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Re: Number sequence transform

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  • bqllpd
    ... doing ... 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
    Message 1 of 21 , May 1, 2004
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      --- 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.

      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
      ... in i ... the ... 97976320, ... 438220881920,... ... the ... above, ... guess c ... consecutively? ... I ... I ... calculator ... for ... honestly, ...
      Message 2 of 21 , 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
      • bqllpd
        ... p. ... p is an arbitrary positive integer, because when n m, then J_h(a_i) = 0 for all (n+p) m. J_h(a_i) is the J-transform for brevity. I solved for h
        Message 3 of 21 , May 1, 2004
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          > 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?

          p is an arbitrary positive integer, because when n>m, then J_h(a_i) =
          0
          for all (n+p)>m. "J_h(a_i)" is the J-transform for brevity. I solved
          for h using my calculator that has n roots for each value of n.

          >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?

          The correction should be
          Sum{(h^(n+p-i)*c(n,k)*a_i, i=0 to n+p}

          The summation does depend on p. It happens that 0 is a common root
          for the J-transform solutions for h when n>m and (n+p)>m


          > 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.

          From what follows below, two equations for the sequence I gave
          without summation is

          (*) -9h^4 - 160h^3 + 627h^2 + 15360h + 42752 = 0
          (**) -9h^6 -240h^5 + 1680h^4 + 76800h^3 + 641280h^2 + 2125824h +
          2519040 = 0

          The value of h that is the same in (*) and (**) and for any n>m is
          the value of h = -k and in this case, -k = -4.

          I hope that clears it up a little.

          Have an excellent day!
          It's so nice out (at least where I am)

          bq


          > >
          > > 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.
          > >
        • bqllpd
          ... Let me rephrase the second part The summation does depend on p. When the J-transform = 0 for values of n m and (n+p) m, then there exists a root for the
          Message 4 of 21 , May 1, 2004
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            > The correction should be
            > Sum{(h^(n+p-i)*c(n,k)*a_i, i=0 to n+p}
            >
            > The summation does depend on p. It happens that 0 is a common root
            > for the J-transform solutions for h when n>m and (n+p)>m
            >


            Let me rephrase the second part

            The summation does depend on p. When the J-transform = 0 for values
            of n>m and (n+p)>m, then there exists a root for the J-transform that
            is constant for n>m and (n+p)>m and this is the solution for h for
            the J-transform where p is an arbitrary positive integer.
          • bqllpd
            The correction should be Sum{(h^(n+p-i)*c(n,i)*(a_i, i=0 to n+p}
            Message 5 of 21 , May 1, 2004
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              The correction should be
              Sum{(h^(n+p-i)*c(n,i)*(a_i, i=0 to n+p}
            • bqllpd
              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...
              Message 6 of 21 , May 1, 2004
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                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
                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 7 of 21 , 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 8 of 21 , 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 9 of 21 , 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 10 of 21 , 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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