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B How to interpret the integral of the absolute value?

  1. Dec 21, 2017 #1

    SeM

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    This is rather basic, and may be a misconception of the notation, however, I can't make the following sum up:

    The following is given:

    x_n(t) = 1 -nt , (if 0 <= t <= 1/n) and 0, (if 1/n < t <= 1)

    However, this part I can't grasp this part in the book:

    \begin{equation}
    ||x_n||^2 = \int_0^1 |x_n(t)|^2 dt = \frac{1}{3n}
    \end{equation}

    I tried it, and got a different answer, where i integrated ##|x_n(t)|^2=(1-nt)^2 = 1-2nt-n^2t^2##:

    \begin{equation}
    ||x_n||^2 = \int_0^1 1-2nt-n^2t^2 dt = t - nt^2 -n^2t^2/3 = 1 - n - n^2/3
    \end{equation}

    The right answer is however given in the first integral. What did I do wrong here?

    Thanks!
     
  2. jcsd
  3. Dec 21, 2017 #2
    It should be $$
    ||x_n||^2 = \int_0^{1/n} 1-2nt+n^2t^2 dt
    $$
     
    Last edited: Dec 21, 2017
  4. Dec 21, 2017 #3

    SeM

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    Ok! Thanks!! Then its a typo.
     

    Attached Files:

  5. Dec 21, 2017 #4
    Wait, how did you get books answer ?
     
  6. Dec 21, 2017 #5

    SeM

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    Attached Untitled.jpg
     
  7. Dec 21, 2017 #6

    fresh_42

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    2017 Award

    Staff: Mentor

    There is a sign error, since ##(1-nt)^2=1-2nt+n^2t^2## and as @Buffu said: ##||x_n||^2 = \int_0^1 = \int_0^{\frac{1}{n}}##.
     
  8. Dec 21, 2017 #7
    Yes, I also copied the error. Now corrected it in my post.
     
  9. Dec 21, 2017 #8

    SeM

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    Thanks! I have corrected it in the book. It's Kreyszig Introduction to Functional Analysis, as part of the discussions with George Jones.
     
    Last edited by a moderator: Dec 21, 2017
  10. Dec 21, 2017 #9

    SeM

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    Hi Buffu, something is odd with the step "and the quotient" yielding the answer >n. Here I got: n/1/3n = 3n^2. but the book says differently again. It appears as an error, can you see if this is yet another error?
     
  11. Dec 21, 2017 #10
    You forgot to take the square root. The integral gives norm squared; you need ratio of norms not of norms squared.
     
    Last edited: Dec 21, 2017
  12. Dec 21, 2017 #11

    SeM

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    Ah! Indeed! Thanks!
     
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