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Re: Number sequence transform

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  • jason1990
    ... in i ... the ... 97976320, ... 438220881920,... ... the ... above, ... guess c ... consecutively? ... I ... I ... calculator ... for ... honestly, ...
    Message 1 of 10 , May 1, 2004
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      --- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
      > --- In mathforfun@yahoogroups.com, jason1990 <no_reply@y...> wrote:
      > > --- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
      > > > I'll try my best
      > > >
      > > > Suppose you have a sequence {a_i} where this is a polynomial
      in i
      > > > times some constant to the i-th power. You don't know what
      the
      > > > polynomial is or what the constant is.
      > > >
      > > > Suppose the sequence is
      > > > -9, -40, 112, 3840, 42752, 354304, 2519040, 16252928,
      97976320,
      > > > 561512448, 8094347776, 16525557760, 86016786432,
      438220881920,...
      > > >
      > > > You could use Newton's useful little formula to perhapse find
      the
      > > > next number, but it would be a little cumbersome given the
      above,
      > > so
      > > > I'll try a better method.
      > > >
      > > > Set the J-transform equal to
      > > > Sum{(h^(n-i))*c(n,i)*a_i, i=0 to n}
      > > > =Sum{(h^(n-i))*c(n+p,i)*a_i, i=0 to n+p}
      > > > =0, n > 0 and solve for h.
      > >
      > > I'm sorry, but I don't understand what you're doing here. I
      guess c
      > > (n,k) is "n choose k". What is p? Are you generating a family of
      > > equations, one for each value of n, and solving them
      consecutively?
      > > Sorry, but could you maybe work through explicitly what you're
      > doing
      > > here. With the actual numbers, I mean. Thanks.
      >
      > c(n,k) is n choose k. p is a positive integer. I did generate a
      > family of equations for each value of n > 0. I guess you can say
      I
      > cheated because I let my calculator do the calculations. The way
      I
      > understand how these calculations are done is to enter the
      > coefficients of a polynomial in h in vector form, then the
      calculator
      > returns the roots also in vector form. If the same root occurs
      for
      > different values of n, then this root is the value of h. I don't
      > know how to do the work with the numbers explicity because
      honestly,
      > I don't know how to do the math my calculator does.

      That's fine, but before I get to the numbers you've listed below,
      I'm still not clear about the setup. For each n, you have an
      equation that you want to solve for h. But that equation involves p.
      If p is just an unknown positive integer, how can you solve for h?
      What is the role of p? When you write this:

      > > > Sum{(h^(n-i))*c(n,i)*a_i, i=0 to n}
      > > > =Sum{(h^(n-i))*c(n+p,i)*a_i, i=0 to n+p}
      > > > =0, n > 0 and solve for h.

      are you claiming that the value of the summation does not depend on
      p? Or are you looking for a value of p so that both summations are
      equal to zero? I guess when I asked you to explicitly work through
      what you're doing here with the actual numbers, I just meant: can
      you write down explicitly the equations you are solving without
      using summation notation. I don't need you to solve them. I trust
      the calculator. Also, just to make sure there's no confusion, it
      would help if each equation you write down has only one "=" in it.
      Thanks.

      >
      > When n=5, the polynomial in vector form to solve for it's roots is
      > [-9, -160, 672, 15360, 42752]
      >
      > for n=7
      > [-9, -240, 1680, 76800, 641280, 2125824, 2519040]
      >
      > For n>3, -4 is the root common to all solutions for the polynomial
      in
      > h.
      >
      > This is all I have. I wish I could tell you more. Any help would
      be
      > appreciated.
      >
      > bq
    • jason1990
      I m sorry, bq. I m totally lost at this point. I could sit down and try to work it out for myself, but I think that defeats the purpose, given that this is
      Message 2 of 10 , May 1, 2004
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        I'm sorry, bq. I'm totally lost at this point. I could sit down and
        try to work it out for myself, but I think that defeats the purpose,
        given that this is something you've already worked through to some
        degree. For what it's worth, I have found that doing mathematics is
        a 3 stage process. First, you have an idea or develop a method;
        second, you write it down with clear and unambiguous notation;
        third, you prove your claims and discoveries. I'm sure you have
        *something* worked out with this transform...I can catch a glimpse
        of it. But I need to see the second stage before I can offer any
        advice or assistance on the third stage -- and I don't think things
        will be cleared up for me in this forum. Maybe someone else here can
        offer some insight.

        --- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
        > Sum{(h^(n+p-i)*c(n+p,i)*(a_i), i=0 to n+p}
        >
        > Sorry. I swear this is it.
        >
        > I'm not a good proof reader at all...
      • jason1990
        Looking again at this message I wrote, it seems a bit curt. Sorry. Just so you know, I *am* interested in figuring out what you re doing here, but I think it s
        Message 3 of 10 , May 1, 2004
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          Looking again at this message I wrote, it seems a bit curt. Sorry.
          Just so you know, I *am* interested in figuring out what you're
          doing here, but I think it's just one of those things that needs to
          be discussed in person.

          (By the way, notice how we reverted to the ASCII math again? :) I'm
          telling you, TeX is where it's at. Learn it if you have a chance, I
          highly recommend it.)

          --- In mathforfun@yahoogroups.com, jason1990 <no_reply@y...> wrote:
          > I'm sorry, bq. I'm totally lost at this point. I could sit down
          and
          > try to work it out for myself, but I think that defeats the
          purpose,
          > given that this is something you've already worked through to some
          > degree. For what it's worth, I have found that doing mathematics
          is
          > a 3 stage process. First, you have an idea or develop a method;
          > second, you write it down with clear and unambiguous notation;
          > third, you prove your claims and discoveries. I'm sure you have
          > *something* worked out with this transform...I can catch a glimpse
          > of it. But I need to see the second stage before I can offer any
          > advice or assistance on the third stage -- and I don't think
          things
          > will be cleared up for me in this forum. Maybe someone else here
          can
          > offer some insight.
          >
          > --- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
          > > Sum{(h^(n+p-i)*c(n+p,i)*(a_i), i=0 to n+p}
          > >
          > > Sorry. I swear this is it.
          > >
          > > I'm not a good proof reader at all...
        • ray chan
          A cube has edges of length 10 inches. The square bases of six identical right pyramids also have side lengths 10 inches and each base is glued to a face of the
          Message 4 of 10 , May 2, 2004
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            A cube has edges of length 10 inches. The square bases of six
            identical right pyramids also have side lengths 10 inches and each
            base is glued to a face of the cube so the squares exactly coincide.
            At each EDGE of the cube, two triangular faces from the two adjoining
            pyramids create a rhombus when seen from the same plane.
            The 24 triangular faces make 12 rhombi in all, thus forming the
            surface of a classic rhombic dodecahedron.

            1) The line joining the tip of the uppermost pyramid to the bottom
            tip of the bottom pyramid is one of the internal diagonals of the
            dodecahedron. Find its length.

            2) Find the lengths of the three other internal diagonals.

            3) Derive a formula for the volume of a rhombic dodecahedron which
            has been formed from a cube with edge of length x inches, in terms of
            x.

            I can visualize the figure, but am still stumped. Can anyone supply a
            lead? Note this needs simple geometric calculations .. anything that
            involves tangents, cosines, calculus, etc, will confuse me further.
            Apologies if this is way too elementary for some of you!

            rc
          • brianscsmith
            This is what I managed to come up with: Let the cube be centered at the origin. Its vertecies are at A (5,5,5), B(5,5,-5), C(5,-5,5), D(5,-5,-5), E(-5,5,5),
            Message 5 of 10 , May 3, 2004
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              This is what I managed to come up with:

              Let the cube be centered at the origin. Its vertecies are at A
              (5,5,5), B(5,5,-5), C(5,-5,5), D(5,-5,-5), E(-5,5,5), F(-5,5,-5), G(-
              5,-5,5), and H(-5,-5,-5). The six pyarmids have their vertecies at I
              (z,0,0), J(-z,0,0), K(0,z,0), L(0,-z,0), M(0,0,z), N(0,0,-z). z
              needs to be chosen so that {A,B,I,K} and others similar sets of
              vertecies are coplanar. Solid ABCDEFGHIJKLMN is the rhombic
              dodecahedron.

              Let x_1*c_1 + x_2*c_2 + x_3*c_3 = 1 be the equation of the plane.
              Then:
              5*c_1 + 5*c_2 + 5*c_3 = 1
              5*c_1 + 5*c_2 - 5*c_3 = 1
              z*c_1 + 0*c_2 + 0*c_3 = 1
              0*c_1 + z*c_2 + 0*c_3 = 1

              c_3 = 0
              c_2 = 1/z
              c_1 = 1/z
              5/z + 5/z = 1
              z = 10

              One of the long diagonals is IJ. The distance from I(10,0,0) to J(-
              10,0,0) is 20.
              The other three different diagonals are represented by IE, AD, and
              AH. IE = 5*sqrt(11), AD = 10*sqrt(2), AH = 10*sqrt(3)

              The volume of this rhombic dodecahedron is cube+6*pyramid = 10^3 + 6*
              (1/3)*(5*5)*5 = 1250.
              The volume of a general rhombic dodecahedron can be found by
              multiplying 1250 by the cube of the ratio of the edges x/10. 1250 *
              (x/10)^3 = 5*x^3/4

              --- In mathforfun@yahoogroups.com, "ray chan" <rayc98@h...> wrote:
              >
              > A cube has edges of length 10 inches. The square bases of six
              > identical right pyramids also have side lengths 10 inches and each
              > base is glued to a face of the cube so the squares exactly coincide.
              > At each EDGE of the cube, two triangular faces from the two
              adjoining
              > pyramids create a rhombus when seen from the same plane.
              > The 24 triangular faces make 12 rhombi in all, thus forming the
              > surface of a classic rhombic dodecahedron.
              >
              > 1) The line joining the tip of the uppermost pyramid to the bottom
              > tip of the bottom pyramid is one of the internal diagonals of the
              > dodecahedron. Find its length.
              >
              > 2) Find the lengths of the three other internal diagonals.
              >
              > 3) Derive a formula for the volume of a rhombic dodecahedron which
              > has been formed from a cube with edge of length x inches, in terms
              of
              > x.
              >
              > I can visualize the figure, but am still stumped. Can anyone supply
              a
              > lead? Note this needs simple geometric calculations .. anything
              that
              > involves tangents, cosines, calculus, etc, will confuse me further.
              > Apologies if this is way too elementary for some of you!
              >
              > rc
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