2019-05-08) `delta = 0.1`
`y/(1+delta) leq z leq y`
`L' = langle 50, 70,82,99 rangle`
@BT@L= langle 50,51,55,70,71,82,83,84,99 rangle@BT@
<br><br>@BT@delta = 0.1@BT@
<br><br> @BT@y/(1+delta) leq z leq y@BT@
<br><br> @BT@L' = langle 50, 70,82,99 rangle@BT@
L=⟨50,51,55,70,71,82,83,84,99⟩. Let δ=0.1. Come up with a δ trimmed list for L.
Trimmed List <50, 70, 82, 99>
Example: y = 55 50 <= z <= 55 z can be 50, so we put 50 into the trimmed list.(Edited: 2019-05-08)
L=⟨50,51,55,70,71,82,83,84,99⟩.
Let δ=0.1. Come up with a δ trimmed list for L.
Trimmed List <50, 70, 82, 99>
Example: y = 55
50 <= z <= 55
z can be 50, so we put 50 into the trimmed list.
L' = <50, 55, 70, 82, 99>
L' = <50, 55, 70, 82, 99>
` L=⟨50,51,55,70,71,82,83,84,99⟩`
and `δ=0.1`
so one trimmed list `L' = <50, 55, 70, 82, 99>`
(Edited: 2019-05-08) @BT@ L=⟨50,51,55,70,71,82,83,84,99⟩@BT@
and @BT@δ=0.1@BT@
so one trimmed list @BT@L' = <50, 55, 70, 82, 99>@BT@
Problem: Suppose we had the list `L=\langle50,51,55,70,71,82,83,84,99\rangle`. Let `\delta=0.1`. Come up with a `\delta` trimmed list for L.
``
``
Solution: A list `L'` is a trimmed list of `L` by `\delta` if it contains only elements from `L` and if for every `y\in L` there is a `z\in L'` such that `\frac{1}{1+\delta} \le z \le y`.
Therefore, let us calculate `\frac{y}{1+\delta} = \frac{y}{1.1}` for all `y\in L`. Therefore the corresponding values are: `T = \langle 45.45, 46.36, 50, 63.64, 64.55, 74.55, 75.45, 76.36, 90\rangle`. Therefore our trimmed list `L'` must include values in between `T[i]` and `L[i]` for all `i`. Therefore the trimmed list is `L' = \langle 50, 55, 70, 82, 99\rangle`
(Edited: 2019-05-08) '''Problem:''' Suppose we had the list @BT@L=\langle50,51,55,70,71,82,83,84,99\rangle@BT@. Let @BT@\delta=0.1@BT@. Come up with a @BT@\delta@BT@ trimmed list for L.
@BT@@BT@
----
@BT@@BT@
'''Solution:''' A list @BT@L'@BT@ is a trimmed list of @BT@L@BT@ by @BT@\delta@BT@ if it contains only elements from @BT@L@BT@ and if for every @BT@y\in L@BT@ there is a @BT@z\in L'@BT@ such that @BT@\frac{1}{1+\delta} \le z \le y@BT@.
Therefore, let us calculate @BT@\frac{y}{1+\delta} = \frac{y}{1.1}@BT@ for all @BT@y\in L@BT@. Therefore the corresponding values are: @BT@T = \langle 45.45, 46.36, 50, 63.64, 64.55, 74.55, 75.45, 76.36, 90\rangle@BT@. Therefore our trimmed list @BT@L'@BT@ must include values in between @BT@T[i]@BT@ and @BT@L[i]@BT@ for all @BT@i@BT@. Therefore the trimmed list is @BT@L' = \langle 50, 55, 70, 82, 99\rangle@BT@
((resource:59812692_371895223422539_2546843841919451136_n.jpg|Resource Description for 59812692_371895223422539_2546843841919451136_n.jpg))
Support we had the list L=⟨50,51,55,70,71,82,83,84,99⟩.
Let δ=0.1. Come up with a δ trimmed list for L.
y1+δ≤z≤y.
So the trimmed list for L: {50,55,70,82}
Support we had the list L=⟨50,51,55,70,71,82,83,84,99⟩.
Let δ=0.1. Come up with a δ trimmed list for L.
y1+δ≤z≤y.
So the trimmed list for L: {50,55,70,82}
since δ = 0.1, therefore calculate y and y/(1+ δ) as follows:
and L' = {50, 70, 82, 99}
| y | z | y/(1+δ) |
| 50 | 50 | 45.45454545 |
| 51 | 46.36363636 | |
| 55 | 50 | |
| 70 | 70 | 63.63636364 |
| 71 | 64.54545455 | |
| 82 | 82 | 74.54545455 |
| 83 | 75.45454545 | |
| 84 | 76.36363636 | |
| 99 | 99 | 90 |
since δ = 0.1, therefore calculate y and y/(1+ δ) as follows:
and L' = {50, 70, 82, 99}
{|
|-
| y || z || y/(1+δ)
|-
| 50 ||50 ||45.45454545
|-
| 51 || ||46.36363636
|-
| 55 || || 50
|-
| 70 ||70|| 63.63636364
|-
| 71 || ||64.54545455
|-
| 82|| 82 ||74.54545455
|-
| 83 || || 75.45454545
|-
| 84 || || 76.36363636
|-
| 99 ||99|| 90
|}
((resource:15573495171761183713200.jpg|Resource Description for 15573495171761183713200.jpg))
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