Loading ...
Sorry, an error occurred while loading the content.

this is interesting!!!

Expand Messages
  • michaelballackachtung
    Let P(x)=ax^3+bx^2+cx+d from R[X] with a 0.If the coeficients of this polinomial are in arithmecical progression, and all of its roots are real then the
    Message 1 of 6 , Jun 2, 2003
    • 0 Attachment
      Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the coeficients of
      this polinomial are in arithmecical progression, and all of its roots
      are real then the following relations holds:




      arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.




      where x1,x2,x3 are its roots.




      Pls help me!!


      bye! thank you!
    • Ricardo Filho
      P(x)=ax^2+bx^2+cx+d. (a,b,c,d) is a arithmetical progression, so P(x)=ax^3+(a+r)x^2+(a+2r)^x+(a+3r) let arctan(1/x1)=m arctan(1/x2)=n arctan(1/x3)=p so
      Message 2 of 6 , Jun 3, 2003
      • 0 Attachment
        P(x)=ax^2+bx^2+cx+d. (a,b,c,d) is a arithmetical progression, so
        P(x)=ax^3+(a+r)x^2+(a+2r)^x+(a+3r)

        let
        arctan(1/x1)=m
        arctan(1/x2)=n
        arctan(1/x3)=p

        so tg(m+n+p)= -1. tg(a+b)=tga+tgb/(1-tga.tgb),
        so tg(m+n+p)=x1x2+x1x3+x2x3 - 1/[x1x2x3 - (x1+x2+x3)]= -1

        by girard relations:
        [(a+2r)/a - 1]/[-(a+3r)/a + (a+r)/a]= -1 ,what is true.


        Ricardo Filho
        Fortaleza - CE - Brasil
        ICQ 23260673

        "Procure dividir-se em alguém."
        Marcelo Camelo
        ----- Original Message -----
        From: michaelballackachtung
        To: mathforfun@yahoogroups.com
        Sent: Monday, June 02, 2003 6:07 PM
        Subject: [MATH for FUN] this is interesting!!!


        Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the coeficients of
        this polinomial are in arithmecical progression, and all of its roots
        are real then the following relations holds:




        arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.




        where x1,x2,x3 are its roots.
      • Lagrangia Louis
        Well the relation is actually: arctan(1/x1)+arctan(1/x2)+arctan(1/x3)=-pi/4 cause other wise it is impossible. try to write it:
        Message 3 of 6 , Jun 3, 2003
        • 0 Attachment
          Well the relation is actually:
          arctan(1/x1)+arctan(1/x2)+arctan(1/x3)=-pi/4
          cause other wise it is impossible.
          try to write it:
          arctan(1/x1)+arctan(1/x2)=-pi/4-arctan(1/x3) and then
          apply tan to both members and then use Viete
          relationships and the fact that the
          coeficients of this polinomial are in arithmecical
          progression and u have the problem.

          Bye!
          Louis.

          --- michaelballackachtung
          <michaelballackachtung@...> wrote:
          > Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the
          > coeficients of
          > this polinomial are in arithmecical progression, and
          > all of its roots
          > are real then the following relations holds:
          >
          >
          >
          >
          > arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.
          >
          >
          >
          >
          > where x1,x2,x3 are its roots.
          >
          >
          >
          >
          > Pls help me!!
          >
          >
          > bye! thank you!
          >
          >
          >


          __________________________________
          Do you Yahoo!?
          Yahoo! Calendar - Free online calendar with sync to Outlook(TM).
          http://calendar.yahoo.com
        • Ricardo Filho
          Valeu =] Ricardo Filho Fortaleza - CE - Brasil ICQ 23260673 Procure dividir-se em alguém. Marcelo Camelo ... From: Carlos Assale To:
          Message 4 of 6 , Jun 3, 2003
          • 0 Attachment
            Valeu =]

            Ricardo Filho
            Fortaleza - CE - Brasil
            ICQ 23260673

            "Procure dividir-se em alguém."
            Marcelo Camelo
            ----- Original Message -----
            From: Carlos Assale
            To: mathforfun@yahoogroups.com
            Sent: Tuesday, June 03, 2003 8:57 PM
            Subject: Re: [MATH for FUN] this is interesting!!!


            Dá-lhe patrício!!!!
            Carlos

            [Non-text portions of this message have been removed]
          • Carlos Assale
            Dá-lhe patrício!!!! Carlos
            Message 5 of 6 , Jun 3, 2003
            • 0 Attachment
              Dá-lhe patrício!!!!
              Carlos

              ----- Original Message -----
              > P(x)=ax^2+bx^2+cx+d. (a,b,c,d) is a arithmetical progression, so
              > P(x)=ax^3+(a+r)x^2+(a+2r)^x+(a+3r)
              >
              > let
              > arctan(1/x1)=m
              > arctan(1/x2)=n
              > arctan(1/x3)=p
              >
              > so tg(m+n+p)= -1. tg(a+b)=tga+tgb/(1-tga.tgb),
              > so tg(m+n+p)=x1x2+x1x3+x2x3 - 1/[x1x2x3 - (x1+x2+x3)]= -1
              >
              > by girard relations:
              > [(a+2r)/a - 1]/[-(a+3r)/a + (a+r)/a]= -1 ,what is true.
              >
              >
              > Ricardo Filho
              > Fortaleza - CE - Brasil
              > ICQ 23260673
              >
              > "Procure dividir-se em alguém."
              > Marcelo Camelo
              > ----- Original Message -----
              > From: michaelballackachtung
              > To: mathforfun@yahoogroups.com
              > Sent: Monday, June 02, 2003 6:07 PM
              > Subject: [MATH for FUN] this is interesting!!!
              >
              >
              > Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the coeficients of
              > this polinomial are in arithmecical progression, and all of its roots
              > are real then the following relations holds:
              >
              >
              >
              >
              > arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.
              >
              >
              >
              >
              > where x1,x2,x3 are its roots.
              >
              >
              >
              >
              > 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/
              >
              >
              >
              >
            • fundoomops
              Let P(x)=ax^3+bx^2+cx+d from R[X] with a 0.If the coeficients of ... roots ... where x1,x2,x3 are its roots. Suggested solution:
              Message 6 of 6 , Jun 4, 2003
              • 0 Attachment
                Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the coeficients of
                > this polinomial are in arithmecical progression, and all of its
                roots
                > are real then the following relations holds:

                > arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.
                where x1,x2,x3 are its roots.

                Suggested solution:

                arctan(1/x1)+arctan(1/x2)=arctan[(1/x1+1/x2)/(1-1/x1x2)]=

                arctan[(x1+x2)/(x1x2-1)]
                Adding arctan(1/x3)
                arctan(1/x1)+arctan(1/x2)+arctan(1/x3)=
                arctan[(x1x3+x2x3+x2x1-1)/(x1x2x3-x3-x1-x2)](1)

                From the theory of equations:

                x1+x2+x3=-b/a

                x1x2+x2x3+x1x3=c/a

                x1x2x3=-d/a

                thus replacing we have in (1):

                x1x3+x2x3+x2x1-1=c/a-1=(c-a)/a

                x1x2x3-x3-x1-x2=-d/a+b/a=(b-d)/a

                But since a,b,c, d are in AP

                a=A-3D
                b=A-D
                c=A+D
                d=A+3D

                c-a=4D
                b-d=-4D

                Thus (1) reduces to

                arctan[(c-a)/(b-d)]=arctan[4D/-4D]=-pi/4

                Note that the solution of arctan must lie within -pi/2 to pi/2


                Saurabh
              Your message has been successfully submitted and would be delivered to recipients shortly.