HW 9: Challenge 2(a,b,c)

Tongzhou, 11/21

(a) Since L(h,P)=\sum_{k=1}^n m_k(x_k-x_{k-1}), m_k needed to be determined firstly.

By definition, m_k =\inf\{h(x): x \in [x_{k-1},x_k]\}.

Since [x_{k-1},x_k] is any partition of [0,1], m_k =\inf \{1,2\}=1.

Therefore, L(h,P) = \sum_{k=1}^n m_k(x_k-x_{k-1}) =\sum_{k=1}^n 1(x_k-x_{k-1})=1(1-0) =1.

(b) Sidework: I want to have U(h,P) < 1+ 1/10 which means 1(x_k-0)+2(1-x_k)<1+1/10. I want to determine how far x_k could be away from 1.

Thus lim_{k \rightarrow n}(x_k-0) + 2(1-x_k)<1.1 so that 1-x_k <0.05.

Now choose P= \{0,0.96,1\}. Again I need to determine M_k before I find U(h,P).

M_1 = \sup\{h(0), h(0.96)\}= 1, M_2 =\sup \{h(0.96), h(1)\}=2.

U(h,P)= 1(0.96)+2(1-0.96) =1.04 < 1.1

(c) By the intuition of (b).

Let \epsilon>0, Choose P_\epsilon = \{0, 1-\frac{\epsilon}{2},1\}

M_1 = \sup \{h(0),h(1-\frac{\epsilon}{2})\}=\sup \{1,1\}=1, M_2 =\sup \{h(1-\frac{\epsilon}{2}),h(1)\} =\sup \{1,2\}=2.

U(h,P_\epsilon) = 1 (1-\frac{\epsilon}{2})+2 (1-1+\frac{\epsilon}{2})=1+\frac{\epsilon}{2}<1+\epsilon.

Posted on by Tongzhou Wang
Categories: Challenges

3 Responses

  1. Jeremy LeCrone says:
    November 23, 2017 at 2:25 pm

    https://richmond.qualtrics.com/jfe/form/SV_72SXTAOV7oVyeYB

  2. Elise Favia says:
    November 26, 2017 at 1:14 pm

    Great work. It would be helpful to the reader to state what h(x) is for a problem like this, as it is not clear what it is since the problem statement wasn’t included.

  3. Gregory Bischoff says:
    November 28, 2017 at 10:12 pm

    All good! Just a note on the notation: for ‘x_k’, the k is indexing its place in the partition. While your sum is considering the values of k up to n, remember that is only to index the left and right endpoints of an interval!

    The step you include the limit on is right after you state the same fact without the limit, and your next result is just a bit of algebra so there’s no need to worry about a limit.

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