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Re: [MATH for FUN] i

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  • Carlos Assale
    Oh yeah.... Ø = pi/2 + k.2.pi....!!!! Carlos, the king of typos.... ... Ø = pi/2 + k.2.pi ? ... for ... http://docs.yahoo.com/info/terms/ ... To unsubscribe
    Message 1 of 7 , May 5, 2003
      Oh yeah....
      Ø = pi/2 + k.2.pi....!!!!
      Carlos, the king of typos....

      ----- Original Message -----

      --- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
      wrote:
      > Analitically:
      >
      > By Euler's Formula
      >
      > e^(i.Ø)=cos Ø + i.sin Ø
      >
      > If we make i = e^(i.Ø) then
      > i = cos Ø + i.sin Ø and Ø = pi/2 + k.pi

      Ø = pi/2 + k.2.pi ?

      >
      > Taking just the first solution:
      >
      > i = e^(i.pi/2)
      >
      >
      > So i^i = [e^(i.pi/2)]^i
      > i^i = e^(i².pi/2)
      > i î = e^(-pi/2) = 0.207879576351......
      >
      > Carlos
      > PS: Please somebody explain the other solutions for k<>0
      >
      > ----- Original Message -----
      > > According to my trusted calculator: 0.207879576351
      > > Carlos
      > >
      > > ----- Original Message -----
      > > From: "grouptheory2002" <no_reply@yahoogroups.com>
      > > > anyone knows the value of i raised to the i th power? thanks
      for
      > > > your help.
      > >
      > >
      > >
      > > To unsubscribe from this group, send an email to:
      > > mathforfun-unsubscribe@yahoogroups.com
      > >
      > >
      > >
      > > Your use of Yahoo! Groups is subject to
      http://docs.yahoo.com/info/terms/
      > >
      > >
      > >
      > >



      To unsubscribe from this group, send an email to:
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    • clooneman
      King of typos? You should see some of mine! ... http://docs.yahoo.com/info/terms/
      Message 2 of 7 , May 6, 2003
        King of typos? You should see some of mine!

        --- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
        wrote:
        > Oh yeah....
        > Ø = pi/2 + k.2.pi....!!!!
        > Carlos, the king of typos....
        >
        > ----- Original Message -----
        >
        > --- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
        > wrote:
        > > Analitically:
        > >
        > > By Euler's Formula
        > >
        > > e^(i.Ø)=cos Ø + i.sin Ø
        > >
        > > If we make i = e^(i.Ø) then
        > > i = cos Ø + i.sin Ø and Ø = pi/2 + k.pi
        >
        > Ø = pi/2 + k.2.pi ?
        >
        > >
        > > Taking just the first solution:
        > >
        > > i = e^(i.pi/2)
        > >
        > >
        > > So i^i = [e^(i.pi/2)]^i
        > > i^i = e^(i².pi/2)
        > > i E= e^(-pi/2) = 0.207879576351......
        > >
        > > Carlos
        > > PS: Please somebody explain the other solutions for k<>0
        > >
        > > ----- Original Message -----
        > > > According to my trusted calculator: 0.207879576351
        > > > Carlos
        > > >
        > > > ----- Original Message -----
        > > > From: "grouptheory2002" <no_reply@yahoogroups.com>
        > > > > anyone knows the value of i raised to the i th power? thanks
        > for
        > > > > your help.
        > > >
        > > >
        > > >
        > > > To unsubscribe from this group, send an email to:
        > > > mathforfun-unsubscribe@yahoogroups.com
        > > >
        > > >
        > > >
        > > > Your use of Yahoo! Groups is subject to
        > http://docs.yahoo.com/info/terms/
        > > >
        > > >
        > > >
        > > >
        >
        >
        >
        > To unsubscribe from this group, send an email to:
        > mathforfun-unsubscribe@yahoogroups.com
        >
        >
        >
        > Your use of Yahoo! Groups is subject to
        http://docs.yahoo.com/info/terms/
      • Carlos Assale
        hehehe....not easy to have 5 thumbs when at the keyboard....hehehe.... Carlos ... From: clooneman King of typos? You should see
        Message 3 of 7 , May 6, 2003
          hehehe....not easy to have 5 thumbs when at the keyboard....hehehe....
          Carlos

          ----- Original Message -----
          From: "clooneman" <no_reply@yahoogroups.com>
          King of typos? You should see some of mine!

          --- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
          wrote:
          > Oh yeah....
          > Ø = pi/2 + k.2.pi....!!!!
          > Carlos, the king of typos....
          >
          > ----- Original Message -----
          >
          > --- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
          > wrote:
          > > Analitically:
          > >
          > > By Euler's Formula
          > >
          > > e^(i.Ø)=cos Ø + i.sin Ø
          > >
          > > If we make i = e^(i.Ø) then
          > > i = cos Ø + i.sin Ø and Ø = pi/2 + k.pi
          >
          > Ø = pi/2 + k.2.pi ?
          >
          > >
          > > Taking just the first solution:
          > >
          > > i = e^(i.pi/2)
          > >
          > >
          > > So i^i = [e^(i.pi/2)]^i
          > > i^i = e^(i².pi/2)
          > > i E= e^(-pi/2) = 0.207879576351......
          > >
          > > Carlos
          > > PS: Please somebody explain the other solutions for k<>0
          > >
          > > ----- Original Message -----
          > > > According to my trusted calculator: 0.207879576351
          > > > Carlos
          > > >
          > > > ----- Original Message -----
          > > > From: "grouptheory2002" <no_reply@yahoogroups.com>
          > > > > anyone knows the value of i raised to the i th power? thanks
          > for
          > > > > your help.
          > > >
          > > >
          > > >
          > > > To unsubscribe from this group, send an email to:
          > > > mathforfun-unsubscribe@yahoogroups.com
          > > >
          > > >
          > > >
          > > > Your use of Yahoo! Groups is subject to
          > http://docs.yahoo.com/info/terms/
          > > >
          > > >
          > > >
          > > >
          >
          >
          >
          > To unsubscribe from this group, send an email to:
          > mathforfun-unsubscribe@yahoogroups.com
          >
          >
          >
          > Your use of Yahoo! Groups is subject to
          http://docs.yahoo.com/info/terms/



          To unsubscribe from this group, send an email to:
          mathforfun-unsubscribe@yahoogroups.com



          Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
        • Adam
          After all the nudists/Mother s Day/SARS/Iragi tees postings, back to math (BTW I am not new to the list) Multiple values of i^i are explained by multivalued
          Message 4 of 7 , May 6, 2003
            After all the nudists/Mother's Day/SARS/Iragi tees postings, back to
            math (BTW I am not new to the list)

            Multiple values of i^i are explained by multivalued complex
            functions. e^z is periodic with period 2*pi*i, so e^(z+2*pi*i)=e^z.
            When you define a^z as e^(z*log(a)) then the logarithm (the inverse
            of e^z) is multivalued and so is a^z. Nothing much else to explain,
            it is like the plus or minus for square roots.

            Adam

            --- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
            wrote:
            > Analitically:
            >
            > By Euler's Formula
            >
            > e^(i.Ø)=cos Ø + i.sin Ø
            >
            > If we make i = e^(i.Ø) then
            > i = cos Ø + i.sin Ø and Ø = pi/2 + k.pi
            >
            > Taking just the first solution:
            >
            > i = e^(i.pi/2)
            >
            >
            > So i^i = [e^(i.pi/2)]^i
            > i^i = e^(i².pi/2)
            > i î = e^(-pi/2) = 0.207879576351......
            >
            > Carlos
            > PS: Please somebody explain the other solutions for k<>0
            >
            > ----- Original Message -----
            > > According to my trusted calculator: 0.207879576351
            > > Carlos
            > >
            > > ----- Original Message -----
            > > From: "grouptheory2002" <no_reply@yahoogroups.com>
            > > > anyone knows the value of i raised to the i th power? thanks
            for
            > > > your help.
            > >
            > >
            > >
            > > To unsubscribe from this group, send an email to:
            > > mathforfun-unsubscribe@yahoogroups.com
            > >
            > >
            > >
            > > Your use of Yahoo! Groups is subject to
            http://docs.yahoo.com/info/terms/
            > >
            > >
            > >
            > >
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