Math with conditions

  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. Sure.
     
  4. I don't known that it's operable. It's cool, will be very important to me in engineering!
     
  5. Could you give a concrete example?
     
  6. Are you intending for the ##x## in your ##\mathrm{d}x## to be different?
     
  7. x, y and z are only a representative symbol...
     
  8. 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. Student100

    Student100 654
    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. 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.
     
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