1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B How to interpret the integral of the absolute value?

  1. Dec 21, 2017 #1

    SeM

    User Avatar

    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

    User Avatar

    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

    User Avatar

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

    fresh_42

    User Avatar
    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

    User Avatar

    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

    User Avatar

    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

    User Avatar

    Ah! Indeed! Thanks!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted