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  • Test (please ignore)

    I wonder whether the horrible emoticons will be suppressed when trying to expose an if condition such as -if x <3 | x == .-

  • #2
    Originally posted by Dirk Enzmann View Post
    I wonder whether the horrible emoticons will be suppressed when trying to expose an if condition such as -if x <3 | x == .-
    Thanks, it works (Skype is showing a throbbing heart if you type <3 in chat mode).

    Comment


    • #3
      Test including code / output snippets with spaces using BBCode "[CODE]" and "[\CODE]":

      Code:
      . sysuse auto, clear
      (1978 Automobile Data)
      
      . set trace off
      
      . sum mpg
      
          Variable |       Obs        Mean    Std. Dev.       Min        Max
      -------------+--------------------------------------------------------
               mpg |        74     21.2973    5.785503         12         41
      
      .

      It should avoid the problem that simply copying will not preserve the spaces:

      . sysuse auto, clear
      (1978 Automobile Data)

      . set trace off

      . sum mpg

      Variable | Obs Mean Std. Dev. Min Max
      -------------+--------------------------------------------------------
      mpg | 74 21.2973 5.785503 12 41

      .

      or pasting it "as plain text" (no difference assumed):

      . sysuse auto, clear
      (1978 Automobile Data)

      . set trace off

      . sum mpg

      Variable | Obs Mean Std. Dev. Min Max
      -------------+--------------------------------------------------------
      mpg | 74 21.2973 5.785503 12 41

      .

      In fact, the preview already shows that BBCode (choosing # from the menu and inserting the output snippet in between "[CODE]" and "[\CODE]") preserves spaces wheras plain text does not.

      Comment


      • #4
        Test the use of mathematical formulas:
        \[ -ln(\frac{1}{p}-1)
        \] or perhaps worse $-ln(\frac{1}{p}-1)$

        Comment


        • #5
          Test again:
          /[
          b_{0} = -ln(\frac{1}{p}-1)
          /]
          and
          /[
          var_{0} = \frac{pi^{2}}{3}
          /]
          /[
          var_{1} = b_{1}^{2}
          /]
          /[
          sf = \sqrt{\frac{var_{0}}{var_{0}+var_{1}}}
          /]
          OK?

          Comment


          • #6
            Test again: \[
            b_{0} = -ln(\frac{1}{p}-1)
            \] and \[
            var_{0} = \frac{pi^{2}}{3}
            \]
            \[ var_{1} = b_{1}^{2} \]
            \[ sf = \sqrt{\frac{var_{0}}{var_{0}+var_{1}}} \] OK?

            Comment


            • #7
              Test: \[
              var_{0} = \frac{\pi^{2}}{3}
              \]

              Comment


              • #8
                Final test: \[ b_{0} = -ln(\frac{1}{p}-1) \] and \[ var_{0} = \frac{\pi^{2}}{3} \], \[ var_{1} = b_{1}^{2} \], \[ sf = \sqrt{\frac{var_{0}}{var_{0}+var_{1}}} \] OK?

                Comment


                • #9
                  Shoudn't it be \[ z ≡ atanhr = atanhρ - ρ/[2(n − 1) \], instead?

                  Comment


                  • #10
                    Shoudn't it be
                    \[ z ≡ atanhr = atanhρ - ρ/[2(n − 1) \]
                    instead?

                    Comment


                    • #11
                      Shoudn't it be

                      $ z ≡ atanhr = atanhρ - ρ/[2(n − 1) $

                      Comment


                      • #12

                        \[ \frac{2}{3}+ \mathbf{\pi} \]

                        Comment

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