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 <br><br>delta = 0.1 <br><br> y/(1+delta) leq z leq y <br><br> L' = langle 50, 70,82,99 rangle

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=⟨50,51,55,70,71,82,83,84,99⟩

and δ=0.1

so one trimmed list L' = <50, 55, 70, 82, 99>

'''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.

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'''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

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}