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

in i

> > --- 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

> > > times some constant to the i-th power. You don't know what

the

> > > polynomial is or what the constant is.

97976320,

> > >

> > > Suppose the sequence is

> > > -9, -40, 112, 3840, 42752, 354304, 2519040, 16252928,

> > > 561512448, 8094347776, 16525557760, 86016786432,

438220881920,...

> > >

the

> > > You could use Newton's useful little formula to perhapse find

> > > next number, but it would be a little cumbersome given the

above,

> > so

guess c

> > > 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

> > (n,k) is "n choose k". What is p? Are you generating a family of

consecutively?

> > equations, one for each value of n, and solving them

> > Sorry, but could you maybe work through explicitly what you're

I

> 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

> cheated because I let my calculator do the calculations. The way

I

> understand how these calculations are done is to enter the

calculator

> coefficients of a polynomial in h in vector form, then the

> 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

honestly,

> know how to do the work with the numbers explicity because

> 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}

are you claiming that the value of the summation does not depend on

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

> > > =0, n > 0 and solve for h.

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.

>

in

> 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

> h.

be

>

> This is all I have. I wish I could tell you more. Any help would

> appreciated.

>

> bq - 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... - 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... - 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 - 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