Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Math with conditions

  1. Dec 13, 2013 #1
    Given
    [tex]x = \left\{\begin{matrix} y \;\;\; case\;A\\ z \;\;\; case\;B\\ \end{matrix}\right.[/tex]
    So, operate x means to operate the 2 cases of right side? For example:
    [tex]\int x\;dx = \left\{\begin{matrix} \int y\;dx \;\;\; case\;A\\ \int z\;dx \;\;\; case\;B\\ \end{matrix}\right.[/tex]
    Correct?
     
  2. jcsd
  3. Dec 13, 2013 #2
    Sure.
     
  4. Dec 13, 2013 #3
    I don't known that it's operable. It's cool, will be very important to me in engineering!
     
  5. Dec 13, 2013 #4
    Could you give a concrete example?
     
  6. Dec 13, 2013 #5
    Are you intending for the ##x## in your ##\mathrm{d}x## to be different?
     
  7. Dec 13, 2013 #6
    x, y and z are only a representative symbol...
     
  8. Dec 13, 2013 #7
    The problem is that you're using x in two different places, and it's not clear whether or not you intend them to mean the same thing.
     
  9. Dec 13, 2013 #8

    Student100

    User Avatar
    Education Advisor
    Gold Member


    [tex]x = \left\{\begin{matrix} y \;\;\; case\;A\\ z \;\;\; case\;B\\ \end{matrix}\right.[/tex]

    [tex]\int x\;dx = \left\{\begin{matrix} \int y\;dy \;\;\; case\;A\\ \int z\;dz\;\;\; case\;B\\ \end{matrix}\right.[/tex]

    Is this what you mean?
     
  10. Dec 14, 2013 #9
    I'm trying find... I think that it's math isn't correct, because:
    [tex]H(x) = \left\{\begin{matrix} 0\;\;\; x<0\\ 1\;\;\; x=0\\ 1\;\;\; x>0\\ \end{matrix}\right.[/tex]
    [tex]\frac{\mathrm{d} }{\mathrm{d} x}H(x) = \left\{\begin{matrix} \frac{\mathrm{d} }{\mathrm{d} x}0\;\;\; x<0\\ \frac{\mathrm{d} }{\mathrm{d} x}1\;\;\; x=0\\ \frac{\mathrm{d} }{\mathrm{d} x}1\;\;\; x>0\\ \end{matrix}\right. = \left\{\begin{matrix} 0\;\;\; x<0\\ 0\;\;\; x=0\\ 0\;\;\; x>0\\ \end{matrix}\right.[/tex]

    is wrong, 'cause contradicts the identity d/dx H(x) = δ(x). So, this "technic" above is not useful, although it seems make sense, theoretically... In other words, I'm trying know how do to integrate and derivative, graphically and algebraically, a function with steps and impulses.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Math with conditions
  1. Eigenvalue Condition (Replies: 5)

  2. Circle conditions (Replies: 1)

  3. Trapezium condition (Replies: 1)

Loading...