- --- In mathforfun@yahoogroups.com, jason1990 <no_reply@y...> wrote:
> --- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:

doing

> > 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

> 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 - 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