-- May 8 In-Class Exercise
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@